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Balbharti Maharashtra State Board 12th Chemistry Textbook Solutions Chapter 6 Chemical Kinetics Textbook Exercise Questions and Answers.

Chemical Kinetics Class 12 Exercise Question Answers Solutions Maharashtra Board

Class 12 Chemistry Chapter 6 Exercise Solutions Maharashtra Board

Chemistry Class 12 Chapter 6 Exercise Solutions

1. Choose the most correct option.

Question i.
The rate law for the reaction aA + bB → P is rate = k[A] [B]. The rate of reaction doubles if
a. concentrations of A and B are both doubled.
b. [A] is doubled and [B] is kept constant
c. [B] is doubled and [A] is halved
d. [A] is kept constant and [B] is halved.
Answer:
b. [A] is doubled and [B] is kept constant

Question ii.
The order of the reaction for which the units of rate constant are mol dm^-3 s^-1 is
a. 1
b. 3
c. 0
d. 2
Answer:
c. 0

Question iii.
The rate constant for the reaction 2N₂O₅(g) → 2N₂O₄(g) + O₂(g) is 4.98 × 10^-4 s^-1. The order of reaction is
a. 2
b. 1
c. 0
d. 3
Answer:
b. 1

Question iv.
Time required for 90 % completion of a certain first order reaction is t. The time required for 99.9 % completion will be
a. t
b. 2t
c. t/2
d. 3t
Answer:
d. 3t

Question v.
Slope of the graph ln[A]_t versus t for first order reaction is
a. -k
b. k
c. k/2. 303
d. -k/2. 303
Answer:
a. -k

Question vi.
What is the half life of a first order reaction if time required to decrease concentration of reactant from 0.8 M to 0.2 M is 12 h?
a. 12 h
b. 3 h
c. 1.5 h
d. 6 h
Answer:
d. 6 h

Question vii.
The reaction, 3ClO^3Θ ClO₃^Θ + 2 Cl^Θ occurs in two steps,
(i) 2 ClO^– → ClO₂^Θ
(ii) ClO₂^Θ + ClO^Θ → ClO₃^Θ + Cl^Θ

The reaction intermediate is
a. Cl^Θ
b. ClO₂^Θ
c. ClO₃^Θ
d. ClO^Θ
Answer:
b. ClO₂^Θ

Question viii.
The elementary reaction O₂(g) + O(g) → 2O₂(g) is
a. unimolecular and second order
b. bimolecular and first order
c. bimolecular and second order
d. unimolecular and first order
Answer:
c. bimolecular and second order

Question ix.
Rate law for the reaction, 2NO + Cl₂ → 2 NOCl is rate = k[NO₂]²[Cl₂]. Thus k would increase with
a. increase of temperature
b. increase of concentration of NO
c. increase of concentration of Cl₂
d. increase of concentrations of both Cl₂ and NO
Answer:
a. increase of temperature

Question x.
For an endothermic reaction, X ⇌ Y. If E f is activation energy of the forward reaction and Er that for reverse reaction, which of the following is correct?
a. E_f = E_r
b. E_f < E_r
c. E_f > E_r
d. ∆H = E_f – E_r is negative
Answer:
(c) Ef → Er

2. Answer the following in one or two sentences.

Question i.
For the reaction,
N₂(g) + 3 H₂(g) → 2NH₃(g), what is the relationship among \(\frac{\mathrm{d}\left[\mathrm{N}_{2}\right]}{\mathrm{dt}}\)\(\frac{\mathrm{d}\left[\mathrm{H}_{2}\right]}{\mathrm{dt}} \text { and } \frac{\mathrm{d}\left[\mathrm{NH}_{3}\right]}{\mathrm{dt}} ?\)
Answer:
N_2(g) + 3H_2(g) → 2NH_3(g)
From the above reaction, when 1 mole of N₂ reacts, 3 moles of H₂ are consumed and 2 moles of NH₃ are formed.

If the instantaneous rate R of the reaction is represented in terms of rate of the consumption of N₂ then, \(R=-\frac{d\left[\mathrm{~N}_{2}\right]}{d t}\)

Hence the rate of reaction in terms of concentration changes in N₂, H₂ and NH₃ may be represented as,

Question ii.
For the reaction,
CH₃Br(aq) + OH-(aq) → CH₃OH^Θ (aq) +Br^Θ (aq), rate law is rate = k[CH₃Br][OH^Θ]
a. How does reaction rate changes if [OH^Θ] is decreased by a factor of 5?
b. What is change in rate if concentrations of both reactants are doubled?
Solution :
Given :
(a) Rate = R = k [CH₃Br] x [OH^–]
If R₁ and R₂ are initial and final rates of reaction then,

Rate will be increased 4 time.

Question iii.
What is the relationship between coeffients of reactants in a balanced equation for an overall reaction and exponents in rate law. In what case the coeffients are the exponents?
Answer:
Explanation : Consider the following reaction, aA + bB → products

If the rate of the reaction depends on the concentrations of the reactants A and B, then, by rate law,
R α [A]^a [B]^b
∴ R = k [A]^a [B^b
where [A] = concentration of A and
[B] = concentration of B

The proportionality constant k is called the velocity constant, rate constant or specific rate of the reaction.

a and b are the exponents or the powers of the concentrations of the reactants A and B respectively when observed experimentally.

The exponents or powers may not be necessarily a and b but may be different x and y depending on experimental observations. Then the rate R will be,
R = k [A]^x [B]^y
For example, if x = 1 and y = 2, then,
R = k [A] x [B]²

Question iv.
Why all collisions between reactant molecules do not lead to a chemical reaction?
Answer:
(i) Collisions of reactant molecules : The basic re-quirements of a reaction is that the reacting species (atoms, ions or molecules) must come together and collide for a reaction to occur. Therefore the rate of the reaction shall depend on the rate and frequency of collisions which increase with the amount of reacting species and temperature.

However it is observed that the rate of reaction is very low as compared to the rate of collisions between reacting species in gaseous phase or liquid phase. This suggests that all the collisions are not fruitful leading to a reaction. Hence it is necessary to consider another factor like energy of colliding species along with collision frequency.

(ii) Energy requirement (Activation energy) : The colliding molecules must possess a certain mini-mum energy called activation energy required far breaking and making bonds resulting in the reaction. This implies that the colliding molecules must have energy equal to or greater than the activation energy. The colliding molecules with less energy do not react.

(iii) Orientation of reactant molecules : The concept of activation energy is satisfactory in case of simple molecules or ions but not in case of complex or higher polyatomic molecules. It is observed that the rates of reaction are less as compared to the rates of collisions between activated molecules with activation energy.

This suggests that in addition, the colliding molecules must have proper orientations relative to each other during collisions. For example, consider the reaction, A – B-l-C → A + B – C. For the reaction to occur, C must collide with B while collisions with A will not be fruitful. Since B has to bond with C.

Question v.
What is the activation energy of a reaction?
Answer:
Activation energy : The energy required to form activated complex or transition state from the reactant molecules is called activation energy.
OR
The height of energy barrier in the energy profile diagram is called activation energy.

Question vi.
What are the units for rate constants for zero order and second order reactions if time is expressed in seconds and concentration of reactants in mol/L?
Answer:
(a) For a zero order reaction, the rate constant has units, molL^-1s^-1.
(b) For second order reaction,
Rate = k x [Reactant]²

Question vii.
Write Arrhenius equation and explain the terms involved in it.
Answer:
Arrhenius equation is represented as k = A x e^-E_a/RT
where
k = Rate constant at absolute temperature T
E_a = Energy of activation R = Gas constant
A = Frequency factor or pre-exponential factor.

Question viii.
What is the rate determining step?
Answer:
Many chemical reactions take place in a series of elementary steps. Among many steps of the reaction, one of the steps is the slowest step compared to other steps.

Rate determining step : The slowest step in the reaction mechanism which involves many steps is called the rate-determining step.

Question ix.
Write the relationships between rate constant and half life of fist order and zeroth order reactions.
Answer:
(a) For first order reaction, half-life period t_1/2 is, \(t_{1 / 2}=\frac{0.693}{k}\) where k is the rate constant.
(b) For zeroth-order reaction, half half period (t_1/2) is, \(t_{1 / 2}=\frac{[\mathrm{A}]_{0}}{2 k}\) where k is the rate constant and [A]₀ is initial concentration of the reactant.

Question x.
How do half lives of the fist order and zero order reactions change with initial concentration of reactants?
Answer:
(A) For the first order reaction, half life, t_1/2 is given by, \(t_{1 / 2}=\frac{0.693}{k}\) where k is rate constant. Hence it is independent of initial concentration of the reactant.

(B) Zero order reaction,
\(t_{1 / 2}=\frac{[\mathrm{A}]_{0}}{2 k}\) where [A]₀ is initial concentration of the reactant.

Hence, half life period increases with the increase in concentration of the reactant.

3. Answer the following in brief.

Question i.
How instantaneous rate of reaction is determined?
Answer:
(1) The instantaneous rate is expressed as an infinite¬simal change in concentration (- dc) of the reactant with the infinitesimal change in time (dt).
For a reaction, A → B, let an infinitesimal change in A be – dc in time dt, then Rate \(=\frac{d[\mathrm{~A}]}{d t}\).

Hence, it is represented as,
∴ Instantaneous rate \(=-\frac{d[\mathrm{~A}]}{d t}\)

The negative sign indicates a decrease in the concentration of A.

It is obtained by drawing a tangent to the curve obtained by plotting the concentration against the time. Hence, the slope at a given point represents the instantaneous rate of the reaction.

(2) The instantaneous rate can also be expressed as an infinitesimal change (or increase) in the concentration of the product with the infinitesimal change in time (dt).

Let dB be an infinitesimal change in the concentration of product B in time dt, then Rate \(=\frac{d[\mathrm{~B}]}{d t}=\frac{d x}{d t}\).

Hence,
Instantaneous rate \(=\frac{d x}{d t}\)

It is obtained from the slope of the curve obtained by plotting the concentration of the product against time.

The instantaneous rate is more useful in obtaining the rate law integrated equations.

Question ii.
Distinguish between order and molecularity of a reaction.
Answer:

Order	Molecularity
1. It is the sum of the exponents to which the concentration terms in rate law expression are raised.	1. ¡t is the number of molecules (or atoms or ions) of the reaCtants taking part in the elementary reaction.
2. It is experimentally determined and indicates the dependence of the reaction rate on the concentration of particular reactants.	2. It is the oretical property and indicátes the number of molecules of reactant in each step of the reaction.
3. It may have values that are integer, fractional, or zero.	3. It is always an integer.
4. Its value depends upon experimental conditions.	4. Its value does not depend upon experimental conditions.
5. It is the property of elementary and complex reactions.	5. It is the property of elementary reactions only.
6. Rate law expression describes the order of the reaction.	6. Rate law does not describe molecularity.

Question iii.
A reaction takes place in two steps,
1. NO(g) + Cl₂(g) NOCl₂(g)
2. NOCl₂(g) + NO(g) → 2NOCl(g)
a. Write the overall reaction.
b. Identify reaction intermediate.
c. What is the molecularity of each step?
Solution :
Given :
(1) NO_(g) + Cl_2(g) → NOCl_2(g)
(2) NOCl_2(g) + NO_(g) → 2NOCl_(g)

(a) Overall reaction is obtained by adding both the reactions
2NO_(g) + Cl_2(g) → 2NOCl_2(g)
(b) The reaction intermediate is NOCl₂, since it is formed in first step and consumed in the second step.
(c) Since the first step is a slow and rate determin­ing step, the molecularity is two.

Since the second step is a fast step its molecularity is not considered.

Question iv.
Obtain the relationship between the rate constant and half-life of a fist order reaction.
Answer:
Consider the following reaction,

If [A]₀ and [A]t are the concentrations of A at start and after time t, then [A]₀ = a and [A]_t = a – x.

The velocity constant or the specific rate constant k for the first order reaction can be represented as,

where, a is the initial concentration of the reactant A, x is the concentration of the product B after time t, so that (a – x) is the concentration of the reactant A after time t.

Half-life of a reaction : The time required to reduce the concentration of the reactant to half of its initial value is called the half-life period or the half-life of the reaction.

If t_1/2 is the half-life of a reaction, then at t = t_1/2, x = a/2, hence a – x = a – a/2 = a/2

Hence, for a first order reaction, the half-life of the reaction is independent of the initial concentration of the reactant.

Question v.
How will you represent zeroth-order reaction graphically?
Answer:
(1) A graph of concentration against time : In case of a zero-order reaction, the rate of reaction is independent of the concentration of the reactant. The concentration [A]_t of the reactant at a time t is given by
[A]_t = – kt + [A]₀ (y = – mx + c)
where [A]₀ is the initial concentration of the reactant and k is a rate constant.

Hence in case of zero order reaction, when the concentration of the reactant is plotted against time, a straight line with the slope equal to – k is obtained. The concentration of the reactants de-crease with time. The intercept on the concentration axis gives the initial concentration, [A]₀.

(2) A graph of rate of a reaction against the concen-tration of the reactant: Rate of a zero order reaction is independent of the concentration of the reactant.

Rate, R = k [A]⁰ = k

Hence even if the concentration of the reactant decreases, the rate of the reaction remains constant.

Therefore if the rate of a zero order reaction is plotted against concentration, then a straight line with zero slope is obtained indicating, no change in the rate of the reaction with a change in the concentration of the reactants.

(3) A graph of half-life period against concentration : The half-life period of a zero-order reaction is given by, \(t_{1 / 2}=\frac{[\mathrm{A}]_{0}}{2 k}\) where [A]₀ is initial con-centration of the reactant and k is the rate constant. Hence the half-life period is directly proportional to the concentration.

When a graph of t_1/2 is plotted against concentration, a straight line passing through origin is obtained, and the slope gives \(\frac{1}{2 k}\), where k is the rate constant.

Question vi.
What are pseudo-fist order reactions? Give one example and explain why it is pseudo-fist order.
Answer:
Pseudo-first-order reaction : A reaction which has higher-order true rate law but is experimentally found to behave as first order is called pseudo first order reaction.
Explanation : Consider an acid hydrolysis reaction of an ester like methyl acetate.
CH₃COOCH_3(aq) + H₂O₍₁₎ \(\stackrel{\mathrm{H}_{(\mathrm{aq})}^{+}}{\longrightarrow}\) CH₃COOH_(aq) + CH₃OH_(aq)
Since the reaction involves two substances, ester and water, it is a bimolecular reaction and the true rate law should be, Rate = k’ [CH₃COOCH₃] x [H₂O]

Hence the reaction is expected to follow second order kinetics. However experimentally it is found that the reaction follows first order kinetics.

This is because solvent water being in a large excess, its concentration remains constant. Hence, [H₂O] = constant = k”
Rate = k [CH₃COOCH₃] x [H₂O]
= k [CH₃COOCH₃] x k”
= k’ x k” x [CH₃COOCH₃]
If k’ x k” = k, then Rate = k [CH₃COOCH₃],

This indicates that second-order true rate law is forced into first order rate law. Therefore this bimolecular reaction which appears of second order is called pseudo first order reaction.

Question vii.
What are the requirements for the colliding reactant molecules to lead to products?
Answer:
Collisions of reactant molecules : The basic re­quirements of a reaction is that the reacting species (atoms, ions or molecules) must come together and collide for a reaction to occur. Therefore the rate of the reaction shall depend on the rate and frequency of collisions which increase with the amount of reacting species and temperature.

However it is observed that the rate of reaction is very low as compared to the rate of collisions between reacting species in gaseous phase or liquid phase. This suggests that all the collisions are not fruitful leading to a reaction. Hence it is necessary to consider another factor like energy of colliding species along with collision frequency.

Energy requirement (Activation energy) : The colliding molecules must possess a certain mini­mum energy called activation energy required far breaking and making bonds resulting in the reac­tion. This implies that the colliding molecules must have energy equal to or greater than the activation energy. The colliding molecules with less energy do not react.

Orientation of reactant molecules : The concept of activation energy is satisfactory in case of simple molecules or ions but not in case of complex or higher polyatomic molecules. It is observed that the rates of reaction are less as compared to the rates of collisions between activated molecules with activa­tion energy.

This suggests that in addition, the colliding mole­cules must have proper orientations relative to each other during collisions. For example, consider the reaction, A – B + C → A + B – C. For the reaction to occur, C must collide with B while collisions with A will not be fruitful. Since B has to bond with C.

Question viii.
How catalyst increases the rate of reaction? Explain with the help of a potential energy diagram for catalyzed and uncatalyzed reactions.
Answer:
(i) A catalyst is a substance, when added to the reactants, increases the rate of the reaction without being consumed. For example, the decomposition of KClO₃ in the presence of small amount of MnO₂ is very fast but very slow in the absence of MnO₂.

2KClO_3(s) \(\frac{\mathrm{MnO}_{2}}{\Delta}\) 2KCl_(s) + 3O_2(g)

(ii) The phenomenon of catalysed reaction is called catalysis and depends on nature of the catalyst. In heterogeneous catalysis, the reactant molecules are adsorbed on the solid catalyst surface while in case of homogeneous catalysis, the catalyst reacts with one of the reactants, forms intermediate and decomposes reforming original catalyst and the products.

(iii) The catalyst provides alternative and lower energy path or mechanism for the reaction.

(iv) In the presence of the catalyst, the activation energy of the reaction is lowered. The height of activation energy barrier is less than that in the uncatalysed reaction.

(v) Due to lowering of energy of activation, (E_a) the number of molecules possessing E_a increases, hence the rate of the reaction increases.

(vi) The rate constant = k = A x e^-E_a/RT where A is a frequency factor and hence the rates of the catalysed reaction are higher than those of un-catalysed reactions.

(vii) The catalyst does not change the extent of the reaction but hastens the reaction.

(viii) The catalyst enters the reaction but does not appear in the balanced equation since it is consumed in one step and regenerated in the another.

Question ix.
Explain with the help of the Arrhenius equation, how does the rate of reaction changes with (a) temperature and (b) activation energy.
Answer:
(a) By Arrhenius equation, k = Ax e^-E_a/RT where k is rate constant, A is a frequency factor and Ed is energy of activation at temperature T. As E_a increases, the rate constant and rate of the reaction decreases.

(b) As temperature increases E_a/RT decreases but due to negative sign, k and rate increase with the increase in temperature.

Question x.
Derive the integrated rate law for first order reaction.
Answer:
Consider following gas phase reaction,

Let initial pressure of A be P₀ at t = 0. If after time t the pressure of a A decreases by jc then the partial pressures of the substances will be, P_A = P_Q – x; P_B = x and P_c = x

Total pressure will be,

P_T + P₀ – x + x + x = Po + x
∴ x = P_T – P_n

The partial pressures at time t will be,

Question xi.
How will you represent first-order reactions graphically.
Answer:
(1) A graph of rate of a reaction and concentra­tion : The differential rate law for first-order reac­tion, A → Products is represented as, Rate = [/latex]-\frac{d[\mathrm{~A}]}{d t}=k[\mathrm{~A}]\(

∴ Rate = k x [A]_t (y = mx). When the rate of a first order reaction is plotted against concentration, [A]_t, a straight line graph is obtained.

With the increase in the concentration [A]_t, rate R, increases. The slope of the line gives the value of rate constant k.

(2) A graph of concentration against time : When the concentration of the reactant is plotted against time t, a curve is obtained. The concentration [A], of the reactant decreases exponentially with time. The variation in the concentration can be represented as,

where [A]₀ and [A]_t are initial and final concentra­tions the reactant and k is the rate constant. The time required to complete the first order reaction is infinity.

(3) A graph of log₁₀ (a – x) against time t :

When log₁₀(a – x) is plotted against time t, a straight line with negative slope is obtained, from which the velocity constant k can be calculated.

(4) A graph of half-life period and concentration : The half-life period, t_1/2 of a first order reaction is given by, where k is the rate constant.

For the given reaction at a constant temperature, t_1/2 is constant and independent of the concentration of the reactant.

Hence when a graph of t_1/2 is plotted against concentration, a straight line parallel to the concen­tration axis (slope = zero) is obtained.

(5) A graph of log₁₀ [latex]\left(\frac{a}{a-x}\right)\) against time : The rate constant, for a first order reaction is represented as, where [A₀] and [A]_t are the respective initial and final concentrations of the reactant after time t.


When \(\log _{10}\left(\frac{a}{a-x}\right)\) is plotted against time t, a straight line graph passing through the origin is obtained and the slope gives the value of k/2.303. From this slope, the rate constant can be calculated.

Question xii.
Derive the integrated rate law for the first order reaction, A(g) → B(g) + C(g) in terms of pressure.
Answer:
Consider following gas phase reaction,

Let initial pressure of A be P₀ at t = 0. If after time t the pressure of a A decreases by jc then the partial pressures of the substances will be, P_A = P_Q – x; P_B = x and P_c = x

Total pressure will be,

P_T + P₀ – x + x + x = Po + x
∴ x = P_T – P_n

The partial pressures at time t will be,

Question xiii.
What is zeroth-order reaction? Derive its integrated rate law. What are the units of rate constant?
Answer:
Definition : Zero order reaction : A reaction in which the rate of the reaction does not depend on the concentration of any reactant taking part in the reaction is called zero order reaction.

Consider a zero order reaction, A → Products
The rate of the reaction is, Rate \(=\frac{-d[\mathrm{~A}]}{d t}\)

By rate law,
Rate = k x [A]⁰ = k
∴ – d[A] = k x dt

If [A]₀ is the initial concentration of the reactant A at t = 0 and [A]_t is the concentration of A present after time t, then by integrating above equation,

This is the integrated rate law expression for rate constant for zero order reaction.
∴ k x t = [A]₀ – [A]_t
∴ [A]_t = – kt + A₀

For a zero order reaction :
The rate of reaction is R = k [A]⁰ = k
Hence, the velocity constant k has the unit of the rate of the reaction, i.e., mol dm^-3 s^-1.

Question xiv.
How will you determine activation energy: (a) graphically using Arrhenius equation (b) from rate constants at two different temperatures?
Answer:
(a) By Arrhenius equation,
Rate constant = = A x e^-E_a/RT where A is a fre-quency factor.

When log₁₀k is plotted against 1/T a straight line with negative slope is obtained. From the slope of the graph, energy of activation E_a, is obtained as follows :
Slope = \(\frac{E_{\mathrm{a}}}{2.303 R}\)
∴ Ea = 2303R x sloPe

(b) For the given reaction, rate constants k₁ and k₂ are measured at two different temperatures T₁ and T₂ respectively. Then \(\log _{10} \frac{k_{2}}{k_{1}}=\frac{E_{\mathrm{a}}\left(T_{2}-T_{1}\right)}{2.303 R \times T_{1} \times T_{2}}\) where E_a is the energy of activation.

Hence by substituting appropriate values, energy of activation E_a for the reaction is determined.

Question xv.
Explain graphically the effect of temperature on the rate of reaction.
Answer:
(i) It has been observed that the rates of chemical reactions increase with the increase in temperature.
(ii) The kinetic energy of the molecules increases with the increase in temperature. The fraction of molecules possessing minimum energy barrier,
i. e. activation energy E_a increases with increase in temperature.
(iii) Hence the fraction of colliding molecules that possess kinetic energy (E_a) also increases, hence the rate of the reaction increases with increase in temperature.

(iv) The above figure shows that the area that represents the fraction of molecules with kinetic energy exceeding E_a is greater at higher temperature T₂ than at lower temperature T₁. This explains that the rate of the reaction increases at higher temperature.
(v) The shaded area to the right of activation energy E_a represents fraction of collisions of activated molecules having energy E_a or greater.

Question xvi.
Explain graphically the effect of catalyst on the rate of reaction.
Answer:
(i) The phenomenon of catalysed reaction is called catalysis and depends on nature of the catalyst. In heterogeneous catalysis, the reactant molecules are adsorbed on the solid catalyst surface while in case of homogeneous catalysis, the catalyst reacts with one of the reactants, forms intermediate and decomposes reforming original catalyst and the products.
(ii) The catalyst provides alternative and lower energy path or mechanism for the reaction.
(iii) In the presence of the catalyst, the activation energy of the reaction is lowered. The height of activation energy barrier is less than that in the uncatalysed reaction.

Question xvii.
For the reaction 2A + B → products, find the rate law from the following data.

[A]/M	[A]/M	rate/M s-1
0.3	0.05	0.15
0.6	0.05	0.30
0.6	0.2	1.20

Solution :
Given : 2A + B → Products
Rates : R₁ = 0.15 Ms^-1 R₂ = 0.3 Ms^-1
[A]₁ = 0.3 M [A]₂ = 0.6 M
[B]₁ = 0.05 M [B]₂ = 0.05 M
(i) If order of the reaction in A is x and in B is y then, by rate law,

∴ y = 1. Hence the reaction has order one in B.
The order of overall reaction = n = n_A + n_B = 1 + 1 = 2

Answer:
(i) Rate law : Rate = fc [A] x [B]
Rate constant = k = 10M^-1s^-1
Order of the reaction = 2

4. Solve

Question i.
In a first order reaction, the concentration of reactant decreases from 20 mmol dm^-3 to 8 mmol dm^-3 in 38 minutes. What is the half life of reaction? (28.7 min)
Solution :
Given: [A]₀ =20 mmol dm^-3;
[A]_t=8 mmol dm^-3; t=38 mm;

Answer:
Half life period = 28.74 min

Question ii.
The half life of a first order reaction is 1.7 hours. How long will it take for 20% of the reactant to react? (32.9 min)
Solution :
Given : t_1/2 = 1.7 hr; [A]₀ = 100;
[A]_t = 100 – 20 = 80; t =?
\(t_{1 / 2}=\frac{0.693}{k}\)

Answer:
Time required = t = 32.86 min

Question iii.
The energy of activation for a first order reaction is 104 kJ/mol. The rate constant at 25 0C is 3.7 × 10^-5 s^-1. What is the rate constant at 30⁰C? (R = 8.314 J/K mol) (7.4 × 10^-5)
Solution :

Answer:
k₂ = 7.382 x 10^-5 s^-1

Question iv.
What is the energy of activation of a reaction whose rate constant doubles when the temperature changes from 303 K to 313 K? (54.66 kJ/mol)
Solution :
Given : k₂ = 2k_t, T₁ = 303 K; T₂ = 313 K; E_a = ?

Answer:
Energy of activation = E_a = 54.66 kJ

Question v.
The rate constant of a reaction at 5000C is 1.6 × 10³ M^-1 s^-1. What is the frequency factor of the reaction if its activation energy is 56 kJ/mol. (9.72 × 106 M^-1 s^-1)
Solution :

Answer:
Frequency factor = A = 9.727 x 10⁶ M^-1s^-1

Question vi.
Show that time required for 99.9% completion of a first order reaction is three times the time required for 90% completion.
Solution :
Given : For 99.9 % completion, if [A]₀ = 100,

If t₁ and t₂ are the times required for 99.9 % and 90 % completion of reaction respectively, then

Answer:
Time required for 99.9 % completion of a first order reaction is three time the time required for 90 % completion of the reaction.

Question vii.
A first order reaction takes 40 minutes for 30% decomposition. Calculate its half life. (77.66 min)
Solution :

Answer:
Half life period = 77.70 min.

Question viii.
The rate constant for the first order reaction is given by log₁₀ k = 14.34 – 1.25 × 10⁴ T. Calculate activation energy of the reaction. (239.3 kJ/mol)
Solution :
Given : log₁₀ k = 14.34 – \(\frac{1.25 \times 10^{4}}{T}\) ……………………. (1)
From Arrhenius equation we can write,
\(\log _{10} k=\log _{10} A-\frac{E_{\mathrm{a}}}{2.303 R \times T}\) ……………………. (2)
By comparing equations (1) and (2),
\(\frac{E_{\mathrm{a}}}{2.303 \times R}\) = 1.25 x 10⁴
∴ Ea = 1.25 x 10⁴ x 2.303 x R
= 1.25 x 10⁴ x 2.303 x 8.314
= 23.93 x 10⁴ = 239.3 kJ mol^-1

[Note : Frequency factor A may also be calculated as follows : log₁₀ A = 14.34
∴ A = Antilog 14.34 = 2.188 x 10⁴
Answer:
Energy of activation = E_a = 239.3 kJ mol^-1.

Question ix.
What fraction of molecules in a gas at 300 K collide with an energy equal to activation energy of 50 kJ/mol? (2 × 10^-9)
Solution :
Given : T = 300 K; E_a = 50 kJ mol^-1
= 50 x 10³ mol^-1
The fraction of molecules undergoing fruitful collisions is

Answer:
Fraction of molecules undergoing collision = 2 x 10^-9

Activity :
1. If you wish to determine the reaction order and rate constant for the reaction, 2AB₂ → A₂ + 2B₂.
a) What data would you collect?
b) How would you use these data to determine whether the reaction is zeroth or first order?

2. The activation energy for two reactions are E_a and E’_a with E_a > E’_a. If the temperature of reacting system increases from T₁ to T₂, predict which of the following is correct?

k values are rate constants at lower temperatures and k values at higher temperatures.

12th Chemistry Digest Chapter 6 Chemical Kinetics Intext Questions and Answers

(Textbook Page No 121)

Question 1.
Write the expressions for rates of reaction for :
2N₂O_5(g) → 4NO_2(g) + O_2(g)?
Answer:
For the given reaction, Rate of reaction =
\(=R=-\frac{1}{2} \frac{d\left[\mathrm{~N}_{2} \mathrm{O}_{5}\right]}{d t}\)
\(\begin{aligned}
&=+\frac{1}{4} \frac{d\left[\mathrm{NO}_{2}\right]}{d t} \\
&=+\frac{d\left[\mathrm{O}_{2}\right]}{d t}
\end{aligned}\)

Problem 6.1: (Textbook Page No 121)

Question 1.
For the reaction,
\(\mathbf{3 I}_{(a q)}^{-}+\mathbf{S}_{2} \mathbf{O}_{8(a q)}^{2-} \longrightarrow \mathbf{I}_{3(\text { (aq) }}^{-}+2 \mathbf{S O}_{4(\mathrm{aq})}^{2-}\)
Calculate (a) the rate of formation of I₃^–,
(b) the rates of consumption of 1 and S₂O and (c) the overall rate of reaction if the rate of formation of \(\mathrm{SO}_{4}^{2-}\) is 0.O22 moles dm^-3 sec^-1.
Answer:


∴ (a) Rate of formation of \(\mathrm{I}_{3}^{-}\) = 0.011 mol dm^-3 s^-1
(b) Rate of consumption of I^– = 0.033 mol dm^-3 s^-1
(c) Rate of consumption of \(\mathrm{S}_{2} \mathrm{O}_{8}^{2-}\) = 0.011 mol dm^-3 s^-1
(d) Overall rate of reaction = Rate of consumption of reactant = Rate of formation of product

Try this….. (Textbook Page No 122)

Question 1.
For the reaction :
NO_2(g) + CO_(g) → NO_(g) + CO_2(g), the rate of reaction is experimentally found to be proportional to the square of the concentration of NO2 and independent that of CO. Write the rate law.
Answer:
Since the rate of the reaction is proportional to [NO₂]² and [CO]⁰, the rate law is R = k[NO₂]² x [CO]⁰
∴ R = k[NO₂]².

Try this….. (Textbook Page No 124)

Question 1.
The reaction,
CHCl_3(g) + Cl_2(g) → CCl_4(g) + HCl_(g) is first order in CHCl₃ and 1/2 order in Cl₂. Write the rate law and overall order of reaction.
Answer:
Since the reaction is first order in CHCl₃ and 1/2 order in Cl₂, the rate law for the reaction will be, Rate = k[CHCl₃] X [Cl₂]^1/2
The overall order (n) of the reaction will be, n = l + = \(\frac{1}{2}=\frac{3}{2}\)

Use your brain power! (Textbook Page No 124)

Question 1.
The rate of the reaction 2A + B → 2C + D is 6 x 10^-4 mol dm^-3 s^-1 when [A] =[B] = O.3 mol dm^-3 If the reaction is of first order in A and zeroth order in B, what is the rate constant?
Answer:
For the reaction,
2A + B → 2C + D,

(Problem 6.7) (Textbook Page No 126)

Question 1.
A reaction occurs in the following steps :
(i) NO_2(g) + F_2(g) → NO₂F_(g) + F_(g) (slow)
(ii) F_(g) + NO_2(g) → NO₂F_(g) (fast)
(a) Write the equation of overall reaction.
(b) Write down rate law.
(c) Identify the reaction intermediate.
Solution :
(a) The addition of two steps gives the overall reaction as
2NO_2(g) + F_2(g) → 2NO₂ F_(g)
(b) Step (i) is slow. The rate law of the reaction is predicted from its stoichiometry. Thus, rate = k [NO₂] [F₂]
(c) F is produced in step (i) and consumed in step (ii) hence F is the reaction intermediate.

Try this….. (Textbook Page No 126)

Question 1.
A complex reaction takes place in two steps :
(i) NO_(g) + O_3(g) → NO_3(g) + O_(g)
(ii) NO_3(g) + O_(g) → NO_2(g) + O_2(g)
The predicted rate law is rate = k [NO] [O₃]. Identify the rate-determining step. Write the overall reaction. Which is the reaction inter-mediate? Why?
Answer:
(i) NO_(g) + O_3(g) → NO_3(g) + O_(g)
(ii) NO_3(g) + O_(g) → NO_2(g) + O_2(g)
(a) The first step is slow and rate determining step since the rate depends on concentrations of NO_(g) and O_3(g). (Given : Rate = k [NO] x [O])
(b) The overall reaction is the combination of two steps.
NO_(g) ⁺ O_3(g) → NO_2(g) + O_2(g)
(c) NO_3(g) and O_(g) are reaction intermediates. They are formed in first step (i) and removed in the second step (ii).

Try this….. (Textbook Page No 129)

Question 1.
The half-life of a first-order reaction is 0.5 min. Calculate (a) time needed for the reactant to reduce to 20% and (b) the amount decomposed in 55 s.
Answer:

Try this….. (Textbook Page No 123)

Question 1.
For the reaction 2A + 2B → 2C + D, if concentration of A is doubled at constant [B] the rate increases by a factor of 4. If the concentration of B is doubled with [A] being constant the rate is doubled. Write the rate law of the reaction.
Answer:
Rate = R₁ = k[A]^x [B]^y
When concentration of A = [2A] and

Hence order with respect to A is 2 and with respect to B is 1. By rate law,
Rate = A: [A]² [B]

Question 2.
The rate law for the reaction A + B → C is found to be rate = k [A]² x [B]. The rate constant of the reaction at 25 °C is 6.25 M^-2 S^-1. What is the rate of reaction when [A] = 1.0 mol dm^-3 and [B] = 0.2 mol dm^-3?
Answer:
Rate = k x [A]² x [B]
= 6.25 x 1² x 0.2
Rate = 1.25 x 10² mol dm^-3 s^-1

12th Std Chemistry Questions And Answers:

• Solid State Class 12 Chemistry Questions And Answers
• Solutions Class 12 Chemistry Questions And Answers
• Ionic Equilibria Class 12 Chemistry Questions And Answers
• Chemical Thermodynamics Class 12 Chemistry Questions And Answers
• Electrochemistry Class 12 Chemistry Questions And Answers
• Chemical Kinetics Class 12 Chemistry Questions And Answers
• Elements of Groups 16, 17 and 18 Class 12 Chemistry Questions And Answers
• Transition and Inner Transition Elements Class 12 Chemistry Questions And Answers

Balbharti Maharashtra State Board 12th Chemistry Textbook Solutions Chapter 11 Alcohols, Phenols and Ethers Textbook Exercise Questions and Answers.

Alcohols, Phenols and Ethers Class 12 Exercise Question Answers Solutions Maharashtra Board

Class 12 Chemistry Chapter 11 Exercise Solutions Maharashtra Board

Chemistry Class 12 Chapter 11 Exercise Solutions

1. Choose the correct option.

Question i.
Which of the following represents the increasing order of boiling points of (1), (2) and (3)?
(1) CH₃ – CH₂ – CH₂ – CH₂ – OH
(2) (CH₃)₂ CH – O – CH₃
(3) (CH₃)₃COH
A. (1) < (2) < (3)
B. (2) < (1) < (3)
C. (3) < (2) < (1)
D. (2) < (3) < (1)
Answer:
(a) (1) < (2) < (3)

Question ii.
Which is the best reagent for carrying out following conversion ?

A. LiAlH₄
B. Conc. H₂SO₄, H₂O
C. H₂/Pd
D. B₂H₆, H₂O₂ – NaOH
Answer:
B. Conc. H₂SO₄, H₂O

Question iii.
Which of the following reaction will give ionic organic product on reaction ?
A. CH₃ – CH₂ – OH + Na
B. CH₃ – CH₂ – OH + SOCl₂
C. CH₃ – CH₂ – OH + PCl₅
D. CH₃ – CH₂ – OH + H₂SO₄
Answer:
C. CH₃ – CH₂ – OH + PCl₅

Question iv.
Which is the most resistant alcohol towards oxidation reaction among the follwoing ?

Answer:
(c)

Question v.
Resorcinol on distillation with zinc dust gives
A. Cyclohexane
B. Benzene
C. Toluene
D. Benzene-1, 3-diol
Answer:
(b) Benzene

Question vi.
Anisole on heating with concerntrated HI gives
A. Iodobenzene
B. Phenol + Methanol
C. Phenol + Iodomethane
D. Iodobenzene + methanol
Answer:
B. Phenol + Methanol

Question vii.
Which of the following is the least acidic compound ?

Answer:
(b)

Question viii.
The compound incapable of hydrogen bonding with water is ……

Answer:
(b)

Question ix.
Ethers are kept in air tight brown bottles because
A. Ethers absorb moisture
B. Ethers evaporate readily
C. Ethers oxidise to explosive peroxide
D. Ethers are inert
Answer:
C. Ethers oxidise to explosive peroxide

Question x.
Ethers reacts with cold and concentrated H₂SO₄ to form
A. oxonium salt
B. alkene
C. alkoxides
D. alcohols
Answer:
A. oxonium salt

2. Answer in one sentence/ word.

Question i.
Hydroboration-oxidation of propene gives…..
Answer:
n-propyl alcohol (CH₃ – CH₂ – CH₂ – OH)

Question ii.
Write the IUPAC name of alcohol having molecular formula C₄H₁₀O which is resistant towards oxidation.
Answer:

Question iii.
Write the structure of optically active alcohol having molecular formula C₄H₁₀O
Answer:

Question iv.
Write name of the electrophile used in Kolbe’s Reaction.
Answer:
Electrophile : Carbon dioxide (O = C = O)

3. Answer in brief.

Question i.
Why phenol is more acidic than ethyl alcohol ?
Answer:
(1) In ethyl alcohol, the -OH group is attached to sp³ – hybridised carbon while in phenols, it is attached to sp² – hybridised carbon.

(2) Due to higher electronegativity of sp² – hybridised carbon, electron density on oxygen decreases. This increases the polarity of O-H bond and results in more ionization of phenol than that of alcohols.

(3) Electron donating inductive effect (+1 effect) of the alkyl group destabilizes alkoxide ion. As a result alcohol does not ionize much in water, therefore alcohol is neutral compound in aqueous medium.

(4) In alkoxide ion, the negative charge is localized on oxygen, while in phenoxide ion the negative charge is delocalized. The delocalization of the negative charge (structure I to V) makes phenoxide ion more stable than that of phenol.

The delocalization of charge in phenol (structures VI to X), the resonating structures have charge separation (where oxygen atom of OH group to be positive and delocalization of negative charge over the ortho and para positions of aromatic ring) due to which phenol molecule is less stable than phenoxide ion. This favours ionization of phenol. Thus phenols are more acidic than ethyl alcohol.

Question ii.
Why p-nitrophenol is a stronger acid than phenol ?
Answer:
(1) In p-nitrophenol, nitro group (NO₂) is an electron withdrawing group present at para position which enhances the acidic strength (-1 effect). The O-H bond is under strain and release of proton (H⁺) becomes easy. Further p-nitrophenoxide ion is more stabilised due to resonance.

(2) Since the absence of electron withdrawing group (like – NO₂) in phenol at ortho and para position, the acidic strength of phenol is less than that of p-nitrophenol.

Question iii.
Write two points of difference between properties of phenol and ethyl alcohol.
Answer:

Question iv.
Give the reagents and conditions necessary to prepare phenol from
a. Chlorobenzene
b. Benzene sulfonic acid.
Answer:
(1) From chlorobenzene : Reagents required : NaOH and dil. HC1 Temperature : 623 K, Pressure : 150 atm
(2) From Benzene sulphonic acid : Reagents required : aq NaOH, caustic soda, dil. HC1 Temperature : 573 K

Question v.
Give the equations of the reactions for the preparation of phenol from isopropyl benzene.
Answer:
Preparation of phenol from cumene (isopropylbenzene) : This is the commercial method of preparation of phenol. When a stream of air is passed through cumene (isopropylbenzene) suspended in aqueous Na₂CO₃ solution in the presence of cobalt naphthenate catalyst, isopropyl benzene hydroperoxide or cumene hydroperoxide is formed. Isopropylbenzene hydroperoxide on warming with dil. H₂SO₄ gives phenol and acetone. Acetone is an important by-product of the reaction and is separated by distillation. The reaction is called auto oxidation.

Question vi.
Give a simple chemical test to distinguish between ethanol and ethyl bromide.
Answer:
When ethyl bromide is heated with aq NaOH; ethyl alcohol is formed whereas ethanol does not react with aq NaOH

4. An ether (A), C₅H₁₂O, when heated with excess of hot HI produce two alkyl halides which on hydrolysis form compound (B)and (C), oxidation of (B) gave and acid (D), whereas oxidation of (C) gave a ketone (E). Deduce the structural formula of (A), (B), (C), (D) and (E).
Answer:

5. Write structural formulae for

a. 3-Methoxyhexane
b. Methyl vinyl ether
c. 1-Ethylcyclohexanol
d. Pentane-1,4-diol
e. Cyclohex-2-en-1-ol
Answer:

6. Write IUPAC names of the following

Answer:

Activity :
• Collect information about production of ethanol as byproduct in sugar industry and its importance in fuel economy.
• Collect information about phenols used as antiseptics and polyphenols having antioxidant activity.

12th Chemistry Digest Chapter 11 Alcohols, Phenols and Ethers Intext Questions and Answers

Use your brain power! (Textbook Page No 235)

Question 1.
Classify the following alcohols as l0/2°/3° and allylic/benzylic

Answer:
(1) Ally lie alcohol (primary)
(2) Allylic alcohol (secondary)
(3) Allylic alcohol (tertiary)
(4) Benzylic alcohol (primary)
(5) Benzylic alcohol (secondary)

Use your brain power ….. (Textbook Page No 236)

Question 1.
Name t-butyl alcohol using carbinol system of nomenclature.
Answer:
Trimethyl carbinol.

Problem 11.1 (Textbook Page No 238)

Question 1.
Draw structures of following compounds:
(i) 2,5-DiethIphenoI
(ii) Prop-2-en-I-oI
(iii) 2-methoxypropane
(iv) Phenylmethanol
Solution :

Try this ….. (Textbook Page No 238)

Write IUPAC names ol (he following compounds.

Do you know (Textbook Page No 238)

Question 1.
The mechanism of hydration of ethylcnc to ethyl alcohol.
Answer:
The mechanism of hydration of ethylene involves three steps:

Step 1: Ethylene gets protonated to form carbocation by electrophilic attack of H₃O (Formation of carbocation intermediate).

Step 2 : Nucleophilic attack of water on carbocation

Step 3 : Deprotonation to form an alcohol

Problem 11.2 : (Textbook Page No 239)

Question 1.
Predict the products for the following reaction.

Solution:
The substrate (A) contains an isolated and an aldehyde group. H₂/Ni can reduce both these functional groups while LiAlH₄ can reduce only – CHO of the two, Hence

Try this ….. (Textbook page 240)

Question 1.
Arrange O – H, C – H and N – H bonds in increasing order of their bond polarity.
Answer:
Increasing order of polarity :C – H, N – H, O – H

Problem 11.3 : (Textbook Page No 241)

Question 1.
The boiling point of n-butyl alcohol, isobutyl alcohol, sec-butyl alcohol and tert-butyl alcohol are 118 °C, 108 °C. 99 °C and 82 °C respectively. Explain.
Solution:
As branching increases, intermolecular van der Waal’s force become weaker and the boiling point decreases. Therefore, n-butyl alcohol has highest boiling point 118 °C and tert-butyl alcohol has lowest boiling point 83 °C. Isobutyl alcohol is a primary alcohol and hence its boiling point is higher than that of sec-butyl alcohol.

Problem 11.4 : (Textbook Page No 242)

The solubility of o-nitrophenol and p-nitrophenol is 0.2 g and 1.7 g/100 g of H₂O respectively. Explain the difference.
Solution :

p-Nitrophenol has strong intermolecular hydrogen bonding with solvent water. On the other hand, o-nitrophenol has strong intramolecular hydrogen bonding and therefore the intermolecular attraction towards solvent water is weak. The stronger the intermolecular attraction between solute and solvent higher is the solubility. Hence p-nitrophenol has higher solubility in water than that of o-nitrophenol.

Problem 11.5 : (Textbook Page No 243 & 244)

Question 1.
Arrange the following compounds in decreasing order of acid strength and justify.
(1) CH₃ – CH₂ – OH
(2) (CH₃)₃ C – OH
(3) C₆H₅ – OH
(4) p-NO₂ – C₆H₄ – OH
Solution :
Compounds (3) and (4) are phenols and therefore are more acidic than the alcohols (1) and (2). The acidic strengths of compounds depend upon stabilization of the corresponding conjugate bases. Hence let us compare electronic effects in the conjugate bases of these compounds :

The conjugate base of the alcohol (1) is destabilized by + 1 effect of one alkyl group, whereas conjugate base of the alcohol (2) is destabilized by +1 effect of three alkyl groups. Hence (2) is weaker acid than (1)

Phenols : The conjugate base of p-nitrophenol (4) is better resonance stabilized due to six resonance structures compared to the five resonance structure of conjugate base of phenol (3). The resonance structure VI has – ve charge on only electronegative oxygens. Hence the phenol (4) is stronger acid than (3). Thus the decreasing order of acid strength is (4), (3), (1), (2).

Use your brain power (Textbook Page No 244)

Question 1.
What are the electronic effects exerted by – OCH₃ and – Cl? Predict the acid strength of

Answer:
The electronic effects exerted by – Cl and – O CH₃ are as follows :
(1) Cl being more electronegative atom it pulls the bonding electrons towards itself. This is known as negative inductive effect (- I).

(2) – OCH₃ is less electronegative group which repels the bonding electrons away from it. This is known as positive inductive effect ( + I).

(3) The relative to parent phenol, is more acidic than .

Problem 11.6 : (Textbook Page No 245)

Question 1.
Mechanism of acid catalyzed dehydration of ethanol to give ethene.
Answer:
The mechanism of dehydration of ethanol involves the following order :
Step 1 : Formation of protonated alcohols : Initially ethyl alcohol gets protonated to form ethyl oxonium ion.

Step 2 : Formation of carbocation : It is the slowest step and hence, the rate determining step of the reaction.

Steps 3: Formation of ethene: Removal of a proton (H⁺) from carbocation.

The acidused in step I is released in step 3, the equilibrium is shifted to the right, ethene is removed as it is formed.

Problem 11.6 : (Textbook Page No 245)

Question 1.
Write the reaction showing major and minor products formed on heating butan-2-ol with concentrrated sulphuric acid.
Solution :
In the reaction described butan-2-ol undergoes dehydration to give but-2-ene (major) and but-l-ene (minor) in accordance with Saytzeff rule.

Problem 11.7 : (Textbook Page No 246)

Question 1.
Write and explain reactions to convert propan-l-ol into propan-2-ol.
Solution :
The dehydration of propane-l-ol to propene is the first step. Markownikoff hydration of propene is the second step to get the product propan-2-ol. This is brought about by reaction with concemtrated H₂SO₄ followed by hydrolysis.

Problem 11.8 : (Textbook Page No 246)

Question 1.
An organic compound gives hydrogen on reaction with sodium metal. It forms an aldehyde having molecular formula C₂H₄O on oxidation with pyridinium chlorochromate. Name the compounds and give equations of these reactions.
Solution :
The given molecular formula C₂H₄O of aldehyde is written as CH₃ – CHO. Hence the formula of alcohol from which this is obtained by oxidation must be CH₃ – CH₂ – OH. The two reactions can, therefore, be represented as follows :

(Do you know? Textbook Page No 248)

Question 90.
Write the mechanism of dehydration of alcohol to give ether.
Answer:
Dehydration of alcohols to form ether is SN₂ reaction. The mechanism of dehydration of ethanol involves the following steps.

Step 1 (Protonation) : Initially ethyl alcohol gets protonated in the presence of acid to form ethyl oxonium ion.

Step 2 (SN² mechanism) : Protonated alcohol species undergoes a backside attack by second molecule of alcohol is a slow step.

Step 3 (Deprotonation) : Formation of diethyl ether by elimination of proton

Problem 11.9 : (Textbook Page No 249)

Question 1.
Ethyl isopropyl ether does not form on reaction of sodium ethoxide and isopropyl chloride.

(i) What would be the main product of this reaction?
(ii) Write another reaction suitable for the preparation of ethyl isopropyl ether.
Solution :
(i) Isopropyl chloride is a secondary chloride. On treating with sodium ethoxide it gives elimination reaction to form propene as the main product.

(ii) Ethyl isopropyl ether can be prepared as follows using ethyl chloride (1⁰ chloride) as substrate.

Do you know? (Textbook Page No 250)

Question 1.
The mechanism of the reaction of HI with methoxy ethane.
Answer:
The reaction mechanism takes place as follows :
Step 1 : Protonation of ether Initially the ether molecule (methoxy ethane) protonated by cone. HI to form oxonium ion.

Step 2 : Iodide is a good nucleophile. It attacks the least substituted carbon of the oxonium ion formed in step 1 and displaces an alcohol molecule by SN² mechanism.

For example :
• Use of excess HI converts the alcohol into alkyl iodide.
• In case of ether having one tertiary alkyl group the reaction with hot HI follows SN1 mechanism, and tertiary iodide is formed rather than tertiary alcohol.

Step 1 :

Step 2 :

12th Std Chemistry Questions And Answers:

• Coordination Compounds Class 12 Chemistry Questions And Answers
• Halogen Derivatives Class 12 Chemistry Questions And Answers
• Alcohols, Phenols and Ethers Class 12 Chemistry Questions And Answers
• Aldehydes, Ketones and Carboxylic Acids Class 12 Chemistry Questions And Answers
• Amines Class 12 Chemistry Questions And Answers
• Biomolecules Class 12 Chemistry Questions And Answers
• Introduction to Polymer Chemistry Class 12 Chemistry Questions And Answers
• Green Chemistry and Nanochemistry Class 12 Chemistry Questions And Answers

Balbharti Maharashtra State Board 12th Chemistry Textbook Solutions Chapter 15 Introduction to Polymer Chemistry Textbook Exercise Questions and Answers.

Introduction to Polymer Chemistry Class 12 Exercise Question Answers Solutions Maharashtra Board

Class 12 Chemistry Chapter 15 Exercise Solutions Maharashtra Board

Chemistry Class 12 Chapter 15 Exercise Solutions

1. Choose the correct option from the given alternatives.

Question i.
Nylon fibers are …………………………………..
A. Semisynthetic fibres
B. Polyamide fibres
C. Polyester fibres
D. Cellulose fibres
Answer:
B. polyamide fibres

Question ii.
Which of the following is naturally occurring polymer?
A. Telfon
B. Polyethylene
C. PVC
D. Protein
Answer:
D. Protein

Question iii.
Silk is a kind of …………………………………. fibre
A. Semisynthetic
B. Synthetic
C. Animal
D. Vegetable
Answer:
C. an animal

Question iv.
Dacron is another name of …………………………………. .
A. Nylon 6
B. Orlon
C. Novolac
D. Terylene
Answer:
D. Terylene

Question v.
Which of the following is made up of polyamides?
A. Dacron
B. Rayon
C. Nylon
D. Jute
Answer:
C. Nylon

Question vi.
The number of carbon atoms present in the ring of ε – caprolactam is
A. Five
B. Two
C. Seven
D. Six
Answer:
D. Six

Question vii.
Terylene is …………………………………. .
A. Polyamide fibre
B. Polyester fibre
C. Vegetable fibre
D. Protein fibre
Answer:
B. Polyester fibre

Question viii.
PET is formed by …………………………………. .
A. Addition
B. Condensation
C. Alkylation
D. Hydration
Answer:
D. Hydration

Question ix.
Chemically pure cotton is …………………………………. .
A. Acetate rayon
B. Viscose rayon
C. Cellulose nitrate
D. Cellulose
Answer:
D. Cellulose

Question x.
Teflon is chemically inert, due to presence of …………………………………. .
A. C-H bond
B. C-F bond
C. H- bond
D. C=C bond
Answer:
A. C-H bond

2. Answer the following in one sentence each.

Question i.
Identify ‘A’ and ‘B’ in the following reaction …………………………………. .

Answer:

Question ii.
Complete the following statements
a. Caprolactam is used to prepare …………………………………. .
b. Novolak is a copolymer of …………………………………. and …………………………………. .
c. Terylene is ………………………………….. polymer of terephthalic acid and ethylene glycol.
d. Benzoyl peroxide used in addtion polymerisation acts as …………………………………. .
e. Polyethene consists of polymerised …………………………………. .
Answer:
a. Nylon-6
b. Phenol, formaldehyde
c. polyester
d. initiator (catalyst)
e. linear or branched-chain

Question iii.
Draw the flow chart diagram to show the classification of polymers based on type of polymerisation.
Answer:

Question iv.
Write examples of Addition polymers and condensation polymers.
Answer:
Addition polymers : Polyvinyl chloride, polythene
Condensation polymers : Bakelite, terylene, Nylon-66

Question v.
Name some chain-growth polymers.
Answer:
Chain growth polymers : Polythene, polyacrylonitrile and polyvinyl chloride.

Question vi.
Define the terms :
1) Monomer
2) Vulcanisation
3) Synthetic fibres
Answer:
1. Monomer is a small and simple molecule and has a capacity to form two chemical bonds with other monomers. Examples : Ethene, Propylene.
2. The process by which a network of cross-links is introduced into an elastomer is called vulcanisation or it can also be defined as the process of heating natural rubber with sulphur to increase the tensile strength, toughness and elasticity of natural rubber is known as vulcanization of rubber.
3. The man-made fibres prepared by polymerization of one monomer or copolymerization of two or more monomers are called synthetic fibres.

Question vii.
What type of intermolecular force leads to high-density polymer?
Answer:
High density polymers have low degree of branching along the hydrocarbon chain. The molecules are closely packed together during crystallization. This closer packing means that the van der Waals attraction between the chains are greater and so the plastic (high density polymer) is stronger and has a melting point.

Question viii.
Give one example each of copolymer and homopolymer.
Answer:
Homopolymer : PVC, Nylon-6
Copolymer : Terylene, Buna-S

Question ix.
Identify Thermoplastic and Thermosetting Plastics from the following …………………………………. .
1. PET
2. Urea-formaldehyde resin
3. Polythene
4. Phenol formaldehyde
Answer:
Thermoplastic plastics : PET, Polythene
Thermosetting plastics : Urea formaldehyde resin, Phenol formaldehyde

3. Answer the following.

Question i.
Write the names of classes of polymers formed according to intermolecular forces and describe briefly their structural characteristics.
Answer:
Molecular forces bind the polymer chains either by hydrogen bonds or Vander Waal’s forces. These forces are called intermolecular forces. On the basis of magnitude of intcrmolccular forces, polymers are further classified as ebstomers, fibres, thermoplastic polymers. thermosetting polymers.

(1) Elastomers: Weak van der Waals type of intermolecular forces of attraction between the polymer chains are observed in cbstomcrs. When polymer is stretched, the polymer chain stretches and when the strain is relieved the chain returns to its odginal position, Thus, polymer shows elasticity and is called elastomers. Elastomers, the elastic polymers, have weak van der Waals type of intermolecular forces which permit the polymer to be stretched. Lilastorners are soft and stretchy and used in making rubber bands. E.g.. neoprene, vulcanized rubber, buna.S, buna-N.

(2) Fibres : It consists of strong intermolecular forces of’ attraction due to hydrogen bonding and strong dipole-dipole forces. These polymers possess high tensile strength. Due to these strong intermolecular forces the fibres are crystalline in nature. They are used in textile industries, strung tyres. etc.. e.g., nylon, terylene.

(3) Thermoplastic polymers: These polymer possess moderately strong intermolecular forces of attraction between those of elastomers and fibres. These polymers arc called thermoplastic because they become soft on heating and hard on cooling. They are either linear or branched chain polymers. They can be remoulded and recycled. E.g. polyethenc, PVC, polystyrene.

(4) Thermosetting polymers: These polymers are cross linked or branched molecules and are rigid polymers. During their formation they have property of being shaped on heating. but they get hardened while hot. Once hardened these become infusible, cannot be softened by heating and therefore, cannot be remoulded and recycled.
This shows extensive cross linking by covalent bonds formed in the moulds during hardening/setting process while hot. E.g. Bakelite, urea formaldehyde resin.

Question ii.
Write reactions of formation of :
a. Nylon 6
b. Terylene
Answer:
Terylene is polyester fibre formed by the polymerization of terephthalic acid and ethylene glycol.

Terylene is obtained by condensation polymerization of ethylene glycol and terephthalic acid in presence of catalyst zinc acetate and antimony trioxide at high temperature.

Properties :

• Terylene has relatively high melting point (265 °C)
• It is resistant to chemicals and water.

Uses :

• It is used for making wrinkle free fabrics by blending with cotton (terycot) and wool (terywool), and also as glass reinforcing materials in safety helmets.
• PET is the most common thermoplastic which is another trade name of the polyester polyethylenetereph- thalate.
• It is used for making many articles like bottles, jams, packaging containers.

Question iii.
Write the structure of natural rubber and neoprene rubber along with the name and structure of thier monomers.
Answer:

Question iv.
Name the polymer type in which the following linkage is present.
Answer:
The linkage is present in terylene or dacron polymer.

Question v.
Write the structural formula of the following synthetic rubbers :
a. SBR rubber
b. Buna-N rubber
c. Neoprene rubber
Answer:

Question vi.
Match the following pairs :
Name of polymer – Monomer
1. Teflon – a. CH₂ = CH₂
2. PVC – b. CF₂ = CF₂
3. Polyester – c. CH₂ = CHCl
4. Polythene – d. C₆H₅OH and HCHO
5. Bakelite – e. Dicarboxylic acid and polyhydoxyglycol
Answer:

1. Teflon – CF₂ = CF₂
2. PVC – CH₂ = CHCI
3. Polyester-Dicarboxylic acid and polyhydoxyglycol
4. Polythene – CH₂ = CH₂
5. Bakelite – C₆H₅OH and HCHO

Question vii.
Draw the structures of polymers formed from the following monomers
1. Adipic acid + Hexamethylenediamine
2. e – Aminocaproic acid + Glycine
Answer:

Question viii.
Name and draw the structure of the repeating unit in natural rubber.
Answer:
Repeating unit of natural rubber (Basic unit : isoprene)

Question ix.
Classify the following polymers as natural and synthetic polymers
a. Cellulose
b. Polystyrene
c. Terylene
d. Starch
e. Protein
f. Silicones
g. Orlon (Polyacrylonitrile)
h. Phenol-formaldehyde resins
Answer:

Natural Polymers	1. Cellulose 4. Starch 5. Protein
Synthetic Polymers	2. Polystyrene 3. Terylene 6. Silicones 7. Orion (Polyacrylonitrile) 8. phenol-formaldehyde resin

Question x.
What are synthetic resins? Name some natural and synthetic resins.
Answer:
Synthetic resins are artificially synthesised high molecular weight polymers. They are the basic raw material of plastic. The main properties of plastic depend on the synthetic resin it is made from.

Examples of natural resins : Rosin, Damar, Copal, Sandarac, Amber, Manila
Examples of synthetic resins : Polyester resin, Phenolic resin, Alkyl resin, Polycarbonate resin, Polyamide resin, Polyurethane resin, silicone resin, Epoxy resin, Acrylic resin.

Question xi.
Distinguish between thermosetting and thermoplastic resins. Write example of both the classes.
Answer:

Thermosetting resin	Thermoplastic resin
(1) They harden when heated. Once hardened it no longer melts.	(1) They soften when heated and harden again when cooled.
(2) They cannot be re-shaped.	(2) They can be reshaped
(3) They are strong, hard.	(3) They are weak, soft.
(4) Thermosetting resin show cross-linking. Examples : Melamine resin Epoxy resins, Bake-lite.	(4) Thermoplastic molecules do not cross link, hence are flexible. Examples : Polythene, polypropylene, nylon, polyester.

Question xii.
Write name and formula of raw material from which bakelite is made.
Answer:
The raw material or monomers used to prepare bakelite are o-hydroxymethyl phenol and formaldehyde (HCHO)

4. Attempt the following :

Question i.
Identify condensation polymers and addition polymers from the following.

Answer:

Question ii.
Write the chemical reactions involved in the manufacture of Nylon 6, 6
Answer:
Nylon-6, 6 is a linear polyamide polymer formed by the condensation polymerisation reaction. The monomers used in the preparation of Nylon-6, 6 are :
(1) Adipic acid : HOOC-(CH₂)₄-COOH
(2) Hexamethylene diamine : H₂N-(CH₂)₆-NH₂

When equimolar aqueous solutions of adipic acid and hexamethylene diamine are mixed and heated, there is neutralization to form a nylon salt. During polymerisation at 553 k nylon salt loses a water molecule to form nylon 6, 6 polymer. Both monomers (hexamethylene diamine and adipic acid) contain six carbon atoms each, hence the polymer is termed as Nylon-6,6.

Properties and uses : Nylon 6,6 is high molecular mass (12000 – 50000 u) linear condensation polymer. It possesses high tensile strength. It does not soak in water. It is used for making sheets, bristles for brushes, surgical sutures, textile fabrics, etc.

Question iii.
Explain the vulcanisation of rubber. Which vulcanizing agents are used for the following synthetic rubber.
a. Neoprene
b. Buna-N
Answer:
The process by which a network of cross links is introduced into an elastomer is called vulcanization.

Vulcanization enhances the properties of natural rubber like tensile strength, stiffness, elasticity, toughness etc. Sulphur forms cross links between polyisoprene chains which results in improved properties of rubber.

• For neoprene vulcanizing agent is MgO.
• For Buna-N vulcanizing agent is sulphur.

Question iv.
Write reactions involved in the formation of …………………………………. .
1) Teflon
2) Bakelite
Answer:
The monomers phenol and formaldehyde undergo polymerisation in the presence of alkali or acid as catalyst.

Phenol reacts with formaldehyde to form ortho or p-hydroxy methyl phenols, which further reacts with phenol to form a linear polymer called Novolac. It is used in paints.

In the third stage, various articles are shaped from novolac by putting it in appropriate moulds and heating at high temperature (138 °C to 176 °C) and at high pressure forms rigid polymeric material called bakelite. Bakelite is insoluble and infusible and has high tensile strength.

Bakelite is used in making articles like telephone instrument, kitchenware, electric insulators frying pans, etc.

2. Teflon is polytetrafluoroethylene. The monomer used in preparation of teflon is tetrafluoroethylene, (CF₂ = CF₂) which is a gas at room temperature. Tetrafluoroethylene is polymerized by using free radical initiators such as hydrogen peroxide or ammonium persulphate at high pressure.

Properties:

• Telflon is tough, chemically inert and resistant to heat and attack by corrosive reagents.
• C – F bond is very difficult to break and remains unaffected by corrosive alkali, organic solvents.
  Uses : Telflon is used in making non-stick cookware, oil seals, gaskets, etc.

Question v.
What is meant by LDP and HDP? Mention the basic difference between the same with suitable examples.
Answer:

• LDP is low density polyethylene and HDP is high density polyethylene.
• LDP is a branched polymer with low density due to chains are loosely held and HDP is a linear polymer with density due to close packing.
• HDP is much stiffer than LDP and has high tensile strength and hardness.

LDP is mainly used in preparation of pipes for agriculture, irrigation and domestic water line connections. HDP is used in manufacture of toys and other household articles like bucket, bottles, etc.

Question vi.
Write preparation, properties and uses of Teflon.
Answer:
Teflon is polytetrafluoroethylene. The monomer used in preparation of teflon is tetrafluoroethylene, (CF₂ = CF₂) which is a gas at room temperature. Tetrafluoroethylene is polymerized by using free radical initiators such as hydrogen peroxide or ammonium persulphate at high pressure.

Properties:

• Telflon is tough, chemically inert and resistant to heat and attack by corrosive reagents.
• C – F bond is very difficult to break and remains unaffected by corrosive alkali, organic solvents.
  Uses : Telflon is used in making non-stick cookware, oil seals, gaskets, etc.

Question vii.
Classify the following polymers as straight-chain, branched-chain and cross-linked polymers.

Answer:

5. Answer the following.

Question i.
How is polythene manufactured? Give their properties and uses.
Answer:
LDP means low density polyethylene. LDP is obtained by polymerization of ethylene under high pressure (1000 – 2000 atm) and temperature (350 – 570 K) in presence of traces of O₂ or peroxide as initiator.

The mechanism of this reaction involves free radical addition and H-atom abstraction. The latter results in branching. As a result the chains are loosely held and the polymer has low density.

Properties of LDP :

• LDP films are extremely flexible, but tough chemically inert and moisture resistant.
• It is poor conductor of electricity with melting point 110 °C.

Uses of LDP :

• LDP is mainly used in preparation of pipes for agriculture, irrigation, domestic water line connections as well as insulation to electric cables.
• It is also used in submarine cable insulation.
• It is used in producing extruded films, sheets, mainly for packaging and household uses like in preparation of squeeze bottles, attractive containers, etc.

HDP means high density polyethylene. It is a linear polymer with high density due to close packing.

HDP is obtained by polymerization of ethene in presence of Zieglar-Natta catalyst which is a combination of triethyl aluminium with titanium tetrachloride at a temperature of 333K to 343K and a pressure of 6-7 atm.

Properties of HDP :

• HDP is crystalline, melting point in the range of 144 – 150 °C.
• It is much stiffer than LDP and has high tensile strength and hardness.
• It is more resistant to chemicals than LDP.

Uses of HDP :

• HDP is used in manufacture of toys and other household articles like buckets, dustbins, bottles, pipes, etc.
• It is used to prepare laboratory wares and other objects where high tensile strength and stiffness is required.

Question ii.
Is synthetic rubber better than natural rubber? If so, in what respect?
Answer:
Yes. Synthetic rubber is more resistant to abrasion than natural rubber and is also superior in resistance to heat and the effects of aging (lasts longer). Many types of synthetic rubber are flame-resistant, so they can be used in insulation for electrical devices.

It also remains flexible at low temperatures and is resistant to grease and oil. It is resistant to heat, light and certain chemicals.

Question iii.
Write main specialities of Buna-S, Neoprene rubber?
Answer:
Buna-S is an elastomer and it is copolymer of styrene with butadiene. Its trade name is SBR. Buna-S is superior to natural rubber, because of its mechanical strength and abrasion resistance. It is used in tyre industry. It is vulcanized with sulphur. Neoprene is a synthetic rubber and it is a condensation polymer of chloroprene (2-chloro-l, 3-butadiene). Vulcanization of neoprene takes place in presence of MgO. It is resistant to petroleum, vegetable oils. Neoprene is used in making hose pipes for transport of gasoline and making gaskets.

Question iv.
Write the structure of isoprene and the polymer obtained from it.
Answer:

Question v.
Explain in detail the free radical mechanism involved during the preparation of the addition polymer.
Answer:
Polymerisation of ethylene is carried out at high temperature and at high pressure in presence of small amount of acetyl peroxide as initiator.

(1) Formation of free radicals : The first step involves clevage of acetyl peroxide to form two carboxy radicals. These carboxy radicals immediately undergo decarboxylation to give methyl initiator free radicals.

(2) Chain initiating step : The methyl radical thus formed adds to ethylene to form a new larger free radical.

(3) Chain propagation step : The larger free radical formed in the chain initiating step reacts with another molecule of ethene to form another big size free radicals and chain grows. This is called chain propagation step.

The chain reaction continues till thousands of ethylene molecules are added.

(4) Chain terminating step : The continuous chain reaction can be terminated by the combination of free radicals to form polyethene.

Activity :
i. Collect the information of the process like extrusion and moulding in Textile Industries.
ii. Make a list of polymers used to make the following articles
a. Photographic film
b. Frames of spectacles
c. Fountain pens
d. Moulded plastic chains
e. Terywool or Terycot fabric
iii. Prepare a report on factors responsible for degradation of polymers giving suitable example.
iv. Search and make a chart/note on silicones with reference to monomers, structure, properties and uses.
v. Collect the information and data about Rubber industry, plastic industry and synthetic fibre (rayon) industries running in India.

12th Chemistry Digest Chapter 15 Introduction to Polymer Chemistry Intext Questions and Answers

Use your brain power! (Textbook Page No 323)

Question 1.
Differentiate between natural and synthetic polymers.
Answer:

Natural polymers	Synthetic polymers
(1) The polymers are obtained either from plants or animals.	(1) The man made fibres prepared by polymerization of monomer or copolymerization of two or more monomers.
(2) They are further divided into two types : (i) plant polymers (ii) Animal polymers.Examples: Cotton, linen, latex	(2) They are further divided into three subtypes : (i) fibres (ii) synthetic rubbers (iii) plastics.Examples : Nylon, terylene Buna-S

Use your brain power! (Textbook Page No 325)

Question 1.
What is the type of polymerization in the following examples?

Answer:
(i) Addition polymerization
(ii) Condensation polymerization

Problem 15.1 : (Textbook Page No 326)

Question 1.
Refer to the following table listing for different polymers formed from respective monomers. Identify from the list whether it is copolymer or homopolymer.

Monomer	Polymers
Ethylene	Polyethene
Vinyl chloride	Polyvinyl chloride
Isobutylene	Polyisobutylene
Acrylonitrile	Polyacrylonitrile
Caprolactam	Nylon 6
Hexamethylene diammonium adipate	Nylon 6, 6
Butadiene + styrene	Buna-S

Solution :
In each of first five cases, there is only one monomer which gives corresponding homopolymer. In the sixth case hexamethylene diamine reacts with adipic acid to form the salt hexamethylene diammonium adipate which undergoes condensation to form Nylon 6, 6. Hence nylon 6, 6 is homopolymer. The polymer Buna-S is formed by polymerization of the monomers butadiene and styrene in presence of each other. The repeating units corresponding to the monomers butadiene and styrene are randomly arranged in the polymer. Hence Buna-S is copolymer.

Use your brain power! (Textbook Page No 328)

Question 1.
(1) From the cis-polyisoprene structure of natural rubber explain the low strength of van der Waals forces in it.
(2) Explain how the vulcanization of natural rubber improves its elasticity. (Hint : consider the intermolecular links.)
Answer:
(1) (i) Natural rubber is cis-polyisoprene. It is obtained by polymerization of isoprene units at 1, 4 positions. In rubber molecule, double bonds are located between C₂ and C₃ of each isoprene unit. These cis-double bonds do not allow the polymer chain to come closer. Therefore, only weak vander Waals’ forces are present. Since the chains are not linear, they can be stretched just like springs and exhibit elastic properties.

(ii) Cis-1, 4 polyisoprene (Natural rubber), due to this cis configuration about the double bonds, has the adjacent chain that do not fit together well (there is no close packing of adjacent chains). The only force that interact is the weak or low strength of van der Waals’ forces.

(iii) Cis-polyisoprene has a coiled structure in which the various polymer chains are held together by weak van der Waals’ forces.

(2) (i) Vulcanization of rubber is a process of improvement of the rubber elasticity and strength by heating it in the presence of sulphur, which results in three dimensional cross-linking of the chain rubber molecules (polyisoprene) bonded to each other by sulphur atoms.

(ii) Vulcanisation makes rubber more elastic and more stiffer. On vulcanisation, sulphur forms cross links at the reactive sites of double bonds and thus rubber get stiffened.

(iii) The improved properties of vulcanised rubber are (i) high elasticity (ii) low water absorption tendency,

(iii) resistance to oxidation.

Use your brain power! (Textbook Page No 334)

Question 1.
Write structural formulae of styrene and polybutadiene.
Answer:

(1) Classify the following polymers as addition or condensation.
(i) PVC (ii) Polyamides
(iii) Polystyrene
(iv) Polycarbonates
(v) Novolac
Answer:
Addition polymers: PVC, Polystyrene
(ondensatlon polymers: Polyamides. Polycarbonates, Novolac

Question 2.
Completed the following table :
Answer:

Use your brain power! (Textbook Page No 335)

(1) Represent the copolymerization reaction between glycine and e aminocaproic acid to form the copolymer nylon 2-nylon 6.
(2) What is the origin of the numbers 2 and 6 in the name of this polymer?
Answer:
(1) It is a copolymer and has polyamide linkages. The monomers glycine and e-amino caproic acid undergo condensation polymerisation to form nylon-2-nylon-6.

Nylon-2-nylon-6 is used in orthopaedic devices and implants.

(2) Monomer glycine contains two carbon atoms and e amino caproic acid contains six carbon atoms, hence the polymer is termed as nylon-2-nylon-6.

12th Std Chemistry Questions And Answers:

• Coordination Compounds Class 12 Chemistry Questions And Answers
• Halogen Derivatives Class 12 Chemistry Questions And Answers
• Alcohols, Phenols and Ethers Class 12 Chemistry Questions And Answers
• Aldehydes, Ketones and Carboxylic Acids Class 12 Chemistry Questions And Answers
• Amines Class 12 Chemistry Questions And Answers
• Biomolecules Class 12 Chemistry Questions And Answers
• Introduction to Polymer Chemistry Class 12 Chemistry Questions And Answers
• Green Chemistry and Nanochemistry Class 12 Chemistry Questions And Answers

Balbharti Maharashtra State Board 12th Chemistry Textbook Solutions Chapter 5 Electrochemistry Textbook Exercise Questions and Answers.

Electrochemistry Class 12 Exercise Question Answers Solutions Maharashtra Board

Class 12 Chemistry Chapter 5 Exercise Solutions Maharashtra Board

Chemistry Class 12 Chapter 5 Exercise Solutions

1. Choose the most correct option.

Question i.
Two solutions have the ratio of their concentrations 0.4 and ratio of their conductivities 0.216. The ratio of their molar conductivities will be
(a) 0.54
(b) 11.574
(c) 0.0864
(d) 1.852
Answer:
(a) 0.54

Question ii.
On diluting the solution of an electrolyte,
(a) both ∧ and κ increase
(b) both ∧ and κ decrease
(c) ∧ increases and κ decreases
(d) ∧ decreases and κ increases
Answer:
(c) ∧ increases and κ decreases

Question iii.
1 S m² mol^-1 is equal to
(a) 10^-4 S m² mol^-1
(b) 10⁴ Ω^-1 cm² mol^-1
(c) 10^-2 S cm² mol^-1
(d) 10² Ω^-1 cm² mol^-1
Answer:
(b) 10⁴Ω^-1 cm² mol^-1

Question iv.
The standard potential of the cell in which the following reaction occurs
H₂₊ (g, 1 atm) + Cu²⁺ (1 M) → 2H (1 M) + Cu_(s), (\(E_{\mathrm{Cu}}^{0}\) = 0.34 V) is
(a) – 0.34 V
(b) 0.34 V
(c) 0.17 V
(d) -0.17 V
Answer:
(b) 0.34 V

Question v.
For the cell, Pb_(s)|Pb²⁺ (1 M)|| Ag⁺ (1 M)|Ag_(s), if concentration of an ion in the anode compartment is increased by a factor of 10, the emf of the cell will
(a) increase by 10 V
(b) increase by 0.0296 V
(c) decrease by 10 V
(d) decrease by 0.0296 V
Answer:
(d) decrease by 0.0296 V

Question vi.
Consider the half reactions with standard potentials

The strongest oxidising and reducing agents respectively are
(a) Ag and Fe²⁺
(b) Ag⁺ and Fe
(c) Pb²⁺ and I^–
(d) I₂ and Fe²⁺
Answer:
(b) Ag⁺ and Fe

Question vii.
For the reaction
Ni_(s) + Cu²⁺ (1 M) → Ni²⁺ (1 M) + Cu_(s), \(E_{\text {cell }}^{0}\) = 0.57 V. Hence ΔG⁰ of the reaction is
(a) 110 kJ
(b) -110 kJ
(c) 55 kJ
(d) -55 kJ
Answer:
(b) -110 kJ

Question viii.
Which of the following is not correct ?
(a) Gibbs energy is an extensive property
(b) Electrode potential or cell potential is an intensive property.
(c) Electrical work = -ΔG
(d) If half reaction is multiplied by a numerical factor, the corresponding E⁰ value is also multiplied by the same factor.
Answer:
(d) If half reaction is multiplied by a numerical factor, the corresponding E⁰ value is also multiplied by the same factor.

Question ix.
The oxidation reaction that takes place in lead storage battery during discharge is

Answer:
(c) \(\mathrm{Pb}_{(\mathrm{s})}+\mathrm{SO}_{4(\mathrm{aq})}{ }^{2-} \longrightarrow \mathrm{PbSO}_{4(\mathrm{~s})}+2 \mathrm{e}^{-}\)

Question x.
Which of the following expressions represent molar conductivity of Al₂(SO₄)₃ ?

Answer:
(b) \(2 \lambda_{\mathrm{Al}^{3+}}^{0}+3 \lambda_{\mathrm{SO}_{4}^{2-}}^{0}\)

2. Answer the following in one or two sentences.

Question i.
What is a cell constant ?
Answer:
(A) Cell constant of a conductivity cell is defined as the ratio of the distance between the electrodes divided by the area of cross section of the electrodes.

In SI units it is expressed as m-1.

Question ii.
Write the relationship between conductivity and molar conductivity and hence unit of molar conductivity.
Answer:
If k is conductivity and ∧_m is molar conductivity then, ∧_m = \(\frac{\kappa \times 1000}{C}\)
Unit of molar conductivity is, Ω^-1 cm² mol^-1 or S cm² mol^-1.

Question iii.
Write the electrode reactions during electrolysis of molten KCl.
Answer:

Question iv.
Write any two functions of salt bridge.
Answer:
The functions of a salt bridge are :

• It maintains the electrical contact between the two electrode solutions of the half cells.
• It prevents the mixing of electrode solutions.
• It maintains the electrical neutrality in both the solutions of two half cells by a flow of ions.
• It eliminates the liquid junction potential.

Question v.
What is standard cell potential for the reaction
3Ni_(s) + 2Al³⁺ (1M) → 3NI²⁺ (1M) + 2Al_(s)
if \(\boldsymbol{E}_{\mathrm{Ni}}^{0}\) = – 0.25 V and \(\boldsymbol{E}_{\mathrm{Al}}^{0}\) = -1.66V?
Solution :
Given : E⁰_Ni²⁺/Ni = -0.25 V
E⁰_Al³⁺/Al = – 1.66 V; E⁰_cell = ?
Since Ni is oxidised and Al³⁺ is reduced,
\(E_{\text {cell }}^{0}=E_{\mathrm{Al}^{3+} / \mathrm{Al}}^{0}-E_{\mathrm{Ni}^{2+} / \mathrm{Ni}}^{0}\)
= – 1.66 – (-0.25)
= – 1.41 V
Ans. \(E_{\text {cell }}^{0}\) = -1.41 V
[Note : Since \(E_{\text {cell }}^{0}\) is negative, the given reaction is not possible but reverse reaction is possible.]

Question vi.
Write Nerst equation. What part of it represents the correction factor for nonstandard state conditions ?
Answer:
(1) Nernst equation for cell potential is,

(2) The part of equation namely,

represents the correction factor for nonstandard state conditions.

Question vii.
Under what conditions the cell potential is called standard cell potential ?
Answer:
In the standard cell, the active masses of the substances taking part in the electrochemical reaction have unit value, i.e., 1 M solution or ions and 1 atm gas.

Question viii.
Formulate a cell from the following electrode reactions :
\(\mathbf{A u}_{(\mathrm{aq})}^{3+}+\mathbf{3 e}^{-} \longrightarrow \mathbf{A} \mathbf{u}_{(\mathrm{s})}\)
\(\mathbf{M g}_{(\mathbf{s})} \longrightarrow \mathbf{M g}_{(\mathrm{aq})}^{2+}+\mathbf{2 e}^{-}\)
Answer:
An electrochemical cell from above electrode reactions is,

Question ix.
How many electrons would have a total charge of 1 coulomb ?
Answer:
Given : 1 Faraday = charge on 1 mol of electrons
= 6.022 × 10²³ electrons and 1 Faraday = 96500 C
∵ 96500 C = 6.022 × 10²³ electrons 6 022 × 10²³
∴ 1 C ≡ \(\frac{6.022 \times 10^{23}}{96500}\) = 6.24 × 10¹⁸ electrons
Ans. Number of electrons = 6.24 × 10¹⁸

Question x.
What is the significance of the single vertical line and double vertical line in the formulation galvanic cell.
Answer:
(i) Consider representation of Daniell cell,

Single vertical line represents separation of two phases, solid Zn_(s) and solution of ions.
(ii) Double vertical lines represent a salt bridge.

3. Answer the following in brief

Question i.
Explain the effect of dilution of solution on conductivity ?
Answer:

• The conductance of a solution is due to the presence of ions in the solution. More the ions, higher is the conductance of the solution.
• Conductivity or the specific conductance is the conductance of unit volume (1 cm³) of the electrolytic solution.
• The conductivity of the electrolytic solution always decreases with the decrease in the concentration of the electrolyte or the increase in dilution of the solution.
• On dilution, the concentration of the solution decreases, hence the number of (current carrying) ions per unit volume decreases. Therefore the conductivity of the solution decreases, with the decrease concentration or increase in dilution. (It is to be noted here that, molar conductivity increases with dilution.)

Question ii.
What is a salt bridge ?
Answer:
A salt bridge is a U-shaped glass tube containing a saturated solution of a strong electrolyte, like KCl, NH₄NO₃, Na₂SO₄ in a solidified agar-agar gel. A hot saturated solution of these electrolytes in 5% agar solution is filled in the U-shaped tube and allowed it to cool and solidify forming a gel.

Fig. 5.9 : Salt bridge
It is used to connect two half cells or electrodes forming a galvanic or voltaic cell.

Question iii.
Write electrode reactions for the electrolysis of aqueous NaCl.
Answer:
Reactions in electrolytic cell :
(i) Reduction half reaction at cathode : There are Na⁺ and H⁺ions but since H⁺ are more reducible than Na⁺, they undergo reduction liberating hydrogen and Na⁺ are left in the solution.
2H₂O_(l) + 2e^– → H_2(g) + 2OH^–_(aq) (reduction) E⁰ = -0.83 V

(ii) Oxidation half reaction at anode : At anode there are Cl^– and OH^–. But Cl^– ions are preferably oxidised due to less decomposition potential.

Net cell reaction : Since two electrons are gained at cathode and two electrons are released at anode for each redox step, the electrical neutrality is maintained. Hence we can write,

Since Na⁺ and OH^– are left in the solution, they form NaOH_(aq).

Question iv.
How many moles of electrons are passed when 0.8 ampere current is passed for 1 hour through molten CaCl₂ ?
Answer:
Given : I = 0.8 A; t = 1 × 60 × 60 = 3600 s
Number of moles of electrons = ?
Q = I × t
= 0.8 × 3600
= 2880 C
1 Faraday = 1 mol electrons
1 Faraday = 96500 C
∵ 96500 C = 1 mol electrons
∴ 2880 C ≡ \(\frac{2880}{96500}\)
= 0.02984 mol electrons
Ans. Number of moles of electrons = 0.02984

Question v.
Construct a galvanic cell from the electrodes Co³⁺|Co and Mn²⁺|Mn. \(\boldsymbol{E}_{\mathrm{Co}}^{0}\) = 1.82 V,
\(\boldsymbol{E}_{\mathrm{Mn}}^{0}\) = – 1.18V. Calculate \(\boldsymbol{E}_{\text {cell }}^{0}\).
Answer:

Question vi.
Using the relationsip between ∆G⁰ of cell reaction and the standard potential associated with it, how will you show that the electrical potential is an intensive property ?
Answer:
(1) For an electrochemical cell involving n number of electrons in the overall cell reaction,
ΔG⁰ = -nF\(E_{\text {cell }}^{0}\)
where ΔG⁰ is standard Gibbs energy change and \(E_{\text {cell }}^{0}\) is a standard cell potential.
(2) ∴ \(E_{\mathrm{cell}}^{0}=\frac{-\Delta G^{0}}{n F}\)
Since ΔG⁰ changes according to number of moles of electrons involved in the cell reaction, the ratio, ΔG⁰/nF remains constant.
(3) Therefore \(E_{\text {cell }}^{0}\) is independent of the amount of substance and it represents the intensive property.

Question vii.
Derive the relationship between standard cell potential and equilibrium constant of cell reaction.
Answer:
For any galvanic cell, the overall cell reaction at equilibrium can be represented as,
Reactants ⇌ Products.
[For example for Daniell cell,
\(\mathrm{Zn}_{(s)}+\mathrm{Cu}_{(\mathrm{aq})}^{2+} \rightleftharpoons \mathrm{Zn}_{(\mathrm{aq})}^{2+}+\mathrm{Cu}_{(\mathrm{s})}\) ]
The equilibrium constant, K for the reversible reaction will be, \(K=\frac{[\text { Products }]}{[\text { Reactants }]}\)

The equilibrium constant is related to the stan-dard free energy change Δ G⁰, as follows,
ΔG⁰ = -RTlnK
If \(E_{\text {cell }}^{0}\) is the standard cell potential (or emf) of the galvanic cell, then ΔG⁰ = -nFE⁰_cell
By comparing above equations,

Question viii.
It is impossible to measure the potential of a single electrode. Comment.
Answer:
(1)

Fig 5.12(a) : Measurement of single electrode potential

Fig 5.12(b) : Measurement of cell potential
According to Nemst theory, electrode potential is the potential difference between the metal and ionic layer around it at equilibrium, i.e. the potential across the electric double layer.

(2) For measuring the single electrode potential, one part of the double layer, that is metallic layer can be connected to the potentiometer but not the ionic layer. Hence, single electrode potential can’t be measured experimentally.

(3) When an electrochemical cell is developed by combining two half cells or electrodes, they can be connected to the potentiometer and the potential difference or cell potential can be measured.
E_cell = E₂ – E₁
where E₁ and E₂ are reduction potentials of two electrodes.

Question ix.
Why do the cell potential of lead accumulators decrease when it generates electricity ? How the cell potential can be increased ?
Answer:
Working of a lead accumulator :
(1) Discharging : When the electric current is withdrawn from lead accumulator, the following reactions take place :

(2) Net cell reaction :
(i) Thus, the overall cell reaction during discharging is

OR
Pb_(s) + PbO_2(s) + 2H₂SO_4(aq) → 2PbSO_4(s) + 2H₂O_(l)
The cell potential or emf of the cell depends upon the concentration of sulphuric acid. During the operation, the acid is consumed and its concentration decreases and specific gravity decreases from 1.28 to 1.17. As a result, the emf of the cell decreases. The emf of a fully charged cell is about 2.0 V.

(ii) Recharging of the cell : When the discharged battery is connected to external electric source and a higher external potential is applied the cell reaction gets reversed generating H₂SO₄.
Reduction at the -ve electrode or cathode :
PbSO4_(s) + 2e^– → Pb_(s) + \(\mathrm{SO}_{4(\mathrm{aq})}^{2-}\)
Oxidation at the + ve electrode or anode :

The emf of the accumulator depends only on the concentration of H₂SO₄.

Question x.
Write the electrode reactions and net cell reaction in NICAD battery.
Answer:
Reactions in the cell :
(i) Oxidation at cadmium anode :
Cd_(s) + 2OH^–_(aq) → Cd(OH)_2(s) + 2e^–
(ii) Reduction at NiO_2(s) cathode :
NiO_2(s) + 2H₂O_(l) + 2e^– → Ni(OH)_2(s) + 2OH^–_(aq)
The overall cell reaction is the combination of above two reactions.
Cd_(s) + NiO_2(s) + 2H₂O_(l) → Cd(OH)_2(s) + Ni(OH)_2(s)

4. Answer the following :

Question i.
What is Kohrausch law of independent migration of ions? How is it useful in obtaining molar conductivity at zero concentration of a weak electrolyte ? Explain with an example.
Answer:
(A) Statement of Kohlrausch’s law : This states that at infinite dilution of the solution, each ion of an electrolyte migrates independently of its co-ions and contributes independently to the total molar conductivity of the electrolyte, irrespective of the nature of other ions present in the solution.

(B) Explanation : Both the ions, cation and anion of the electrolyte make a definite contribution to the molar conductivity of the electrolyte at infinite dilution or zero concentration (∧₀).
If \(\lambda_{+}^{0}\) and \(\lambda_{-}^{0}\) are the molar ionic conductivities of cation and anion respectively at infinite dilution, then
∧₀ = \(\lambda_{+}^{0}\) + \(\lambda_{-}^{0}\).
This is known as Kohlrausch’s law of independent migration of ions.
For an electrolyte, B_xA_y giving x number of cations and y number of anions,
∧₀ = x\(\lambda_{+}^{0}\) + y\(\lambda_{-}^{0}\).

(C) Applications of Kohlrausch’s law :
(1) With this law, the molar conductivity of a strong electrolyte at zero concentration can be determined. For example,
\(\wedge_{0(\mathrm{KCl})}=\lambda_{\mathrm{K}^{+}}^{0}-\lambda_{\mathrm{Cl}^{-}}^{0}\)
(2) ∧₀ values of weak electrolyte with those of strong electrolytes can be obtained. For example,

Molar conductivity of a weak electrolyte at infinite dilution or zero concentration cannot be measured experimentally.
Consider the molar conductivity (∧₀) of a weak acid, CH₃COOH at zero concentration. By Kohlrausch s law,

where λ⁰CH₃COO^– and λ⁰_H⁺ are the molar ionic conductivities of CH₃COO^– and H⁺ ions respectively.
If ∧_0CH3COONa, ∧_0HCl and ∧_0NaCl are molar conductivities of CH₃COONa, HCl and NaCl respectively at zero concentration, then by
Kohlrausch’s law,

Hence, from ∧₀ values of strong electrolytes, ∧₀ of a weak electrolyte CH₃COOH, at infinite dilution can be calculated.

Question ii.
Explain electrolysis of molten NaCl.
Answer:
(1) Construction of an electrolytic cell : It consists of a vessel containing molten (fused) NaCl. Two graphite (carbon) inert electrodes are dipped in it, and connected to an external source of direct electric current (battery). The electrode connected to a negative terminal of the battery is a cathode and that connected to a positive terminal is an anode.

(2) Working of the cell :
(A) In the external circuit, the electrons flow through the wires from anode to cathode of the cell.
(B) The fused NaCl dissociates to form cations (Na⁺) and anions (Cl^–).
\(\mathrm{NaCl}_{\text {(fused) }} \longrightarrow \mathrm{Na}_{(\mathrm{l})}^{+}+\mathrm{Cl}_{(\mathrm{l})}^{-}\)
Na⁺ migrate towards cathode and Cl^– migrate towards anode.

Fig. 5.7 : Electrolysis of fused sodium chloride

(C) Reactions in electrolytic cell :
(i) Reduction half reaction at cathode : The Na⁺ ions get reduced by accepting electrons from a cathode supplied by a battery and form metallic sodium.
\(\mathrm{Na}^{+}+\mathrm{e}^{-} \longrightarrow \mathrm{Na}_{(\mathrm{s})} \text { (reduction) }\)

(ii) Oxidation half reaction at anode : The Cl^– ions get oxidised by giving up electrons to the anode forming neutral Cl atoms in the primary process, and these Cl atoms combine forming Cl₂ gas in the secondary process.

The released electrons in the anodic oxidation half reaction return to battery through the metallic wires.

Net cell reaction : In order to maintain the electrical neutrality, the number of electrons gained at cathode must be equal to the number of electrons released at anode. Hence the reduction half reaction is multiplied by 2 and both reactions, oxidation half reaction and reduction half reaction are added to obtain a net cell reaction.

Results of electrolysis :

• A molten silvery white Na is formed at cathode which floats on the surface of molten NaCl.
• A pale green Cl₂ gas is liberated at anode.

Question iii.
What current strength in amperes will be required to produce 2.4g of Cu from CuSO₄ solution in 1 hour ? Molar mass of Cu = 63.5 g mol^-1.
Answer:
Given : W_Cu = 2.4 g; t = 1 hr = 1 × 60 × 60 s
M_Cu = 63.5 g mol^-1; I = ?

Ans. Current strength = I = 2.026 A

Question iv.
Equilibrium constant of the reaction,
2Cu⁺_(aq) → Cu²⁺_(aq) + Cu_(s)
is 1.2 × 10⁶. What is the standard potential of the cell in which the reaction takes place ?
Answer:
For the cell reaction, n = 1

Question v.
Calculate emf of the cell
Zn_(s)|Zn²⁺ (0.2M)||H⁺(1.6M)|H₂(g, 1.8 atm)|Pt at 25°C.
Answer:
Given : Zn_(s)|Zn²⁺(0.2M)||H⁺(1.6M)|H₂(g, 1.8 atm)|Pt


= 0.763 – 0.0296 × (- 0.8521)
= 0.763 + 0.02522
= 0.7882
Ans. \(E_{\text {cell }}^{0}\) = 0.7882 V

Question vi.
Calculate emf of the following cell at 25°C.
Zn_(s)| Zn²⁺(0.08M)||Cr³⁺(0.1M)|Cr
E⁰_Zn = – 0.76 V, E⁰_Cr = – 0.74 V
Answer:

Question vii.
What is a cell constant ? What are its units? How is it determined experimentally?
Answer:
(A) Cell constant of a conductivity cell is defined as the ratio of the distance between the electrodes divided by the area of cross section of the electrodes.

In SI units it is expected as m^-1.

The resistance of an electrolytic solution is measured by using a conductivity cell and Wheatstone

Fig. 5.6 : Measurement of conductance
The measurement of molar conductivity of a solution involves two steps as follows :
Step I : Determination of cell constant of the conductivity cell :
KCl solution (0.01 M) whose conductivity is accurately known (κ = 0.00141 Ω^-1 cm^-1) is taken in a beaker and the conductivity cell is dipped. The two electrodes of the cell are connected to one arm while the variable known resistance (R) is placed in another arm of Wheatstone bridge.

A current detector D’ which is a head phone or a magic eye is used. J is the sliding jockey (contact) that slides on the arm AB which is a wire of uniform cross section. A source of A.C. power (alternating power) is used to avoid electrolysis of the solution.

By sliding the jockey on wire AB, a balance point (null point) is obtained at C. Let AC and BC be the lengths of wire.

If R_solution is the resistance of KCl solution and R_x is the known resistance then by Wheatstone’s bridge principle,
\(\frac{R_{\text {solution }}}{\mathrm{BC}}=\frac{R_{x}}{\mathrm{AC}}\)
∴ \(R_{\text {solution }}=\mathrm{BC} \times \frac{R_{x}}{\mathrm{AC}}\)
Then the cell constant ‘ b ’ of the conductivity cell is obtained by, b = κ_Kcl × R_solution.

Step II : Determination of conductivity of the given solution :
KCl solution is replaced by the given electrolytic solution and its resistance (R_s) is measured by Wheatstone bridge method by similar manner by obtaining a null point at D.
The conductivity (κ) of the given solution is,
κ = \(\frac{\text { cell constant }}{R_{\mathrm{s}}}=\frac{b}{R_{\mathrm{s}}}\)

Step III: Calculation of molar conductivity :
The molar conductivity (∧_m) is given by,

Since the concentration of the solution is known, ∧_m can be calculated.

Question viii.
How will you calculate the moles of electrons passed and mass of the substance produced during electrolysis of a salt solution using reaction stoichiometry.
Answer:
Calculation of moles of electrons passed : The charge carried by one mole of electrons is referred to as one faraday (F). If total charge passed is Q C, then moles of electrons passed = \(\frac{Q(\mathrm{C})}{F\left(\mathrm{C} / \mathrm{mol} \mathrm{e}^{-}\right)}\)

Calculation of mass of product : Mass, W of product formed is given by,
W = moles of product × molar mass of product (M)
= \(\frac{Q}{96500}\) × mole ratio × M
= \(\frac{I \times t}{96500}\) × mole ratio × M 96500
When two electrolytic cells containing different electrolytes are connected in series so that same quantity of electricity is passed through them, then the masses W₁ and W₂ of products produced are given by,

Question ix.
Write the electrode reactions when lead storage cell generates electricity. What are the anode and cathode and the electrode reactions during its recharging?
Answer:
Recharging of the cell : When the discharged battery is connected to external electric source and a higher external potential is applied the cell reaction gets reversed generating H₂SO₄.
Reduction at the – ve electrode or cathode :
\(\mathrm{PbSO}_{4(\mathrm{~s})}+2 \mathrm{e}^{-} \rightarrow \mathrm{Pb}^{(s)}+\mathrm{SO}_{4(\mathrm{aq})}^{2-}\)
Oxidation at the + ve electrode or anode :

The emf of the accumulator depends only on the concentration of H₂SO₄.

Question x.
What are anode and cathode of H₂-O₂ fuel cell ? Name the electrolyte used in it. Write electrode reactions and net cell reaction taking place in the fuel cell.
Answer:
Construction :
(i) In fuel cell the anode and cathode are porous electrodes with suitable catalyst like finely divided platinum.

(iii) H₂ is continuously bubbled through anode while O, gas is bubbled through cathode.

Working (cell reactions) :
(i) Oxidation at anode : At anode, hydrogen gas is oxidised to H₂O.
2H_2(g) + 4OH^–_(aq) → 4H₂O_(l) + 4e^– (oxidation half reaction)
(ii) Reduction at cathode : The electrons released at anode travel to cathode through external circuit and reduce oxygen gas to OH^–.
O_2(g) + 2H₂O_(l) + 4e^– → 4OH^–_(aq) (reduction half reaction)

(iii) Net cell reaction: Addition of both the above reactions at anode and cathode gives a net cell reaction.
2H_2(g) + O_2(g) → 2H₂O_(l) (overall cell reaction)

Question xi.
What are anode and cathode for Leclanche’ dry cell ? Write electrode reactions and overall cell reaction when it generates electricity.
Answer:
A dry cell has zinc vessel as anode and graphite rod as cathode and moist paste of ZnCl₂, MnO₂, NH₄Cl as electrolytes.
At anode :
Zn_(s) → \(\mathrm{Zn}_{(\mathrm{aq})}^{2+}\) + 2e^– (Oxidation half reaction)
At graphite (c) cathode :
\(2 \mathrm{NH}_{4(\mathrm{e})}^{+}\) + 2e^– → 2NH_3(aq) + H_2(g) (Reduction half reaction)
2MnO_2(s) + H₂ → Mn₂O_3(s) + H₂O_(l)
There is a side reaction inside the cell, between Zn²⁺ ions and aqueous NH₃.
\(\mathrm{Zn}_{(\mathrm{aq})}^{2+}+4 \mathrm{NH}_{3(\mathrm{aq})} \longrightarrow\left[\mathrm{Zn}\left(\mathrm{NH}_{3}\right)_{4}\right]_{(\mathrm{aq})}^{2+}\)

Question xii.
Identify oxidising agents and arrange them in order of increasing strength under standard state conditions. The standard potentials are given in parenthesis.
Al(- 1.66 V), Cl₂ (1.36 V), Cd²⁺ (-0.4 V), Fe (-0.44 V), I₂ (0.54 V), Br^– (1.09 V).
Answer:
The oxidising agents are I₂, Br^– and Cl₂. The increasing strength is

(Note : Actually Br₂ acts as an oxidising agent but not Br^–.)

Question xiii.
Which of the following species are reducing agents? Arrange them in order of increasing strength under standard state conditions. The standard potentials are given in parenthesis.
K (-2.93V), Br₂(1.09V), Mg(-2.36V), Co³⁺(1.61V), Ti²⁺(-0.37V), Ag⁺(0.8V), Ni (-0.23V).
Answer:
Lower the standard reduction potential, higher is reducing power. The reducing agents are Ni, Mg and K. Their increasing strength is,

(Note : Cations don’t act as reducing agent since they are already in oxidised state.)

Question xiv.
Predict whether the following
reactions would occur spontaneously
under standard state conditions.
a. Ca_(s) + Cd²⁺_(aq) → Ca²⁺_(aq) + Cd_(s)
b. 2 Br-_(s) + Sn²⁺(aq) → Br_2(l) + Sn_(s)
c. 2Ag_(s) + Ni²⁺_(aq) → 2 Ag⁺_(aq) + Ni_(s)
(use information of Table 5.1)
Answer:

12th Chemistry Digest Chapter 5 Electrochemistry Intext Questions and Answers

Question 1.
How does electrical resistance depend on the dimensions of an electronic (metallic) conductor?
Answer:
The electrical resistance of an electronic conductor is linearly proportional to its length (l) and inversely proportional to its cross section area a.

Fig. 5.3 : Electronic conductor
Thus, R ∝ l; R ∝ \(\frac{1}{a}\)
∴ R ∝ \(\frac{l}{a}\) or R = ρ × \(\frac{l}{a}\)
where the proportionality constant p is called specific resistance. IUPAC recommends the term resistivity for specific resistance.

Question 2.
What are the units of resistivity ?
Answer:
For an electronic conductor of length l, and cross section area a, the resistance R is represented as
R = ρ × \(\frac{l}{a}\)
where ρ is the resistivity of the conductor.
∴ ρ = R × \(\frac{a}{l}\)
If l = 1 m, a = 1 m², ρ = R

Hence, resistivity is the resistance of a conductor of volume of 1 m³.
(In C.G.S. units, the units of ρ are ohm cm. Hence, ρ is the resistance of a conductor of unit volume or 1 cm³.)

Question 3.
Define resistivity. What are its units ?
Answer:
Resistivity (or specific resistance) : It is the resistance of a conductor that is 1 m in length and 1 m² in cross section area in SI units. (In C.G.S. units, it is the resistance of a conductor that is 1 cm in length and 1 cm² in cross section area.) Hence, the resistivity is the resistance of a conductor of unit volume. (In case of electrolytic solution, ρ is the resistivity i.e., resistance of a solution of unit volume.)
It has SI units, ohm m and C.G.S. units, ohm cm.

Question 4.
Why is alternating current used in the measurement of conductivity of the solution ?
Answer:
If direct current (D.C.) by battery is used, there will be electrolysis and the concentration of the solution is changed. Hence alternating current (A.C.) with high frequency is used.

Try this… (Textbook page No. 93)

Question 1.
What must be the concentration of a solution of silver nitrate to have the molar conductivity of 121.4 Ω^-1 cm² mol^-1 and the conductivity of 2.428 × 10^-3 Ω^-1 cm^-1 at 25 °C ?
Answer:

∴ Concentration of a Solution = 0.02 M

Try this… (Textbook page No. 96)

Question 1.
Obtain the expression for dissociation constant in terms of ∧_c and ∧₀ using Ostwald’s dilution law.
Answer:
Consider a solution of a weak electrolyte, BA having concentration C mol dm^-3. If α is the degree of dissociation, then by Ostwald’s theory of weak electrolytes,

If K is the dissociation constant of the weak electrolyte, then by Ostwald’s dilution law,
K = \(\frac{\alpha^{2} C}{(1-\alpha)}\)
If ∧_m is the molar conductivity of the electrolyte BA at the concentration C and ∧₀ is the molar conductivity at zero concentration or infinite dilution, then

Hence by measuring ∧_m at the concentration C and knowing ∧₀, the dissociation constant can be calculated.
If \(\lambda_{+}^{0}\) and \(\lambda_{-}^{0}\) are the ionic conductivities, then by Kohlrauseh’s law, ∧₀ = \(\lambda_{+}^{0}\) + \(\lambda_{-}^{0}\).

Learn this as well…

Question 1.
How is the cell constant of a conductivity cell determined?
Answer:
The cell constant of a given conductivity cell is obtained by measuring the resistance (R) (or the conductance) of a standard solution whose conductivity (f_c) is accurately known by using Wheatstone’s bridge (discussed in Q. 37). For this purpose, KCl solution of accurately known conductivity is used.
\(\kappa_{\mathrm{KCl}}=\frac{1}{R_{\mathrm{KCl}}} \times \frac{l}{a}\) where \(\frac{l}{a}\) is a cell constant, represented by b.
∴ \(\kappa_{\mathrm{KCl}}=\frac{b}{R_{\mathrm{KCl}}}\)
or b = κ_KCl × R_KCl
For example, the conductivity of 0.01 M KCl is 0.00141 Ω^-1 cm^-1 (S cm^-1). Hence by measuring R KCl the cell constant b can be obtained.

Try this… (Textbook page No. 95)

Question 1.
Calculate ∧₀ (CH₂ClCOOH) if ∧₀ values for HCl, KCl and CH₂ClCOOK are respectively, 4.261, 1.499 and 1.132 Ω^-1 m² mol^-1.
Solution :

Adding equations (i) and (ii) and subtracting equation (iii) we get equation (I).

Can you tell ? (Textbook page No. 103)

Question 1.
You have learnt Daniel cell in XIth standard. Write notations for anode and cathode. Write the cell formula.
Answer:
Daniel cell is represented as,

Try this… (Textbook page No. 104)

Question 1.
Write electrode reactions and overall cell reaction for Daniel cell you learnt in standard XI.
Answer:
Reactions for Daniell cell:

Question 1.
Describe different types of reversible electrodes with examples. (1 mark for each type)
Answer:
A reversible electrochemical cell or a galvanic cell consists of two reversible half cells or electrodes. There are four types of reversible electrodes according to their compositions.
(1) Metal-metal ion electrode : This electrode is set up by dipping a metal in a solution containing its own ions, e.g. Zn rod dipped into ZnSO₄ solution containing Zn⁺⁺ ions of concentration C.
It is represented as,
\(\mathrm{Zn}^{2+}{ }_{(\mathrm{aq})} \mid \mathrm{Zn}_{(\mathrm{s})}\)
The reduction reaction at the electrode is,
Zn⁺⁺_(aq) + 2e^– → Zn_(s)

(2) Metal-sparingly soluble salt electrode : This electrode consists of a metal coated with one of its sparingly soluble salts and immersed in a solution containing an electrolyte having a common anion as that of the salt. For example, silver electrode coated with sparingly soluble AgCl dipped in KCl solution with common anion Cl^–. This electrode is represented as,
Cl^–_(aq) | AgCl_(s) | Ag_(s)
The reduction reaction is,
AgCl_(s) + e → Ag_(s) + Cl^–_(aq)

(3) Gas electrode : This is developed by bubbling pure and dry gas around a platinised platinum foil dipped in the solution containing ions (of the gas) reversible with respect to the gas bubbled.
The gas is adsorbed on the surface of platinum foil and establishes an equilibrium with its ions in the solution. Pt electrode provides electrical contact and also acts as a catalyst.
Some of the gas electrodes are represented as follows :
(i) Hydrogen gas electrode :
H⁺_(aq) | H₂(g, P_H2) | Pt
Reduction reaction : H⁺_(aq) + e^– → \(\frac {1}{2}\)H_2(g)
(ii) Chlorine gas electrode :
Cl^–_(aq) | Cl₂(g, PCl₂) | Pt
Reduction reaction : \(\frac {1}{2}\)Cl_2(g) + e- → Cl^–_(aq)

(4) Redox electrode (Oxidation reduction electrode) : This electrode consists of a platinum wire dipped in a solution containing the ions of the same metal (or a substance) in two different oxidation states, like Fe²⁺ – Fe³⁺, Sn²⁺ – Sn⁴⁺, Mn⁺⁺ – MnO^–₄, etc.
A platinum electrode which provides an electrical contact and acts as catalyst aquires an equilibrium between two ions in the solution, due to their tendency to undergo a change from one oxidation state to another. The electrodes are represented as,
Fe²⁺_(aq), Fe³⁺_(aq) | Pt
Reduction reaction : Fe³⁺_(aq) + e^– → Fe²⁺_(aq)
SnCl_2(aq), SnCl_4(aq) | Pt
Reduction reaction : Sn⁴⁺_(aq) + 2e^– →Sn²⁺_(aq)

Use your brain power! (Textbook page No. 98)

Question 1.
Distinguish between electrolytic and galvanic cells.
Answer:
Electrolytic cell:

1. This device is used to bring about a non-spontaneous chemical reaction by passing an electric current.
2. It is used to bring about a chemical reaction generally for the dissociation (electrolysis) of compounds.
3. In this cell, electrical energy is converted into chemical energy.
4. In this cell, the cathode is negative and the anode is positive.
5. Electrolytic cells are irreversible.
6. Oxidation takes place at the positive electrode and reduction at the negative electrode.
7. The electrons are supplied by the external source and enter through cathode and come out through anode.
8. It is used for electroplating, electrorefining, etc.

Electrochemical cell (Galvanic cell or Voltaic cell):

1. This device is used to produce electrical energy by a spontaneous chemical reaction.
2. It is used to generate electricity.
3. In this cell, chemical energy is converted into electrical energy.
4. In this cell, the cathode is positive and the anode is negative.
5. Electrochernical cells are reversible.
6. Oxidation takes place at the negative electrode and reduction at the positive electrode.
7. The electrons move from anode to cathode in the external circuit.
8. It is used as a source of electric current.

Try this… (Textbook page No. 107)

Question 1.
Write expressions to calculate equilibrium constant from
i. Concentration data
ii. Thermochemical data
iii. Electrochemical data
Answer:
(i) Consider following a reversible cell reaction.
aA + bB ⇌ cC + dD
If [A], [B], [C] and [D] represent concentrations of reactants and products then the equilibrium constant K is,
K = \(\frac{[\mathrm{C}]^{c} \times[\mathrm{D}]^{d}}{[\mathrm{~A}]^{a} \times[\mathrm{B}]^{b}}\)
(ii) If ΔG⁰ is the standard Gibbs free energy change at temperature T then,
ΔG⁰ = – RTlnK = – 2.303 RTlog₁₀K
(iii) From electrochemical data,
if \(E_{\text {cell }}^{0}\) is the standard cell potential and K is the equilibrium constant for the cell reaction at a temperature T, then,
\(E_{\text {cell }}^{0}=\frac{0.0592}{n} \log _{10} K\)

Learn this as well…

Question 1.
The construction and working of the calomel electrode.
Answer:
(1) Since standard hydrogen electrode (SHE) is not convenient for experimental use, a secondary reference electrode like calomel electrode is used.
(2) Construction : It consists of a glass vessel with side arm B for dipping in a desired solution of another electrode like, ZnSO_4(aq) for an electric contact. The vessel is filled with mercury, a paste of Hg and Hg₂Cl₂ (calomel) and saturated KCl solution.

Fig. 5.15 : Determination of standard electrode potential using calomel electrode
(3) The potential developed depends upon the concentration of Cl^– or KCl solution. When saturated KCl solution is used, its reduction potential is 0.242 V.
(4) Consider following cell :
Zn_(s) | ZnSO_4(aq) || KCl_(aq) | Hg₂Cl_2(s) | Hg
OR Zn_(s) | ZnSO_4(aq) || Calomel electrode
Reduction reaction for calomel electrode :
Hg₂Cl_2(s) + 2e^– → 2Hg_(l) + 2Cl^–_(aq)
Hence potential of calomel electrode depends on the concentration of Cl^– or KCl solution.

Can you tell ? (Textbook page No. 114)

Question 1.
In what ways are fuel cells and galvanic cells similar and in what ways are they different ?
Answer:
Similarity between fuel cells and galvanic cells :

• In both the cells, there is oxidation at anode and j reduction at cathode.
• The cell potential is developed due to net redox reactions.
• Both are galvanic cells.

Difference in fuel cells and galvanic cells :

• Fuel cells involve electrodes with large surface area while galvanic cells involve electrodes with j compact surface area.
• Fuel cells involve gaseous materials on a large scale while galvanic cells involve gaseous materials at a definite pressures along with electrolytes or there may not be gases.
• In fuel cells, the cell potential is developed due to exothermic combustion reactions while in galvanic cell, cell potential is developed due to normal redox reactions.
• In fuel cells gaseous electrode materials are continuously supplied from outside while in galvanic cells electrode materials have constant concentration or may change due to reactions.

Use your brain power (Textbook page No. 114)

Question 1.
Indentify the strongest and the weakest oxidizing agents from the electrochemical series.
Answer:
From the electrochemical series,
(a) The strongest oxidising agent is fluorine since it has the highest standard reduction potential (\(E_{\mathrm{F}_{2} / \mathrm{F}^{-}}^{0}\) = + 2.87 V).
(b) The weakest oxidising agent (or the strongest reducing agent) is lithium since it has the lowest standard reduction potential, (\(E_{\mathrm{Li}^{+} / \mathrm{Li}}^{0}\) = -3.045 V).

Use your brain power (Textbook page No. 115)

Question 1.
Identify the strongest and the weakest reducing agents from the electrochemical series.
Answer:
(a) From the electrochemical series, the strongest reducing agent is lithium since it has the lowest standard reduction potential (\(E_{\mathrm{Li}^{+} / \mathrm{Li}}^{0}\) = -3.045 V).
(b) The weakest reducing agent is fluorine since it has the highest standard reduction potential,
(\(E_{\mathrm{F}_{2} / \mathrm{F}^{-}}^{0}\) = +2.87 V).

Question 2.
From E° values given in Table 5.1, predict whether Sn can reduce I₂ or Ni²⁺.
Answer:
From electrochemical series,

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Balbharti Maharashtra State Board 12th Physics Textbook Solutions Chapter 16 Semiconductor Devices Textbook Exercise Questions and Answers.

Semiconductor Devices Class 12 Exercise Question Answers Solutions Maharashtra Board

Class 12 Physics Chapter 16 Exercise Solutions Maharashtra Board

Physics Class 12 Chapter 16 Exercise Solutions

1. Choose the correct option

i.
In a BJT, the largest current flow occurs
(A) in the emitter
(B) in the collector
(C) in the base
(D) through CB junction.
Answer:
(A) in the emitter

ii.
A series resistance is connected in the Zener diode circuit to
(A) properly reverse bias the Zener
(B) protect the Zener
(C) properly forward bias the Zener
(D) protect the load resistance.
Answer:
(A) properly reverse bias the Zener

iii.
An LED emits visible light when its
(A) junction is reverse biased
(B) depletion region widens
(C) holes and electrons recombine
(D) junction becomes hot.
Answer:
(C) holes and electrons recombine

iv.
Solar cell operates on the principle of
(A) diffusion
(B) recombination
(C) photovoltaic action
(D) carrier flow.
Answer:
(C) photovoltaic action

v.
A logic gate is an electronic circuit which
(A) makes logical decisions
(B) allows electron flow only in one direction
(C) works using binary algebra
(D) alternates between 0 and 1 value.
Answer:
(A) makes logical decisions

2 Answer in brief.

i.
Why is the base of a transistor made thin and is lightly doped?
Answer:
The base of a transistor is lightly doped than the emitter and is made narrow so that virtually all the electrons injected from the emitter (in an npn tran-sistor) diffuse right across the base to the collector junction without recombining with holes. That is, the base width is kept less than the recombination distance. Also, the emitter is much heavily doped than the base to improve emitter efficiency and common-base current gain a.

ii.
How is a Zener diode different than an ordinary diode?
Answer:
A Zener diode is heavily doped-the doping con-centrations for both p- and n-regions is greater than 1018 cm-3 while those of an ordinary diode are voltage (PIV) of an ordinary diode is higher than a Zener diode and the breakdown occurs by impact ionization (avalanche process). Their I-V characteristics are otherwise similar.

iii.
On which factors does the wavelength of light emitted by a LED depend?
Answer:
The intensity of the emitted light is directly propor-tional to the recombination rate and hence to the diode forward current. The colour of the light emitted by an LED depends on the compound semiconductor material used and its composition (and doping levels) as given below :
Table: Typical semiconductor materials and emitted colours of LEDs

Material	Emitted colour(s)
Gallium arsenide (GaAs), Indium gallium arsenide phosphide (InGaAsP)	Infrared
Aluminum gallium arsenide (AlGaAs)	Deep red, also IR laser
Indium gallium phosphide (InGaP)	Red
Gallium arsenide phosphide (GaAsP), aluminum indium gallium phosphide (AlInGaP)	Orange, red or yellow
Gallium phosphide (GaP)	Green or yellow
Aluminium gallium phosphide (AlGaP), zinc selenide (ZnSe), zinc selenide telluride (ZnSeTe), nitrogen impregnated gallium phosphide (GaP:N)	Green
Indium gallium nitride (InGaN), gallium nitride (GaN), sine sulphide (ZnS)	Blue and violet Longer wave lengths (green and yellow) are obtained by increasing the indium (In) content. Phosphor encapsulation produce white light.
Aluminium gallium nitride (AlGaN)	Ultraviolet

iv.
Why should a photodiode be operated in reverse biased mode?
Answer:
A photodiode is operated in a reverse biased mode because as photodetector or photosensor, it must conduct only when radiation is incident on it. In the reverse biased mode, the dark current for zero illumination is negligibly small—of the order of few picoamperes to nanoamperes. But when illuminated, the photocurrent is several orders of magnitude greater.

v.
State the principle and uses of a solar Cell.
Answer:
A solar cell is an unbiased pn-junction that converts the energy of sunlight directly into electricity with a high conversion efficiency.

Principle : A solar cell works on the photovoltaic effect in which an emf is produced between the two layers of a pn-junction as a result of irradiation.

Question 3.
Draw the circuit diagram of a half wave rectifier. Explain its working. What is the frequency of ripple in its output?
Answer:
A device or a circuit which rectifies only one-half of each. cycle of an alternating voltage is called a half-wave rectifier.
Electric circuit : The alternating voltage to be rectified is applied across the primary coil (P₁P₂) of a transformer. The secondary coil (S₁S₂) of the transformer is connected in series with the junction diode and a load resistance R_L, as shown in below figure. The alternating voltage across the secondary coil is the ac input voltage V_i. The dc voltage across the load resistance is called the output voltage V₀.

Working : Due to the alternating voltage V_i, the p-region of the diode becomes alternatively positive and negative with respect to the n-region.
During the half-cycle when the p-region is positive, the diode is forward biased and conducts. A current I_L passes through the load resistance RL in the direction shown.

During the next half cycle, when the p-region is negative, the diode is reverse biased and the forward current drops to zero.

Thus, the diode conducts only during one-half of the input cycle and thus acts as a half-wave rectifier. The intermittent output voltage V₀ has a fixed polarity but changes periodically with time between zero and a maximum value. I_L is unidirectional. Above figure shows the input and output voltage waveforms.

The pulsating dc output voltage of a half-wave rectifier has the same frequency as the input.

Question 4.
Why do we need filters in a power supply?
Answer:
A rectifier-half-wave or full-wave – outputs a pul-sating dc which is not directly usable in most electronic circuits. These circuits require something closer to pure dc as produced by batteries. Unlike pure dc waveform of a battery, a rectifier output has an ac ripple riding on a dc waveform.

The circuit used in a dc power supply to remove the ripple is called a filter. A filter circuit can produce a very smooth waveform that approximates the waveform produced by a battery. The most common technique used for filtering is a capacitor connected across the output of a rectifier.

Question 5.
Draw a neat diagram of a full wave rectifier and explain it’s working.
Answer:
A device or a circuit which rectifies both halves of each cycle of an alternating voltage is called a full-wave rectifier.
Electric circuit : The alternating voltage to be rectified is applied across the primary coil (P₁P₂) of a transformer with a centre-tapped secondary coil (S₁S₂). The terminals and S2 of the secondary are connected to the two p-regions of two junction diodes D₁ and D₂, respectively. The centre-tap T is connected to the ground. The load resistance RL is connected across the common n-regions and the

P₁P₂, S₁S₂ : Primary and secondary of transformer,
T : Centre-tap on secondary; D₁ D₂ : Junction diodes,
R_L : Load resistance, I_L : Load current,
V_i: AC input voltage, V₀ : DC output voltage
Above Figure : Full-wave rectifier circuit

Working : During one half cycle of the input, terminal S₁ of the secondary is positive while S₂ is negative with respect to the ground (the centre-tap T). During this half cycle, diode D₁ is forward biased and conducts, while diode D₂ is reverse biased and does not conduct. The direction of current Z_L through R_L is in the sense shown.

During the next half cycle of the input voltage, S2 becomes positive while S, is negative with respect to T. Diode D2 now conducts sending a current IL through RL in the same sense as before. Dt now does not conduct. Thus, the current through RL flows in the same direction, i.e., it is unidirectional, for both halves or the full-wave of the input. This is called full-wave rectification.

The output voltage has a fixed polarity but varies periodically with time between zero and a maximum value. Above figure shows the input and output voltage waveforms. The pulsating dc output voltage of a full-wave rectifier has twice frequency of the input.

Question 6.
Explain how a Zener diode maintains constant voltage across a load.
Answer:
Principle : In the breakdown region of a Zener diode, for widely changing Zener current, the voltage across the Zener diode remains almost constant.

Electric circuit : The circuit for regulating or stabilizing the voltage across a load resistance RL against change in load current and supply voltage is shown in above figure. The Zener diode is connected parallel to load Rh such that the current through the Zener diode is from the n to p region. The series resistance Rs limits the current through the diode below the maximum rated value.
From the circuit, I = I_Z + I_L and V = IR_s + V_Z
= (I_Z + I_L)R_s + V_Z
Working: When the input unregulated dc voltage V across the Zener diode is greater than the Zener voltage V_Z in magnitude, the diode works in the Zener breakdown region. The voltage across the diode and load Rh is then V_Z. The corresponding current in the diode is I_Z.

As the load current (I) or supply voltage (V) changes, the diode current (7Z) adjusts itself at constant V_Z. The excess voltage V-V_Zappears across the series resistance Rs.

For constant supply voltage, the supply current I and the voltage drop across R_s remain constant. If the diode is within its regulating range, an increase in load current is accompanied by a decrease in Iz at constant V_Z.
Since the voltage across R_L remains constant at V_Z, the Zener diode acts as a voltage stabilizer or voltage regulator.

Question 7.
Explain the forward and the reverse characteristic of a Zener diode.
Answer:
The forward bias region of a Zener diode is identical to that of a regular diode. There is forward current only after the barrier potential of the pn- junction is overcome. Beyond this threshold or cut in voltage, there is an exponential upward swing.

The typical forward voltage at room temperature with a current of around 1 mA is around 0.6 V.

In the reverse bias condition the Zener diode is an open circuit and only a small reverse saturation current flows as shown with change of scale. At the reverse breakdown voltage there is an abrupt rapid increase in the current-the knee is very sharp, followed by an almost vertical increase in current. The voltage across the Zener diode in the breakdown region is very nearly constant with only a small increase in voltage with increasing current. There is a minimum Zener current, I_Z (min), that places the operating point in the desired breakdown region. At some high current level, I_ZM, the power dissipation of the diode becomes excessive beyond which the diode can be damaged.

Zener diode characteristics

The I-V characteristics of a Zener diode is not totally vertical in the breakdown region. This means that for slight changes in current, there will be a small change in the voltage across the diode. The voltage change for a given change in current is the resistance R_Z of the Zener diode.

Question 8.
Explain the working of a LED.
Answer:
Working :
An LED is forward-biased with about 1.2 V to 3.6 V at 12 mA to 20 mA. Majority carriers electrons from n-type layer and holes from p-type layer are injected into the active layer. Electrons cross the junction into the p-layer. In the active p-layer, some of these excess minority carriers electrons, recombine radiatively with majority carriers, holes, thereby emitting photons. The resulting photon has an energy approximately equal to the bandgap of the active layer material. Modifying the bandgap of the active layer creates photons of diferent energies.

In the energy band diagram this recombination is equivalent to a transition of the electron from a higher energy state in the conduction band to a lower energy state in the valence band. The energy difference is emitted as a photon of energy hv.
[Note : The photons originate primarily in the p-side of the junction which has a bandgap E_Gp narrower than that of the n-side, E_Gn. Thus, with hv < E_Gn, the photons are emitted through the wide-bandgap n-region with essentially no absorption.]

Question 9.
Explain the construction and working of solar cell.
Answer:
Construction :
A simple pn-junction solar cell con-sists of a p-type semiconductor substrate backed with a metal electrode back contact. A thin n-layer (less than 2.5 pm, for silicon) is grown over the p-type substrate by doping with suitable donor impurity. Metal finger electrodes are prepared on top of the n-layer so that there is enough space between the fingers for sunlight to reach the n-layer and, subsequently, the underlying pn-junction.

Working : When exposed to sunlight, the absorption of incident radiation (in the range near-UV to infrared) creates electron-hole pairs in and near the depletion layer.

Consider light of frequency v incident on the pn-junction such that the incident photon energy hv is greater than the band gap energy EG of the semiconductor. The photons excite electrons from the valence band to the conduction band, leaving vacancies or holes in the valence band, thus generating electron-hole pairs.

The photogenerated electrons and holes move towards the n side and p side, respectively. If no external load is connected, these photogenerated charges get collected at the two sides of the junction and give rise to a forward photovoltage. In a closed- circuit, a current I passes through the external load as long as the solar cell is exposed to sunlight.

A solar cell module consists of several solar cells connected in series for a higher voltage output. For outdoor use with higher power output, these modules are connected in different series and parallel combinations to form a solar cell array.

[Note : Currently most of the crystalline solar cells are p-type as described above. This is because of a lower cost of production of p-type. But performance wise, n-type solar cells (a thin p-layer over an n-type substrate by doping with suitable acceptor impurity) can give much better efficiency compared to p-type solar cells.]

Question 10.
Explain the principle of operation of a photodiode.
Answer:
Construction:
A photodiode consists of an n-type silicon substrate with a metal electrode back contact. A thin p-type layer is grown over the n-type substrate by diffusing a suitable acceptor dopant.

The area of the p-layer defines the photodiode active area. An ohmic contact pad is deposited on the active area. The rest of the active area is left open with a protective antireflective coating of silicon nitride to minimize the loss of photons. The nonactive area is covered with an insulating opaque SiO₂ coating.

Depending on the required spectral sensitivity, i.e., the operating wavelength range, typical photodiode materials are silicon, germanium, indium gallium arsenide phosphide (InGaAsP) and indium gallium arsenide (InGaAs), of which silicon is the cheapest while the last two are expensive.

Working : The band gap energy of silicon is E_G = 1.12 eV at room temperature. Thus, photons or particles with energies greater than or equal to 1.12 eV, which corresponds to 110 nm, can transfer electrons from the valence band into the conduction band.

A photodiode is operated in the reverse bias mode which results in a wider depletion region. When operated in the dark (zero illumination), there is a reverse saturation current due solely to the thermally generated minority charge carriers. This is called the dark current. Depending on the minority carrier concentrations, the dark current in an Si photodiode may range from 5 pA to 10 nA.

When exposed to radiation of energy hv ≥ E_G (in the range near-UV to near-IR), electron-hole pairs are created in the depletion region. The electric field in the depletion layer accelerates these photogenerated electrons and holes towards the n side and p side, respectively, constituting a photocurrent l in the external circuit from the p side to the n side. Due to the photogeneration, more charge carriers are available for conduction and the reverse current is increased. The photocurrent is directly propor-tional to the intensity of the incident light. It is independent of the reverse bias voltage.
[Notes : Typical photodiode materials are :
(1) silicon (Si) : low dark current, high speed, good sensitivity between ~ 400 nm and 1000 run (best around 800 nm-900 nm)
(2) germanium (Ge) : high dark current, slow speed, good sensitivity between ~ 900 nm and 1600 nm (best around 1400 nm-1500 nm)
(3) indium gallium arsenide phosphide (InGaAsP) : expensive, low dark current, high speed, good sensitivity between ~ 1000 nm and 1350 nm (best around 1100 nm- 1300 nm)
(4) indium gallium arsenide (InGaAs): expensive, low dark current, high speed, good sensitivity between ~ 900 nm and 1700 nm (best around 1300 nm-1600 nm],

Question 11.
What do you mean by a logic gate, a truth table and a Boolean expression?
Answer:
A logic gate is a basic switching circuit used in digital circuits that determines when an input pulse can pass through to the output. It generates a single output from one or more inputs.

Explanation/Uses :
Any digital computation process consists in performing a sequence of arithmetical operations on the data of the problem. At each stage in the computation, the nature of the operation to be performed is decided partly by the pre-determined program and partly by the outcome of earlier stages in the process. We therefore need switches with multiple inputs to perform logical operations, i.e., the outputs of these switches are determined in specified ways by the condition (binary state) of their inputs. These arrangements are known as logic gates, and mostly they are extension of a simple transistor switch.

(1) Boolean expression : An equation expressing a logical compound statement in Boolean algebra is called a Boolean expression. A Boolean expression for a logic gate expresses the relation between input(s) and output of a logic gate.

(2) Truth table : The table which shows the truth values of a Boolean expression for a logic gate for all possible combinations of its inputs is called the truth table of logic gate.

The truth table contains one row for each input combination. Since a logical variable can assume only two possible values, 0 and 1, there are 2N combinations of N inputs so that the table has 2N rows.
[Note : Boolean algebra is a form of symbolic logic developed in 1847 by George Boole (1815-64) British mathematician.]

Question 12.
What is logic gate? Write down the truth table and Boolean expression for ‘AND’ gate.
Answer:
A logic gate is a basic switching circuit used in digital circuits that determines when an input pulse can pass through to the output. It generates a single output from one or more inputs.

The AND gate : It is a circuit with two or more inputs and one output in which the output signal is HIGH if and only if all the inputs are HIGH simultaneously.

The AND operation represents a logical multiplication.

Below figure shows the 2-input AND gate logic symbol and the Boolean expression and the truth table for the AND function.
Logic symbol:

Truth table:

Boolean expression:
Y = A ∙ B

Question 13.
What are the uses of logic gates? Why is a NOT gate known as an inverter?
Answer:
Explanation/Uses :
Any digital computation process consists in performing a sequence of arithmetical operations on the data of the problem. At each stage in the computation, the nature of the operation to be performed is decided partly by the pre-determined program and partly by the outcome of earlier stages in the process. We therefore need switches with multiple inputs to perform logical operations, i.e., the outputs of these switches are determined in specified ways by the condition (binary state) of their inputs. These arrangements are known as logic gates, and mostly they are extension of a simple transistor switch.

The NOT gate or INVERTER : It is a circuit with one input whose output is HIGH if the input is LOW and vice versa.

The NOT operation outputs an inverted version of the input. Hence, a NOT gate is also known as an INVERTER.

The small invert bubble on the output side of the inverter logic symbol, below figure and the over bar (^–) in the Boolean expression represent the invert function.

Question 14.
Write the Boolean expression for (i) OR gate, (ii) AND gate, and (iii) NAND Gate.
Answer:
(i) The OR gate : It is a circuit with two or more inputs and one output in which the output signal is HIGH if any one or more of the inputs is HIGH.
The OR operation represents a logical addition.
Below figure shows the 2-input OR gate logic sym-bol, and the Boolean expression and the truth table for the OR function.

(ii) The AND gate : It is a circuit with two or more inputs and one output in which the output signal is HIGH if and only if all the inputs are HIGH simultaneously.

The AND operation represents a logical multiplication.

Below figure shows the 2-input AND gate logic symbol and the Boolean expression and the truth table for the AND function.
Logic symbol:

Truth table:

Boolean expression:
Y = A ∙ B

(iii) The NAND gate : It is a circuit with two or more inputs and one output, whose output is HIGH if any one or more of the inputs is LOW; the output is LOW if all the inputs are HIGH.

The NAND gate is a combination of an AND gate followed by a NOT gate so that the truth table of the NAND function is obtained by inverting the outputs of the AND gate.

Question 15.
Why is the emitter, the base and the collector of a BJT doped differently?
Answer:
A BJT being a bipolar device, both electrons and holes participate in the conduction process. Under the forward-biased condition, the majority carriers injected from the emitter into the base constitute the largest current component in a BJT. For these carriers to diffuse across the base region with t negligible recombination and reach the collector junction, these must overwhelm the majority carriers of opposite charge in the base. The total emitter current has two components, that due to majority carriers in the emitter and that due to minority carriers diffused from base into emitter. The ratio of the current component due to the injected majority carriers from the emitter to the total emitter current is a measure of the emitter efficiency. To improve the emitter efficiency and the common-base current gain (a), it can be shown that’ the emitter should be much heavily doped than the base.

Also, the base width is a function of the base- collector voltage. A low doping level of the collector increases the size of the depletion region. This increases the maximum collector-base voltage and reduces the base width. Further, the large depletion region at the collector-base junction-extending mainly into the collector-corresponds to a smaller electric field and avoids avalanche breakdown of the reverse-biased collector-base junction.
[Note : Effective dopant concentrations of (a) npn transistor (b) pnp transistor are shown below.

The base doping is less than the emitter doping but greater than the collector doping. Contrary to the impression stressed in the Board’s and NCERT textbooks, collector doping is typically an order of magnitude lower than base doping. {Ref. : Semiconductor Devices Physics and Technology (3rd Edition), Simon M. Sze and M. K. Lee, p. 125}]

Question 16.
Which method of biasing is used for operating transistor as an amplifier?
Answer:
For use as an amplifier, the transistor should be in active mode. Therefore, the emitter-base junction is forward biased and the collector-base junction is reverse biased. Also, an amplifier uses an emitter bias rather than a base bias.

Question 17.
Define α and β. Derive the relation between then.
Answer:
The dc common-base current ratio or current gain (α_dc) is defined as the ratio of the collector current to emitter current.
α_dc = \(\frac{I_{C}}{I_{\mathrm{E}}}\)
The dc common-emitter current ratio or current gain (β_dc) is defined as the ratio of the collector current to base current.
β_dc = \(\frac{I_{C}}{I_{B}}\)
Since the emitter current I_E = I_B + I_C
\(\frac{I_{\mathrm{E}}}{I_{C}}=\frac{I_{\mathrm{B}}}{I_{\mathrm{C}}}+1\)
∴ \(\frac{1}{\alpha_{\mathrm{dc}}}=\frac{1}{\beta_{\mathrm{dc}}}+1\)
Therefore, the common-base current gain in terms of the common-emitter current gain is
α_dc = \(\frac{\beta_{\mathrm{dc}}}{1+\beta_{\mathrm{dc}}}\)
and the common-emitter current gain in terms of the common-base current gain is
β_dc = \(\frac{\alpha_{\mathrm{dc}}}{1-\alpha_{\mathrm{dc}}}\)
For a transistor, α_dc is close to but always less than 1 (about 0.92 to 0.98) and β_dc ranges from 20 to 200 for most general purpose transistors.

Question 18.
The common-base DC current gain of a transistor is 0.967. If the emitter current is 10mA. What is the value of base current?
Answer:
Data : α_dc = 0.967, I_E = 10 mA
α_dc = \(\frac{I_{C}}{I_{\mathrm{E}}}\) and I_E = I_B + I_C
The collector current,
I_C = α_dcI_E = 0.967 × 10 = 9.67 mA
Therefore, the base current,
I_B = I_E – I_C = 10 – 9.67 = 0.33 mA

Question 19.
In a comman-base connection, a certain transistor has an emitter current of 10mA and collector current of 9.8 mA. Calculate the value of the base current.
Answer:
DATA : I_E = 10 mA, I_C = 9.8 mA
I_E = I_B + I_C
Therefore, the base current,
I_B = I_E – I_C – 10 – 9.8 = 0.2 mA

Question 20.
In a common-base connection, the emitter current is 6.28mA and collector current is 6.20 mA. Determine the common base DC current gain.
Answer:
Data : I_E = 6.28 mA, I_C = 6.20 mA
α_dc = \(\frac{I_{C}}{I_{\mathrm{E}}}\) and β_dc = \(\frac{I_{\mathrm{C}}}{I_{\mathrm{B}}}=\frac{\alpha_{\mathrm{dc}}}{1-\alpha_{\mathrm{dc}}}\)
Common-emitter current gain, α_dc = \(\frac{6.20}{6.28}\) = 0.9873
Therefore, common-base current gain,
β_dc = \(\frac{0.9873}{1-0.9873}=\frac{0.9873}{0.0127}\) = 77.74
OR
I_E = I_B + I_C
∴ I_B = I_E – I_C = 6.28 – 6.20 = 0.08 mA
∴ β_dc = \(\frac{6.20}{0.08}\) = 77.5
[Note : The answer given in the textbook obviously refers to the common-emitter current gain.]

12th Physics Digest Chapter 16 Semiconductor Devices Intext Questions and Answers

Remember this (Textbook Page No. 346)

Question 1.
A full wave rectifier utilises both half cycles of AC input voltage to produce the DC output.
Answer:
A half-wave rectifier rectifies only one half of each cycle of the input ac wave while a full-wave rectifier rectifies both the halves. Hence the pulsating dc output voltage of a half-wave rectifier has the same frequency as the input but that of a full-wave rectifier has double the frequency of the ac input.

Do you know (Textbook Page No. 346)

Question 1.
The maximum efficiency of a full wave rectifier is 81.2% and the maximum efficiency of a half wave rectifier is 40.6%. It is observed that the maximum efficiency of a full wave rectifier is twice that of half wave rectifier.
Answer:
The ratio of dc power obtained at the output to the applied input ac power is known as rectifier efficiency. A half-wave rectifier can convert maximum 40.6% of ac power into dc power, and the remaining power of 59.4% is lost in the rectifier circuit. In fact, 50% power in the negative half cycle is not converted and the remaining 9.4% is lost in the circuit. Hence, a half wave rectifier efficiency is 40.6%. The maximum efficiency of a full-wave rectifier is 81.2%, i.e., twice that of a half-wave rectifier.

Do you know (Textbook Page No. 349)

Question 1.
The voltage stabilization is effective when there is a minimum Zener current. The Zener diode must be always operated within its breakdown region when there is a load connected in the circuit. Similarly, the supply voltage V_s must be greater than V_z.
Answer:
A Zener diode is operated in the breakdown region. There is a minimum Zener current, I_z, that places the desired operating point in the breakdown region. There is a maximum Zener current, I_zM, at which the power dissipation drives the junction temperature to the maximum allowed. Beyond that current the diode can be damaged. Hence, the supply voltage must be greater than V_z and the current-limiting resistor must limit the diode current to less than the rated maxi mum, I_zM.

Remember this (Textbook Page No. 350)

Question 1.
Zener effect occurs only if the diode is heavily doped, because when the depletion layer is thin, breakdown occurs at low reverse voltage and the field strength will be approximately 3 × 10⁷ V/m. It causes an increase in the flow of free carriers and increase in the reverse current.
Answer:
Zener breakdown occurs only in heavily doped pn junctions (doping concentrations for both p- and n-regions greater than 1018 cm3) and can take place only if the electric field in the depletion region of the reverse-biased junction is very high. It is found that the critical field at which tunneling becomes probable, i.e., at which Zener breakdown commences, is approximately 10⁶ V/cm. [“internal Field Emissiot at Narrow Silicon and Germanium PN-Junctions,” Phys. Rev., 118, 425 (1960).]

Can you tell (Textbook Page No. 350)

Question 1.
How does a cell phone charger produce a voltage of 5.0 V form the line voltage of 230V?
Answer:
A phone charger is usually a 5 V power supply. A 4-diode bridge input rectifier rectifies the ac mains voltage a provide a high voltage dc. A transistor chopper switches this on and off at high frequency. This stage is required because this high frequency allows the transformer to be smaller, lighter and much lower in cost.

A small transformer steps this down to a low voltage high-frequency ac. An output rectifier and filter convert this to low-voltage (5 V) dc and smooths out the ripple. A chopper controller provides a feedback to the chopper through an optoisohitor and adjusts the chopping cycle to maintain the output voltage at 5 V.

Question 2.
Why is a resistance connected in series with a Zener diode when used in a circuit?
Answer:
The I-V characteristics in the breakdown region of a Zener diode is almost vertical. That is, the current I_Z can rapidly increase at constant V_Z. To prevent damage due to excessive heating, the Zener current should not exceed the rated maximum current, I_ZM. Hence, a current-limiting resistor R_s is connected in series with the diode.

I_Z and the power dissipated in the Zener diode will be large for I _L = 0 (no-load condition) or when I_L is less than the rated maximum (when R_s is small and R_L is large). The current-limiting resistor R_s is so chosen that the Zener current does not exceed the rated maximum reverse current, I_ZM when there is no load or when the load is very high.
The rated maximum power of a Zener diode is
P_ZM = I_ZM = V_Z

At n-load condition, the current through R is I = I_ZM and the voltage drop across it is V – V_Z, where V is the unregulated source voltage. The diode current will be maximum when V is maxi mum at V_max and I = I_ZM. Then, the minimum value of the series resistance should be
R_{s, min} = \(\frac{V_{\max }-V_{\mathrm{Z}}}{I_{\mathrm{ZM}}}\)

Question 3.
The voltage across a Zener diode does not remain strictly constant with the changes in the Zener current. This is due to R_Z, the Zener impedance, or the internal resistance of the Zener diode. R_Z acts like a small resistance in series with the Zener. Changes in I_Z cause small changes in V_Z .
Answer:
The I-V characteristics of a Zener diode in the breakdown region is not strictly vertical. Its slope is 1/R_Z, where R_Z is the Zener impedance.

Can you know (Textbook Page No. 354)

Question 1.
What is the difference between a photo diode and a solar cell?
Answer:
Both are semiconductor photovoltaic devices. A photodiode is a reverse-biased pn-junction diode while a solar cell is an unbiased pn-junction diode. Photod iodes, however, are optimized for light detection while solar cells are optimized for energy conversion efficiency.

Question 2.
When the intensity of light incident on a photo diode increases, how is the reverse current affected?
Answer:
The photocurrent increases linearly with increasing illuminance, limited by the power dissipation of the photodiode.

Do you know (Textbook Page No. 355)

Question 1.
LED junction does not actually emit that much light so the epoxy resin body is constructed in such a way that the photons emitted by the junction are reflected away from the surrounding substrate base to which the diode is attached and are focused upwards through the domed top of the LED, which itself acts like a lens concentrating the light. This is why the emitted light appears to be brightest at the top of the LED.
Answer:
The pn-junction of an LED is encased in a transparent, hard plastic (epoxy resin), not only for shock protection but also for enhancing the brightness in one direction. Light emitted by the pn-junction is not directional. The hemispherical epoxy lens focuses the light in the direction of the hemispherical part. This is why the emitted light appears to be brightest at the top of the LED.

Question 2.
The current rating of LED is of a few tens of milli-amps. Hence it is necessary to connect a high resistance in series with it. The forward voltage drop of an LED is much larger than an ordinary diode and is around 1.5 to 3.5 volts.
Answer:
Most common LEDs require a forward operating voltage of between approximately 1.2 V (for a standard red LED) to 3.6 V (for a blue LED) with a forward current rating of about 10 mA to 30 mA, with 12 mA to 20 mA being the most common range. Like any diode, the forward current is approximately an exponential function of voltage and the forward resistance is very small. A small voltage change may result in a large change in current. If the current exceeds the rated maximum, an LED may overheat and get destroyed. LEDs are current driven devices and a current-limiting series resistor is required to prevent burning up the LED.

Do you know (Textbook Page No. 356)

Question 1.
White Light LEDs or White LED Lamps:
Shuji Nakamura, a Japanese – born American electronic engineer invented the blue LED. He was awarded the Nobel prize for physics for 2014. He was also awarded the global energy prize in the year 2015. His invention of blue LED made the fabrication of white LED possible.
LED lamps, bulbs, street lighting are becoming very popular these days because of the very high efficiency of LEDs in terms of light output per unit input power(in milliWatts), as compared to the incandescent bulbs. So for general purpose lightings, white light is preferred.
Commercially available white LEDs are normally manufactured by using the technique of wavelength conversion. It is a process which partly or completely converts the radiation of a LED into white light. There are many ways of wavelength conversion. One of these methods uses blue LED and yellow phosphor. In this method of wavelength conversion, a LED which emits blue colour is used to excite a yellow colour phosphor. This results in the emission of yellow and blue light and this mixture of blue and yellow light gives the appearance of white light. This method is the least expensive method for producing white light.
Answer:

The all important blue LEDs
The development of LEDs has made more efficient light sources possible. Creating white light that can be used for lighting requires a combination of red, green and blue light. Blue LEDs proved to be much more difficult to create than red and green LEDs. During the 1980s and 1990s Isamu Akasaki, Hiroshi Amano, and Shuji Nakamura successfully used the difficult-to-handle semiconductor gallium nitride to create efficient blue LEDs. Isamu Akasaki is known for invent ing the bright gallium nitride (CaN) pn-junction blue LED in 1989 and subsequently the high-brightness CaN blue LED.

Using blue LEDs, highly efficient white light sources. became possible by converting part of the blue light emitted from an LED to yellow using a phosphor. To the human eye, the combination of blue and yellow light is perceived as white. A white LED can be created by embedding phosphors in the plastic cap which surrounds a blue LED. Higher quality white light can also be created by mixing blue light with other colors as well, including red and green

Isamu Akasaki, together with Shuji Nakamura and Hiroshi Amano, received the 2014 Nobel Prize in Physics for the invention of efficient blue light-emitting diodes which has enabled bright and energy saving white light sources.

Use your brain power (Textbook Page No. 357)

Question 1.
What would happen if both junctions of a BJT are forward biased or reverse biased?
Answer:
A BJT has four regimes of operation, depending on the four combinations of the applied biases (voltage polarities) to the emitter-base junction and the collector-base junction, as shown in the following table; ‘F’ and ‘R’ indicate forward bias and reverse bias, respectively.

Remember This (Textbook Page No. 358)

Question 1.
The lightly doped, thin base region sandwiched between the heavily doped emitter region and the intermediate doped collector region plays a crucial role in the transistor action.
Answer:
If the two junctions in a BJT are physically close compared with the minority carrier diffusion length (i.e., the distance within which recombination will take place), the careers injected from the emitter can diffuse through the base to reach the base-collector junction. The narrow width of the base is thus crucial for transistor action.

Use your brain power (Textbook Page No. 361)

Question 1.
If a transistor amplifies power, explain why it is not used to generate power.
The term ‘amplification’ is used as an abstraction of the transistor properties so that we have few equations which are useful for a large number of practical problems. Transistors use a small power to control a power supply which can output a huge power. The large output comes from the power supply, while the input signal valves the transistor on and off. The increased power comes from the power supply so that a transistor does not violate the law of conservation of energy.

12th Std Physics Questions And Answers:

• Current Electricity Class 12 Physics Questions And Answers
• Magnetic Fields due to Electric Current Class 12 Physics Questions And Answers
• Magnetic Materials Class 12 Physics Questions And Answers
• Electromagnetic Induction Class 12 Physics Questions And Answers
• AC Circuits Class 12 Physics Questions And Answers
• Dual Nature of Radiation and Matter Class 12 Physics Questions And Answers
• Structure of Atoms and Nuclei Class 12 Physics Questions And Answers
• Semiconductor Devices Class 12 Physics Questions And Answers

Balbharti Maharashtra State Board 12th Physics Textbook Solutions Chapter 15 Structure of Atoms and Nuclei Textbook Exercise Questions and Answers.

Structure of Atoms and Nuclei Class 12 Exercise Question Answers Solutions Maharashtra Board

Class 12 Physics Chapter 15 Exercise Solutions Maharashtra Board

Physics Class 12 Chapter 15 Exercise Solutions

In solving problems, use m_e = 0.00055 u = 0.5110 MeV/c², m_p = 1.00728 u, m_n = 1.00866u, m_H = 1.007825 u, u = 931.5 MeV, e = 1.602 × 10^-19 C, h = 6.626 × 10^-34 Js, ε₀ = 8.854 × 10^-12 SI units and me = 9.109 × 10^-31 kg.

1. Choose the correct option.

i) In which of the following systems will the radius of the first orbit of the electron be the smallest?
(A) hydrogen
(B) singly ionized helium
(C) deuteron
(D) tritium
Answer:
(D) tritium

ii) The radius of the 4th orbit of the electron will be smaller than its 8th orbit by a factor of
(A) 2
(B) 4
(C) 8
(D) 16
Answer:
(B) 4

iii) In the spectrum of the hydrogen atom which transition will yield the longest wavelength?
(A) n = 2 to n = 1
(B) n = 5 to n = 4
(C) n = 7 to n = 6
(D) n = 8 to n = 7
Answer:
(D) n = 8 to n = 7

iv) Which of the following properties of a nucleus does not depend on its mass number?
(A) radius
(B) mass
(C) volume
(D) density
Answer:
(D) density

v) If the number of nuclei in a radioactive sample at a given time is N, what will be the number at the end of two half-lives?
(A) \(\frac{N}{2}\)
(B) \(\frac{N}{4}\)
(C) \(\frac{3N}{4}\)
(D) \(\frac{N}{8}\)
Answer:
(B) \(\frac{N}{4}\)

2. Answer in brief.

i) State the postulates of Bohr’s atomic model.
Answer:
The postulates of Bohr’s atomic model (for the hydrogen atom) :

1. The electron revolves with a constant speed in acircular orbit around the nucleus. The necessary centripetal force is the Coulomb force of attraction of the positive nuclear charge on the negatively charged electron.
2. The electron can revolve without radiating energy only in certain orbits, called allowed or stable orbits, in which the angular momentum of the electron is equal to an integral multiple of h/2π, where h is Planck’s constant.
3. Energy is radiated by the electron only when it jumps from one of its orbits to another orbit having lower energy. The energy of the quantum of elec-tromagnetic radiation, i.e., the photon, emitted is equal to the energy difference of the two states.

ii) State the difficulties faced by Rutherford’s atomic model.
Answer:
(1) According to Rutherford, the electrons revolve in circular orbits around the atomic nucleus. The circular motion is an accelerated motion. According to the classical electromagnetic theory, an accelerated charge continuously radiates energy. Therefore, an electron during its orbital motion, should go on radiating energy. Due to the loss of energy, the radius of its orbit should go on decreasing. Therefore, the electron should move along a spiral path and finally fall into the nucleus in a very short time, of the order of 10^-16 s in the case of a hydrogen atom. Thus, the atom should be unstable. We exist because atoms are stable.

(2) If the electron moves along such a spiral path, the radius of its orbit would continuously decrease. As a result, the speed and frequency of revolution of the electron would go on increasing. The electron, therefore, would emit radiation of continuously changing frequency, and hence give rise to a con-tinuous spectrum. However, atomic spectrum is a line spectrum.

iii) What are alpha, beta and gamma decays?
Answer:
(a) A radioactive transformation in which an α-particle is emitted is called α-decay.
In an α-decay, the atomic number of the nucleus decreases by 2 and the mass number decreases by 4.
Example : \({ }_{92}^{238} \mathrm{U} \rightarrow{ }_{90}^{234} \mathrm{Th}+{ }_{2}^{4} \alpha\)
Q = [m_u – m_Th – m_α]c²

(b) A radioactive transformation in which a β-particle is emitted is called β-decay.
In a β^–-decay, the atomic number of the nucleus increases by 1 and the mass number remains unchanged.
Example : \({ }_{90}^{234} \mathrm{Th} \rightarrow{ }_{91}^{234} \mathrm{~Pa}+{ }_{-1}^{0} e+\bar{v}_{\mathrm{e}}\)
where \(\bar{v}_{\mathrm{e}}\) is the neutrino emitted to conserve the momentum, energy and spin.
Q = [m_u – m_Th – m_α]c²
In a β⁺-decay, the atomic number of the nucleus decreases by 1 and the mass number remains unchanged.
Example : \(\begin{aligned}
&30 \\
&15
\end{aligned} P \rightarrow{ }_{14}^{30} \mathrm{Si}+{ }_{+1}^{0} e+v_{\mathrm{e}}\)
where v_e is the neutrino emitted to conserve the momentum, energy and spin.
Q = [m_P – m_Si – m_e]c²
[Note : The term fi particle refers to the electron (or positron) emitted by a nucleus.]

A given nucleus does not emit α and β-particles simultaneously. However, on emission of α or β-particles, most nuclei are left in an excited state. A nucleus in an excited state emits a γ-ray photon in a transition to the lower energy state. Hence, α and β-particle emissions are often accompanied by γ-rays.

iv) Define excitation energy, binding energy and ionization energy of an electron in an atom.
Answer:
(1) Excitation energy of an electron in an atom : The energy required to transfer an electron from the ground state to an excited state (a state of higher energy) is called the excitation energy of the electron in that state.

(2) Binding energy of an electron in an atom is defined as the minimum energy that should be provided to an orbital electron to remove it from the atom such that its total energy is zero.

(3) Ionization energy of an electron in an atom is defined as the minimum energy required to remove the least strongly bound electron from a neutral atom such that its total energy is zero.

v) Show that the frequency of the first line in Lyman series is equal to the difference between the limiting frequencies of Lyman and Balmer series.
Answer:
For the first line in the Lyman series,
\(\frac{1}{\lambda_{\mathrm{L} 1}}=R\left(\frac{1}{1^{2}}-\frac{1}{2^{2}}\right)=R\left(1-\frac{1}{4}\right)=\frac{3 R}{4}\)
∴ v_L1 = \(\frac{c}{\lambda_{\mathrm{L} 1}}=\frac{3 R_{c}}{4}\), where v denotes the frequency,
c the speed of light in free space and R the Rydberg constant.
For the limit of the Lyman series,

Hence, the result.

Question 3.
State the postulates of Bohr’s atomic model and derive the expression for the energy of an electron in the atom.
Answer:
The postulates of Bohr’s atomic model (for the hydrogen atom) :
(1) The electron revolves with a constant speed in acircular orbit around the nucleus. The necessary centripetal force is the Coulomb force of attraction of the positive nuclear charge on the negatively charged electron.
(2) The electron can revolve without radiating energy only in certain orbits, called allowed or stable orbits, in which the angular momentum of the electron is equal to an integral multiple of h/2π, where h is Planck’s constant.
(3) Energy is radiated by the electron only when it jumps from one of its orbits to another orbit having lower energy. The energy of the quantum of elec-tromagnetic radiation, i.e., the photon, emitted is equal to the energy difference of the two states.

Consider the electron revolving in the nth orbit around the nucleus of an atom with the atomic number Z. Let m and e be the mass and the charge of the electron, r the radius of the orbit and v the linear speed of the electron.

According to Bohr’s first postulate, centripetal force on the electron = electrostatic force of attraction exerted on the electron by the nucleus
∴ \(\frac{m v^{2}}{r}=\frac{1}{4 \pi \varepsilon_{0}} \cdot \frac{Z e^{2}}{r^{2}}\) ……………. (1)
where ε₀ is the permittivity of free space.
∴ Kinetic energy (KE) of the electron
= \(\frac{1}{2} m v^{2}=\frac{Z e^{2}}{8 \pi \varepsilon_{0} r}\) ………….. (2)
The electric potential due to the nucleus of charge +Ze at a point at a distance r from it is
V = \(\frac{1}{4 \pi \varepsilon_{0}} \cdot \frac{\mathrm{Ze}}{r}\)
∴ Potential energy (PE) of the electron
= charge on the electron × electric potential
= – e × \(\frac{1}{4 \pi \varepsilon_{0}} \frac{Z e}{r}=-\frac{Z e^{2}}{4 \pi \varepsilon_{0} r}\) …………….. (3)
Hence, the total energy of the electron in the nth orbit is
E = KE + PE = \(\frac{-Z e^{2}}{4 \pi \varepsilon_{0} r}+\frac{Z e^{2}}{8 \pi \varepsilon_{0} r}\)
∴ E = \(-\frac{Z e^{2}}{8 \pi \varepsilon_{0} r}\) ………….. (4)
This shows that the total energy of the electron in the nth orbit of the atom is inversely proportional to the radius of the orbit as Z, ε₀ and e are constants. The radius of the nth orbit of the electron is
r = \(\frac{\varepsilon_{0} h^{2} n^{2}}{\pi m Z e^{2}}\) …………….. (5)
where h is Planck’s constant.
From Eqs. (4) and (5), we get,
E_n = \(-\frac{Z e^{2}}{8 \pi \varepsilon_{0}}\left(\frac{\pi m Z e^{2}}{\varepsilon_{0} h^{2} n^{2}}\right)=-\frac{m Z^{2} e^{4}}{8 \varepsilon_{0}^{2} h^{2} n^{2}}\) ……………… (6)
This gives the expression for the energy of the electron in the nth Bohr orbit. The minus sign in the expression shows that the electron is bound to the nucleus by the electrostatic force of attraction.
As m, Z, e, ε₀ and h are constant, we get
E_n ∝ \(\frac{1}{n^{2}}\)
i.e., the energy of the electron in a stationary energy state is discrete and is inversely proportional to the square of the principal quantum number.
[ Note : Energy levels are most conveniently expressed in electronvolt. Hence, substituting the values of m, e, £0 and h, and dividing by the conversion factor 1.6 × 10^-19 J/eV,
E_n ≅ \(-\frac{13.6 Z^{2}}{n^{2}}\) (in eV)
For hydrogen, Z = 1
∴ E_n ≅ \(-\frac{13.6}{n^{2}}\) (in eV).

Question 4.
Starting from the formula for energy of an electron in the n^th orbit of hydrogen atom, derive the formula for the wavelengths of Lyman and Balmer series spectral lines and determine the shortest wavelengths of lines in both these series.
Answer:
According to Bohr’s third postulate for the model of the hydrogen atom, an atom radiates energy only when an electron jumps from a higher energy state to a lower energy state and the energy of the
quantum of electromagnetic radiation emitted in this process is equal to the energy difference between the two states of the electron. This emission of radiation gives rise to a spectral line.

The energy of the electron in a hydrogen atom,
when it is in an orbit with the principal quantum
number n, is
E_n = \(-\frac{m e^{4}}{8 \varepsilon_{0}^{2} h^{2} n^{2}}\)
where m = mass of electron, e = electronic charge, h = Planck’s constant and = permittivity of free space.

Let E_m be the energy of the electron in a hydrogen atom when it is in an orbit with the principal quantum number m and E, its energy in an orbit with the principal quantum number n, n < m. Then
E_m = \(-\frac{m e^{4}}{8 \varepsilon_{0}^{2} h^{2} m^{2}}\) and E_n = \(-\frac{m e^{4}}{8 \varepsilon_{0}^{2} h^{2} m^{2}}\)
Therefore, the energy radiated when the electron jumps from the higher energy state to the lower energy state is
E_m – E_n = \(\frac{-m e^{4}}{8 \varepsilon_{0}^{2} h^{2} m^{2}}-\left(-\frac{m e^{4}}{8 \varepsilon_{0}^{2} h^{2} n^{2}}\right)\)
= \(\frac{m e^{4}}{8 \varepsilon_{0}^{2} h^{2}}\left(\frac{1}{n^{2}}-\frac{1}{m^{2}}\right)\)
This energy is emitted in the form of a quantum of radiation (photon) with energy hv, where V is the frequency of the radiation.
∴ E_m – E_n = hv
∴ v = \(\frac{E_{m}-E_{n}}{h}=\frac{m e^{4}}{8 \varepsilon_{0}^{2} h^{3}}\left(\frac{1}{n^{2}}-\frac{1}{m^{2}}\right)\)
The wavelength of the radiation is λ = \(\frac{c}{v^{\prime}}\)
where c is the speed of radiation in free space.
The wave number, \(\bar{v}=\frac{1}{\lambda}=\frac{v}{c}\)
\(\bar{v}=\frac{1}{\lambda}=\frac{m e^{4}}{8 \varepsilon_{0}^{2} h^{3} c}\left(\frac{1}{n^{2}}-\frac{1}{m^{2}}\right)=R\left(\frac{1}{n^{2}}-\frac{1}{m^{2}}\right)\)
where \(R\left(=\frac{m e^{4}}{8 \varepsilon_{0}^{2} h^{3} c}\right)\) is a constant called the Ryd berg constant.

This expression gives the wave number of the radiation emitted and hence that of a line in hydrogen spectrum.

For the Lyman series, n = 1,m = 2, 3, 4, ………… ∞
∴ \(\frac{1}{\lambda_{\mathrm{L}}}=R\left(\frac{1}{1^{2}}-\frac{1}{m^{2}}\right)\) and for the shortest wavelength line m this series, \(\frac{1}{\lambda_{\mathrm{Ls}}}=R\left(\frac{1}{1^{2}}\right)\) as m = ∞.
For the Balmer series, n = 2, m = 3, 4, 5, … ∞.
∴ \(\frac{1}{\lambda_{\mathrm{B}}}=R\left(\frac{1}{4}-\frac{1}{m^{2}}\right)\) and for the shortest wavelength line in this series, \(\frac{1}{\lambda_{\mathrm{Bs}}}=R\left(\frac{1}{4}\right)\) as m = ∞
[Note: Johannes Rydberg (1854—1919), Swedish spectroscopist, studied atomic emission spectra and introduced the idea of wave number. The empirical formula \(\bar{v}=\frac{1}{\lambda}=R\left(\frac{1}{n^{2}}-\frac{1}{m^{2}}\right)\) where m and n are simple integers, is due to Rydberg. When we consider the finite mass of the nucleus, we find that R varies slightly from element to element.]

Question 5.
Determine the maximum angular speed of an electron moving in a stable orbit around the nucleus of hydrogen atom.
Answer:
The radius of the ,ith Bohr orbit is
r = \(\frac{\varepsilon_{0} h^{2} n^{2}}{\pi m Z e^{2}}\) ………….. (1)
and the linear speed of an electron in this orbit is
ν = \(\frac{Z e^{2}}{2 \varepsilon_{0} n h}\) …………… (2)
where ε ₀ permittivity of free space, h ≡ Planck’s constant, n ≡ principal quantum number, m ≡ electron mass, e electronic charge and Z ≡ atomic number of the atom.
Since angular speed ω = \(\frac{v}{r}\), then from Eqs. (1) and (2), we get,
ω = \(\frac{v}{r}=\frac{Z e^{2}}{2 \varepsilon_{0} n h} \cdot \frac{\pi m Z e^{2}}{\varepsilon_{0} h^{2} n^{2}}=\frac{\pi m Z^{2} e^{4}}{2 \varepsilon_{0}^{2} h^{3} n^{3}}\) ………………. (3)
which gives the required expression for the angular speed of an electron in the nth Bohr orbit.
From Eq. (3), the frequency of revolution of the electron,
f = \(\frac{\omega}{2 \pi}=\frac{1}{2 \pi} \times \frac{\pi m Z^{2} e^{4}}{2 \varepsilon_{0}^{2} h^{3} n^{3}}=\frac{m Z^{2} e^{4}}{4 \varepsilon_{0}^{2} h^{3} n^{3}}\) …………….. (4)
as required.
[Note : From Eq. (4), the period of revolution of the electron, T = \(\frac{1}{f}=\frac{4 \varepsilon_{0}^{2} h^{3} n^{3}}{m Z e^{4}}\). Hence, f ∝ \(\frac{1}{n^{3}}\) and T ∝ n³].

Obtain the formula for ω and continue as follows :

This is required quantity.

Question 6.
Determine the series limit of Balmer, Paschen and Bracket series, given the limit for Lyman series is 912 Å.
Answer:
Data : λ_L∞ = 912 Å

as n = 5 and m = ∞
From Eqs. (1) and (2), we get,
\(\frac{\lambda_{\mathrm{Pa} \infty}}{\lambda_{\mathrm{L} \infty}}=\frac{R_{\mathrm{H}}}{R_{\mathrm{H}} / 9}\) = 9
∴ λ_Pa∞ = 9λ_L∞ = (9) (912) = 8202 Å
\(\frac{\lambda_{\mathrm{Pf} \infty}}{\lambda_{\mathrm{L} \infty}}=\frac{R_{\mathrm{H}}}{R_{\mathrm{H}} / 25}\) = 25
∴ λ_Pf∞ = 25λ_L∞ = (25) (912) = 22800 Å
This is the series limit of the pfund series.

Question 7.
Describe alpha, beta and gamma decays and write down the formulae for the energies generated in each of these decays.
Answer:
(a) A radioactive transformation in which an α-particle is emitted is called α-decay.
In an α-decay, the atomic number of the nucleus decreases by 2 and the mass number decreases by 4.
Example : \({ }_{92}^{238} \mathrm{U} \rightarrow{ }_{90}^{234} \mathrm{Th}+{ }_{2}^{4} \alpha\)
Q = [m_u – m_Th – m_α]c²

(b) A radioactive transformation in which a β-particle is emitted is called β-decay.
In a β^–-decay, the atomic number of the nucleus increases by 1 and the mass number remains unchanged.
Example : \({ }_{90}^{234} \mathrm{Th} \rightarrow{ }_{91}^{234} \mathrm{~Pa}+{ }_{-1}^{0} e+\bar{v}_{\mathrm{e}}\)
where \(\bar{v}_{\mathrm{e}}\) is the neutrino emitted to conserve the momentum, energy and spin.
Q = [m_u – m_Th – m_α]c²
In a β⁺-decay, the atomic number of the nucleus decreases by 1 and the mass number remains unchanged.
Example : \(\begin{aligned}
&30 \\
&15
\end{aligned} P \rightarrow{ }_{14}^{30} \mathrm{Si}+{ }_{+1}^{0} e+v_{\mathrm{e}}\)
where v_e is the neutrino emitted to conserve the momentum, energy and spin.
Q = [m_P – m_Si – m_e]c²
[Note : The term fi particle refers to the electron (or positron) emitted by a nucleus.]

A given nucleus does not emit α and β-particles simultaneously. However, on emission of α or β-particles, most nuclei are left in an excited state. A nucleus in an excited state emits a γ-ray photon in a transition to the lower energy state. Hence, α and β-particle emissions are often accompanied by γ-rays.

Question 8.
Explain what are nuclear fission and fusion giving an example of each. Write down the formulae for energy generated in each of these processes.
Answer:
Nuclear fission is a nuclear reaction in which a heavy nucleus of an atom, such as that of uranium, splits into two or more fragments of comparable size, either spontaneously or as a result of bombardment of a neutron on the nucleus (induced fission). It is followed by emission of two or three neutrons.
The mass of the original nucleus is more than the sum of the masses of the fragments. This mass difference is released as energy, which can be enormous as in the fission of ²³⁵U.
Nuclear fission was discovered by Lise Meitner, Otto Frisch, Otto Hahn and Fritz Strassmann in 1938.

The products of the fission of ²³⁵U by thermal neutrons are not unique. A variety of fission fragments are produced with mass number A ranging from about 72 to about 138, subject to the conservation of mass-energy, momentum, number of protons (Z) and number of neutrons (N). A few typical fission equations are

A type of nuclear reaction in which lighter atomic nuclei (of low atomic number) fuse to form a heavier nucleus (of higher atomic number) with the’ release of enormous amount of energy is called nuclear fusion.

Very high temperatures, of about 107 K to 108 K, are required to carry out nuclear fusion. Hence, such a reaction is also called a thermonuclear reaction.

Example : The D-T reaction, being used in experimental fusion reactors, fuses a deuteron and a triton nuclei at temperatures of about 10⁸ K.

(2) The value of the energy released in the fusion of two deuteron nuclei and the temperature at which the reaction occurs mentioned in the textbook are probably misprints.]

Question 9.
Describe the principles of a nuclear reactor. What is the difference between a nuclear reactor and a nuclear bomb?
Answer:
In a nuclear reactor fuel rods are used to provide a suitable fissionable material such as \(\begin{gathered}
236 \\
92
\end{gathered}\)U. Control rods are used to start or stop the reactor. Moderators are used to slow down the fast neutrons ejected in a nuclear fission to the appropriate lower speeds. Material used as a coolant removes the energy released in the nuclear reaction by converting it into thermal energy for production of electricity.

In a nuclear reactor, a nuclear fission chain reaction is used in a controlled manner, while in a nuclear bomb, the nuclear fission chain reaction is not controlled, releasing tremendous energy in a very short time interval.
[Note : The first nuclear bomb (atomic bomb) was dropped on Hiroshima in Japan on 06 August 1945. The second bomb was dropped on Nagasaki in Japan on 9 August 1945.]

Question 10.
Calculate the binding energy of an alpha particle given its mass to be 4.00151 u.
Answer:
Data : M = 4.00151 u, = 1.00728 u,
m_n = 1.00866 u, 1 u = 931.5 MeV/c²
The binding energy of an alpha particle
(Zm_p + N_n -M)c²
=(2m_p + 2m_n -M)c²
= [(2)(1.00728u) + 2(1.00866 u) – 4.00151 u]c²
= (2.01456 + 2.01732 – 4.00151)(931.5) MeV
= 28.289655 MeV
= 28.289655 × 10⁶ eV × 1.602 × 10^-19 J
= 4.532002731 × 10^-12 J

Question 11.
An electron in hydrogen atom stays in its second orbit for 10^-8 s. How many revolutions will it make around the nucleus in that time?
Answer:
Data : z = 1, m = 9.1 × 10^-31 kg, e = 1.6 × 10^-19 C, ε₀ = 8.85 × 10^-12 C² / N.m², h = 6.63 × 10 ^-34 J.s, n = 2, t = 10^-8 s
The periodic time of the electron in a hydrogen atom,

Let N be the number of revolutions made by the electron in time t. Then, t = NT.
∴ N = \(\frac{t}{T}=\frac{10^{-8}}{3.898 \times 10^{-16}}\) = 2.565 × ⁷

Question 12.
Determine the binding energy per nucleon of the americium isotope \(_{95}^{244} \mathrm{Am}\) , given the mass of \({ }_{95}^{244} \mathrm{Am}\) to be 244.06428 u.
Answer:
Data : Z = 95, N = 244  – 95 = 149,
m_p = 1.00728 u, m_n = 1.00866 u,
M = 244.06428 u, 1 u = 931.5 MeV/c²
The binding energy per nucleon,

= 7.3209 MeV/nucleon

Question 13.
Calculate the energy released in the nuclear reaction \({ }_{3}^{7} \mathrm{Li}\) + p → 2α given mass of \({ }_{3}^{7} \mathrm{Li}\) atom and of helium atom to be 7.016 u and 4.0026 u respectively.
Answer:
Data: M₁ (\({ }_{3}^{7} \mathrm{Li}\) Li atom)= 7.016 u, M₂ (He atom)
= 4.0026 u, m_p = 1.00728 u, 1 u = 931.5 MeV/c²
∆M = M₁ + m_p – 2M₂
= [7.016 + 1.00728 – 2(4.0026)]u
= 0.01808 u = (0.01808)(931.5) MeV/c²
= 16.84152 MeV/c²
Therefore, the energy released in the nuclear reaction = (∆M) c² = 16.84152 MeV

Question 14.
Complete the following equations describing nuclear decays.

Answer:
(a) \({ }_{88}^{226} \mathrm{Ra} \rightarrow{ }_{2}^{4} \alpha+{ }_{86}^{222} \mathrm{Em}\)
Em (Emanation) ≡ Rn (Radon)
Here, α particle is emitted and radon is formed.

(b) \({ }_{8}^{19} \mathrm{O} \rightarrow e^{-}+{ }_{9}^{19} \mathrm{~F}\)
Here, e^– ≡ \({ }_{-1}^{0} \beta\) is emitted and fluorine is formed.

(c) \(\underset{90}{228} \mathrm{Th} \rightarrow{ }_{2}^{4} \alpha+{ }_{88}^{224} \mathrm{Ra}\)
Here, α particle is emitted and radium is formed.

(d) \({ }_{7}^{12} \mathrm{~N} \rightarrow{ }_{6}^{12} \mathrm{C}+{ }_{1}^{0} \beta\)
\({ }_{1}^{0} \beta\) is e⁺ (positron)
Here, β⁺ is emItted and carbon is formed.

Question 15.
Calculate the energy released in the following reactions, given the masses to be \({ }_{88}^{223} \mathrm{Ra}\) : 223.0185 u, \({ }_{82}^{209} \mathrm{~Pb}\) : 208.9811, \({ }_{6}^{14} C\) : 14.00324, \({ }_{92}^{236} \mathrm{U}\) : 236.0456, \({ }_{56}^{140} \mathrm{Ba}\) : 139.9106, \({ }_{36}^{94} \mathrm{Kr}\) : 93.9341, \({ }_{6}^{11} \mathrm{C}\) : 11.01143, \({ }_{5}^{11} \mathrm{~B}\) : 11.0093. Ignore neutrino energy.

Answer:

(a) \({ }_{88}^{223} \mathrm{Ra} \rightarrow{ }_{82}^{209} \mathrm{~Pb}+{ }_{6}^{14} \mathrm{C}\)
The energy released in this reaction = (∆M) c²
= [223.0185 – (208.9811 + 14.00324)j(931 .5) MeV
= 31.820004 MeV

(b) \({ }_{92}^{236} \mathrm{U} \rightarrow{ }_{56}^{140} \mathrm{Ba}+{ }_{36}^{94} \mathrm{Kr}+2 \mathrm{n}\)
The energy released in this reaction =
(∆M) c² = [236.0456 – (139.9106 + 93.9341 + (2)(1 .00866)1(93 1 .5)MeV
= 171.00477 MeV

(c) \({ }_{6}^{11} \cdot \mathrm{C} \rightarrow{ }_{5}^{11} \mathrm{~B}+e^{+}\) + neutrino
The energy released in this reaction = (∆M) c²
= [11.01143 – (11.0093 + O.00055)](931.5) MeV
= 1.47177 MeV

Question 16.
Sample of carbon obtained from any living organism has a decay rate of 15.3 decays per gram per minute. A sample of carbon obtained from very old charcoal shows a disintegration rate of 12.3 disintegrations per gram per minute. Determine the age of the old sample given the decay constant of carbon to be 3.839 × 10^-12 per second.
Answer:
Data: 15.3 decays per gram per minute (living organism), 12.3 disintegrations per gram per minute (very old charcoal). Hence, we have,
\(\frac{A(t)}{A_{0}}=\frac{12.3}{15.3}\), λ = 3.839 × 10^-12 per second

Question 17.
The half-life of \({ }_{38}^{90} \mathrm{Sr}\) is 28 years. Determine the disintegration rate of its 5 mg sample.
Answer:
Data: T_1/2 = 28 years = 28 × 3.156 × 10⁷ s
=8.837 × 10⁸s, M = 5 mg =5 × 10^-3g
90 grams of \({ }_{38}^{90} \mathrm{Sr}\) contain 6.02 × 10²³ atoms
Hence, here, N = \(\frac{\left(6.02 \times 10^{23}\right)\left(5 \times 10^{-3}\right)}{90}\)
= 3.344 × 10¹⁹ atoms
∴ The disintegration rate = Nλ = N\(\frac{0.693}{T_{1 / 2}}\)
= \(\frac{\left(3.344 \times 10^{19}\right)(0.693)}{8.837 \times 10^{8}}\)
= 2.622 × 10¹⁰ disintegrations per second

Question 18.
What is the amount of \({ }_{27}^{60} \mathrm{Co}\) necessary to provide a radioactive source of strength 10.0 mCi, its half-life being 5.3 years?
Answer:
Data : Activity = 10.0 mCi = 10.0 × 10^-3 Ci
= (10.0 × 10^-3)(3.7 × 10¹⁰) dis/s = 3.7 × 10⁸ dis/s
T_1/2 = 5.3 years = (5.3)(3.156 × 10⁷) s
= 1.673 × 10⁸ s
Decay constant, λ = \(\frac{0.693}{T_{1 / 2}}=\frac{0.693}{1.673 \times 10^{8}} \mathrm{~s}^{-1}\)
=4.142 × 10^-9 s^-1
Activity = Nλ
∴ N = \(\frac{\text { activity }}{\lambda}=\frac{3.7 \times 10^{8}}{4.142 \times 10^{-9}} \text { atoms }\)
= 8.933 × 10¹⁶ atoms
=60 grams of \({ }_{27}^{60} \mathrm{Co}\) contain 6.02 × 10²³ atoms
Mass of 8.933 × 10¹⁶ atoms of \({ }_{27}^{60} \mathrm{Co}\)
= \(\frac{8.933 \times 10^{16}}{6.02 \times 10^{23}} \times 60 \mathrm{~g}\)
= 8.903 × 10^-6 g = 8.903 µg
This is the required amount.

Question 19.
Disintegration rate of a sample is 10¹⁰ per hour at 20 hrs from the start. It reduces to 6.3 × 10⁹ per hour after 30 hours. Calculate its half life and the initial number of radioactive atoms in the sample.
Answer:
Data : A (t₁) = 10¹⁰ per hour, where t₁ = 20 h,
A (t₂) = 6.3 × 10⁹ per hour, where t₂ = 30 h
A(t) = A₀e^-λt ∴ A(t₁) = A₀e^-λt₁ and A(t₂) = A_oe^-λt₂

∴ 1.587 e^10λ ∴ 10λ =2.3031og₁₀(1.587)
∴ λ = (0.2303)(0.2007) = 0.04622 per hour
The half life of the material, T_1/2 = \(\frac{0.693}{\lambda}=\frac{0.693}{0.04622}\)
= 14.99 hours
Now, A₀ = A (t₁)e^λt₁ = 10¹⁰e^{(0.04622)(20)}
= 10¹⁰ e^0.9244
Let x = e^0.9244 ∴ 2.3031og₁₀x = 0.9244
∴ 1og₁₀x = \(\frac{0.9244}{2.303}\) = 0.4014
∴ x = antilog 0.4014 = 2.52
∴ A₀ = 2.52 × 10¹⁰ per hour
Now A₀ = N₀λ ∴ N₀ = \(\frac{A_{0}}{\lambda}=\frac{2.52 \times 10^{10}}{0.04622}\)
= 5.452 × 10¹¹
This is the initial number of radioactive atoms in the sample.

Question 20.
The isotope ⁵⁷Co decays by electron capture to ⁵⁷Fe with a half-life of 272 d. The ⁵⁷Fe nucleus is produced in an excited state, and it almost instantaneously emits gamma rays.
(a) Find the mean lifetime and decay constant for ⁵⁷Co.
(b) If the activity of a radiation source ⁵⁷Co is 2.0 µCi now, how many ⁵⁷Co nuclei does the source contain?
(c) What will be the activity after one year?
Answer:
Data: T_1/2 = 272d = 272 × 24 × 60 × 60s = 2.35 × 10⁷ s,
A₀ = 2.0uCi = 2.0 × 10^-6 × 3.7 × 10¹⁰
= 7.4 × 10⁴ dis/s
t = 1 year = 3.156 × 10⁷ s
(a) T_1/2 = \(\frac{0.693}{\lambda}\) = 0.693 τ ∴ The mean lifetime for
⁵⁷Co = τ = \(\frac{T_{1 / 2}}{0.693}=\frac{2.35 \times 10^{7}}{0.693}\) = 3391 × 10⁷ s
The decay constant for ⁵⁷Co = λ = \(\frac{1}{\tau}\)
= \(\frac{1}{3.391 \times 10^{7} \mathrm{~s}}\)
= 2949 × 10^-8 s^-1

(b)A₀ = N₀A ∴ N₀ = \(\frac{A_{0}}{\lambda}\) = A₀τ
= (7.4 × 10⁴)(3.391 × 10⁷)
= 2.509 × 10¹² nuclei
This is the required number.

(c) A(t) = A₀e^-λt = 2e^{-(2.949 × 10-8)(3.156 × 107)}
= 2e^-0.9307 = 2 / e^0.9307
Let x = e^0.9307 ∴ Iog_ex = 0.9307
∴ 2.303log₁₀x = 0.9307
∴ log₁₀x = \(\frac{0.9307}{2.303}\) = 0.4041
∴ x = antilog 0.4041 = 2.536
∴ A (t) = \(\frac{2}{2.536}\) μCi = 0.7886 μCi

Question 21.
A source contains two species of phosphorous nuclei, \({ }_{15}^{32} \mathrm{P}\) (T_1/2 = 14.3 d) and \({ }_{15}^{32} \mathrm{P}\) (T_1/2 = 25.3 d). At time t = 0, 90% of the decays are from \({ }_{15}^{32} \mathrm{P}\) . How much time has to elapse for only 15% of the decays to be from \({ }_{15}^{32} \mathrm{P}\) ?
Answer:

∴ (0.04846 – 0.02739) t = 2.303 (2.1847 – 0.4771)
∴ t = \(\frac{(2.303)(1.7076)}{0.02107}\) = 186.6 days
This is the required time.

Question 22.
Before the year 1900 the activity per unit mass of atmospheric carbon due to the presence of ¹⁴C averaged about 0.255 Bq per gram of carbon. (a) What fraction of carbon atoms were ¹⁴C? (b) An archaeological specimen containing 500 mg of carbon, shows 174 decays in one hour. What is the age of the specimen, assuming that its activity per unit mass of carbon when the specimen died was equal to the average value of the air? Half-life of 14C is 5730 years?
Answer:
0.693
Data: T_1/2 = 5730y ∴ λ = \(\frac{0.693}{5730 \times 3.156 \times 10^{7}} \mathrm{~s}^{-1}\)
= 3.832 × 10^-12 s^-1, A = 0.255 Bq per gram of carbon in part (a); M = 500 mg = 500 × 10^-3 g,
174 decays in one hour \(\frac{174}{3600}\) dis/s = 0.04833 dis/s in part (b) (per 500 mg].

(a) A = Nλ ∴ N = \(\frac{A}{\lambda}=\frac{0.255}{3.832 \times 10^{-12}}\)
= 6.654 × 10¹⁰
Number of atoms in 1 g of carbon = \(\frac{6.02 \times 10^{23}}{12}\)
=5.017 × 10²²
\(\frac{5.017 \times 10^{22}}{6.654 \times 10^{10}}\) = 0.7539 × 10¹²
∴ 1 ¹⁴C atom per 0.7539 × 10¹² atoms of carbon
∴ 4 ¹⁴C atoms per 3 × 10¹² atoms of carbon

(b) Present activity per gram = \(\)
= 0.09666 dis/s per gram
A₀ = 0.255 dis/s per gram
Now, A(t) = A₀e^-λt

This is the required quantity.

Question 23.
How much mass of ²³⁵U is required to undergo fission each day to provide 3000 MW of thermal power? Average energy per fission is 202.79 MeV
Answer:
Data: Power = 3000 MW = 3 × 10⁹ J/s
∴ Energy to be produced each day
=3 × 10⁹ × 86400 J each day
= 2.592 × 10¹⁴ J each day
Energy per fission = 202.79 MeV
= 202.79 × 10⁶ × 1.6 × 10^-19 J = 3,245 × 10^-11 J
∴ Number of fissions each day
= \(\frac{2.592 \times 10^{14}}{3.245 \times 10^{-11}}\) × 10²⁴ each day
0.235 kg of ²³⁵U contains 6.02 × 10²³ atoms
7988 x 1024
∴ M = \(\left(\frac{7.988 \times 10^{24}}{6.02 \times 10^{23}}\right)\) (o.235) = 3.118 kg
This is the required quantity.

Question 24.
In a periodic table the average atomic mass of magnesium is given as 24.312 u. The average value is based on their relative natural abundance on earth. The three isotopes and their masses are \({ }_{12}^{24} \mathrm{Mg}\) (23.98504 u), \({ }_{12}^{25} \mathrm{Mg}\) (24.98584 u) and \({ }_{12}^{26} \mathrm{Mg}\) (25.98259 u). The natural abundance of \({ }_{12}^{24} \mathrm{Mg}\) is 78.99% by mass. Calculate the abundances of other two isotopes.
[Answer: 9.3% and 11.7%]
Answer:
Data : Average atomic mass of magnesium =

12th Physics Digest Chapter 15 Structure of Atoms and Nuclei Intext Questions and Answers

Use your brainpower (Textbook Page No. 336)

Question 1.
Why don’t heavy nuclei decay by emitting a single proton or a single neutron?
Answer:
According to quantum mechanics, the probability for these emissions is extremely low.

12th Std Physics Questions And Answers:

• Current Electricity Class 12 Physics Questions And Answers
• Magnetic Fields due to Electric Current Class 12 Physics Questions And Answers
• Magnetic Materials Class 12 Physics Questions And Answers
• Electromagnetic Induction Class 12 Physics Questions And Answers
• AC Circuits Class 12 Physics Questions And Answers
• Dual Nature of Radiation and Matter Class 12 Physics Questions And Answers
• Structure of Atoms and Nuclei Class 12 Physics Questions And Answers
• Semiconductor Devices Class 12 Physics Questions And Answers

Balbharti Maharashtra State Board 12th Physics Textbook Solutions Chapter 14 Dual Nature of Radiation and Matter Textbook Exercise Questions and Answers.

Dual Nature of Radiation and Matter Class 12 Exercise Question Answers Solutions Maharashtra Board

Class 12 Physics Chapter 14 Exercise Solutions Maharashtra Board

Physics Class 12 Chapter 14 Exercise Solutions

1. Choose the correct answer.

i) A photocell is used to automatically switch on the street lights in the evening when the sunlight is low in intensity. Thus it has to work with visible light. The material of the cathode of the photocell is
(A) zinc
(B) aluminum
(C) nickel
(D) potassium
Answer:
(D) potassium

ii) Polychromatic (containing many different frequencies) radiation is used in an experiment on the photoelectric effect. The stopping potential
(A) will depend on the average wavelength
(B) will depend on the longest wavelength
(C) will depend on the shortest wavelength
(D) does not depend on the wavelength
Answer:
(C) will depend on the shortest wavelength

iii) An electron, a proton, an α-particle and a hydrogen atom are moving with the same kinetic energy. The associated de Broglie wavelength will be longest for
(A) electron
(B) proton
(C) α-particle
(D) hydrogen atom
Answer:
(A) electron

iv) If N_Red and N_Blue are the number of photons emitted by the respective sources of equal power and equal dimensions in unit time, then
(A) N_Red < N_Blue
(B) N_Red = N_Blue
(C) N_Red > N_Blue
(D) N_Red ≈ N_Blue
Answer:
(C) N_Red > N_Blue

v) The equation E = pc is valid
(A) for all sub-atomic particles
(B) is valid for an electron but not for a photon
(C) is valid for a photon but not for an electron
(D) is valid for both an electron and a photon
Answer:
(C) is valid for a photon but not for an electron

2. Answer in brief.

i) What is photoelectric effect?
Answer:
The phenomenon of emission of electrons from a metal surface when electromagnetic radiation of appropriate frequency is incident on it is known as photoelectric effect.

ii) Can microwaves be used in the experiment on photoelectric effect?
Answer:
No

iii) Is it always possible to see photoelectric effect with red light?
Answer:
No

iv) Using the values of work function given in Table 14.1, tell which metal will require the highest frequency of incident radiation to generate photocurrent.

Answer:
Gold.
[ Note : W₀ = hv₀, where h is Planck’s constant. The larger the work function (W₀), the higher is the threshold frequency (v₀). ]

v) What do you understand by the term wave-particle duality? Where does it apply?
Answer:
Depending upon experimental conditions or structure of matter, electromagnetic radiation and material particles exhibit wave nature or particle nature. This is known as wave-particle duality.

It applies to all phenomena. The wave nature and particle nature are liked by the de Broglie relation λ = h/p, where λ is the wavelength of matter waves, also called de Broglie waves / Schrodinger waves, p is the magnitude of the momentum of a particle or quantum of radiation and h is the universal constant called Planck’s constant.

[Note : It is the smallness of h (= 6.63 × 10^-34 J∙s) that is very significant in wave-particle duality.]

Question 3.
Explain the inverse linear dependence of stopping potential on the incident wavelength in a photoelectric effect experiment.
Answer:
We have V₀e = \(\frac{h c}{\lambda}\) – Φ, where V₀ is the stopping potential, e is the magnitude of the charge on the electron, h is Planck’s constant, c is the speed of light in free space, λ is the wavelength of the electromagnetic radiation incident on a metal surface and Φ is the work function for the metal, h, c and e are constants. Φ is constant for a particular metal.
Hence, it follows that as \(\frac{1}{\lambda}\) increases, V₀ increases.
The plot of V₀ verses \(\frac{1}{\lambda}\) is linear. This is because the energy associated with a quantum of radiation (photon) is directly proportional to the frequency of radiation and hence inversely proportional to the wavelength of radiation.

Question 4.
It is observed in an experiment on photoelectric effect that an increase in the intensity of the incident radiation does not change the maximum kinetic energy of the electrons. Where does the extra energy of the incident radiation go? Is it lost? State your answer with explanatory reasoning.
Answer:
When electromagnetic radiation with frequency greater than the threshold frequency is incident on a metal surface, there is emission of electrons. It is observed that not every incident photon is effective in liberating an electron. In fact, the number of electrons emitted per second is far less than the number of photons incident per second. The photons that are not effective in liberation of electrons are reflected (or scattered) or absorbed resulting in rise in the temperature of the metal surface. The maximum kinetic energy of a photoelectron depends on the frequency of the incident radiation and the threshold frequency for the metal. It has nothing to do with the intensity of the incident radiation. The increase in intensity results in increase in the number of electrons emitted per second.

Question 5.
Explain what do you understand by the de Broglie wavelength of an electron. Will an electron at rest have an associated de Broglie wavelength? Justify your answer.
Answer:
Under certain conditions an electron exhibits wave nature. Waves associated with a moving electron are called matter waves or de Broglie waves or- Schrodinger waves. The de Broglie wavelength of these matter waves is given by X = h/p, where h is Planck’s constant and p is the magnitude of the momentum of the electron.

If an electron is at rest, its momentum would be zero, and hence the corresponding de Broglie wavelength would be infinite indicating absence of a matter wave. However, according to quantum mechanics/wave mechanics, this is not possible.

Question 6.
State the importance of Davisson and Germer experiment.
Answer:
The Davisson and Germer experiment directly indicated the wave nature of material particles and quantitatively verified the de Broglie hypothesis for the existence of matter waves.

[Note : The aim of the experiment was not to verify wave like properties of electrons. The realisation came only later, an example of serendipity.]

[Note : Like X-rays, electrons exhibit wave nature under suitable conditions. When the wavelength of matter waves associated with moving electrons is comparable to the inter-atomic spacing in a crystal, electrons show diffraction effects. In 1927, Sir George Thomson (1892 – 1975), British physicist, with his student Alex Reid, observed electron diffraction with a metal foil. It is found that neutrons, atoms, molecules, Œ-particles, etc. show wave nature under suitable conditions.]

Question 7.
What will be the energy of each photon in monochromatic light of frequency 5 × 10¹⁴ Hz?
Answer:
Data: y = 5 × 10¹⁴ Hz, h = 6.63 × 10^-34 Js,
1eV=1.6 × 10^-19 J
The energy of each photon,
E = hv = (6.63 × 10^-34 J.s)(5 × 10¹⁴ Hz)
= 3.315 × 10^-19 J
= \(\frac{3.315 \times 10^{-19} \mathrm{~J}}{1.6 \times 10^{-19} \mathrm{~J} / \mathrm{eV}}\) = 2.072 eV

Question 8.
Observations from an experiment on photoelectric effect for the stopping potential by varying the incident frequency were plotted. The slope of the linear curve was found to be approximately 4.1 × 10^-15 V s. Given that the charge of an electron is 1.6 × 10^-19 C, find the value of the Planck’s constant h.
Answer:
Data : Slope=4.1 × 10^-15 V∙s, e = 1.6 ×10^-19 C
V₀e = hv – hv₀
∴ V₀ =\(\left(\frac{h}{e}\right) v-\left(\frac{h v_{0}}{e}\right)\)
∴ Slope = \(\frac{h}{e}\) ∴ Planck’s constant,
h = (slope) (e)=(4.1 × 10^-15 V∙s)(1.6 × 10^-19 C)
= 6.56 × 10 34J. (as 1 V = \(\frac{1 \mathrm{~J}}{1 \mathrm{C}}\))

Question 9.
The threshold wavelength of tungsten is 2.76 × 10^-5 cm. (a) Explain why no photoelectrons are emitted when the wavelength is more than 2.76 × 10^-5 cm.(b) What will be the maximum kinetic energy of electrons ejected in each of the following cases
(i) if ultraviolet radiation of wavelength λ = 1.80 × 10^-5 cm and
(ii) radiation of frequency 4 × 10¹⁵ Hz is made incident on the tungsten surface.
Answer:
Data: λ₀ = 2.76 × 10^-5 cm = 2.76 × 10^-7 m,
λ =1.80 × 10^-5 cm = 1.80 × 10^-7 m,
v = 4 × 10¹⁵ Hz, h = 6.63 × 10^-34 J∙s,c = 3 × 10⁸ m/s
(a) For λ > λ₀, v < v₀ (threshold frequency).
∴ hv < hv₀. Hence, no photoelectrons are emitted.

(b) Maximum kinetic energy of electrons ejected
= hc\(\left(\frac{1}{\lambda}-\frac{1}{\lambda_{0}}\right)\)
=(6.63 × 10^-34)(3 × 10⁸)\(\left(\frac{10^{7}}{1.8}-\frac{10^{7}}{2.76}\right)\)J
= (6.63 × 10^-19)(0.5555 – 0.3623)
= (6.63)(0.1932 × 10^-19)J = 1.281 × 10^-19 J
= \(\frac{1.281 \times 10^{-19} \mathrm{~J}}{1.6 \times 10^{-19} \mathrm{~J} / \mathrm{eV}}\) = 0.8006 eV

(c) Maximum kinetic energy of electrons ejected
= hv – \(\frac{h c}{\lambda_{0}}\)
=(6.63 × 10^-34(4 × 10¹⁵) – \(\frac{\left(6.63 \times 10^{-34}\right)\left(3 \times 10^{8}\right)}{2.76 \times 10^{-7}}\)
= 26.52 × 10^-19 – 7.207 × 10^-19
= 19.313 × 10^-19 J
= \(\frac{19.313 \times 10^{-19} \mathrm{~J}}{1.6 \times 10^{-19} \mathrm{~J} / \mathrm{eV}}\) = 12.07eV

Question 10.
Photocurrent recorded in the micro ammeter in an experimental set-up of photoelectric effect vanishes when the retarding potential is more than 0.8 V if the wavelength of incident radiation is 4950 Å. If the source of incident radiation is changed, the stopping potential turns out to be 1.2 V. Find the work function of the cathode material and the wavelength of the second source.
Answer:
Data: V₀ = 0.8 V, λ = 4950 Å = 4.950 × 10^-7 m,
V₀‘ = 1.2V, h = 6.63 × 10^-34 J∙s, c = 3 × 10⁸ m/s.
(i) V₀e = hv – Φ = \(\frac{h c}{\lambda}\) – Φ
∴ The work function of the cathode material,

Question 11.
Radiation of wavelength 4500 Å is incident on a metal having work function 2.0 eV. Due to the presence of a magnetic field B, the most energetic photoelectrons emitted in a direction perpendicular to the field move along a circular path of radius 20 cm. What is the value of the magnetic field B?
Answer:
Data: λ = 4500Å = 4.5 × 10^-7 m,
Φ = 2.0eV = 2 × 1.6 × 10^-19 J = 3.2 × 10^-19 J,
h = 6.63 × 10^-34 J∙s, c = 3 × 10⁸ m/s,
r = 20 cm = 0.2 m, e= 1.6 × 10^-19 C,
m = 9.1 × 10^-31kg

This is the value of the magnetic field.

Question 12.
Given the following data for incident wavelength and the stopping potential obtained from an experiment on photoelectric effect, estimate the value of Planck’s constant and the work function of the cathode material. What is the threshold frequency and corresponding wavelength? What is the most likely metal used for emitter?

Answer:
Data: λ = 2536Å = 2.536 × 10^-7 m,
λ’ = 3650Å = 3.650 ×10^-7 m, V₀ = 1.95V, V₀‘ = 0.5V,
c = 3 × 10⁸ mIs, e = 1.6 × 10^-19 C

(i) V₀e = \(\frac{h c}{\lambda}\) – Φ and V₀‘e =\(\frac{h c}{\lambda^{\prime}}\) – Φ
∴ (V₀ – V₀‘)e = hc\(\left(\frac{1}{\lambda}-\frac{1}{\lambda^{\prime}}\right)\)
∴ (1.95 – 0.5(1.6 × 10^-19)
= h (3 × 10⁸\(\left(\frac{10^{7}}{2.536}-\frac{10^{7}}{3.650}\right)\)
∴ 2.32 × 10^-19 = h(3 × 10¹⁵)(0.3943 – 0.2740)
∴ h = \(\frac{2.32 \times 10^{-34}}{0.3609}\) = 6.428 × 10^-34 J∙s
This is the value of Planck’s constant.

(ii) Φ = \(\frac{h c}{\lambda}\) – V₀e

This is the work function of the cathode material.

(iii) Φ = hv₀
∴ The threshold frequency, v₀ = \(\frac{\phi}{h}\)
= \(\frac{4.484 \times 10^{-19} \mathrm{~J}}{6.428 \times 10^{-34} \mathrm{~J} \cdot \mathrm{s}}\) = 6.976 × 10¹⁴ Hz

(iv) v₀ = \(\frac{c}{\lambda_{\mathrm{o}}}\) ∴ The threshold frequency, λ₀ = \(\frac{c}{v_{\mathrm{o}}}\)
= \(\frac{3 \times 10^{8}}{6.976 \times 10^{14}}\) = 4.300 × 10^-7 m = 4300 Å

(v) The most likely metal used for emitter : calcium

Question 13.
Calculate the wavelength associated with an electron, its momentum and speed
(a) when it is accelerated through a potential of 54 V
Answer:
Data : V = 54 V, m = 9.1 × 10^-31 kg, e
e = 1.6 × 10^-19 C, h = 6.63 × 10^-34 J.s, KE = 150 eV
(a) We assume that the electron is initially at rest.
∴ Ve = \(\frac{1}{2}\)mv²
∴ v = \(\sqrt{\frac{2 V e}{m}}=\sqrt{\frac{2(54)\left(1.6 \times 10^{-19}\right)}{9.1 \times 10^{-31}}}\)
= \(\sqrt{19 \times 10^{12}}\) = 4.359 × 10⁶ m/5
This is the speed of the electron.
p = mv = (9.1× 10^-31)(4.359 × 10⁶)
= 3.967 × 10^-24 kg∙m/s
This is the momentum of the electron. The wavelength associated with the electron,
λ = \(\frac{h}{p}=\frac{6.63 \times 10^{-34}}{3.967 \times 10^{-24}}\) = 1.671 × 10^-10 m
= 1.671 Å = 0.1671 nm

(b) when it is moving with kinetic energy of 150 eV.
Answer:
As KE ∝ \(\sqrt{V}\), we get
\(\frac{v^{\prime}}{v}=\sqrt{\frac{150}{54}}\) = 1.666
∴ v’ = 1.666v = (1.666)(4.356 × 10⁶)
= 7.262 × 10⁶ m/s
This is the speed of the electron.
p’ = mv’’=(9.1 × 10^-31)(7.262 × 10⁶)
= 6.608 × 10^-24 kg∙m/s
This is the momentum of the electron. The
wavelength associated with the electron,
λ = \(\frac{h}{p^{\prime}}=\frac{6.63 \times 10^{-34}}{6.608 \times 10^{-24}} \) = 1.003 × 10^-10 m
= 1.003 Å = 0.1003 nm

Question 14.
The de Broglie wavelengths associated with an electron and a proton are same. What will be the ratio of
(i) their momenta
(ii) their kinetic energies?
Answer:
Data : λ (electron) = λ (proton)
m (proton) = 1836 m (electron)
(i) λ = \(\frac{h}{p}\) As λ (electron) = λ (proton),
\(\frac{p(\text { electron })}{p \text { (proton) }}\) = 1, where p denotes the magnitude of momentum.

(ii) Assuming v «c,
KE = \(\frac{1}{2}\)mv² = \(\frac{1}{2} \frac{m^{2} v^{2}}{m}=\frac{p^{2}}{2 m}\)
∴ \(\frac{\mathrm{KE} \text { (electron) }}{\mathrm{KE} \text { (proton) }}=\frac{m \text { (proton) }}{m \text { (electron) }}\) = 1836 as p is the same for the electron and the proton.

Question 15.
Two particles have the same de Broglie wavelength and one is moving four times as fast as the other. If the slower particle
is an α-particle, what are the possibilities for the other particle?
Answer:
Data : λ₁ = λ₂, v₁ = 4v₂
λ = \(\frac{h}{p}=\frac{h}{m v}\) ∴ λ₁ = \(\frac{h}{m_{1} v_{1}}\), λ₂ = \(\frac{h}{m_{2} v_{2}}\)
∴ m₁ = m₂ \(\frac{v_{2}}{v_{1}}\) = m₂\(\left(\frac{1}{4}\right)=\frac{m_{2}}{4}\)
As particle 2 is the a-particle, particle 1 (having the mass \(\frac{1}{4}\) times that of the a-particle) may be a proton or neutron.

Question 16.
What is the speed of a proton having de Broglie wavelength of 0.08 Å?
Answer:
Data : λ = 0.08 Å = 8 × 10^-12m, h = 6.63 × 10^-34 J∙s, m = 1.672 × 10^-27 kg
λ = \(\frac{h}{m v}\) ∴ v = \(\frac{h}{\lambda m}=\frac{6.63 \times 10^{-34}}{\left(8 \times 10^{-12}\right)\left(1.672 \times 10^{-27}\right)}\)
∴ v = 4.957 × 10⁴ m/s
This is the speed of the proton.

Question 17.
In nuclear reactors, neutrons travel with energies of 5 × 10^-21 J. Find their speed and wavelength.
Answer:
Data : KE = 5 × 10^-21 J, m = 1.675 × 10^-27 kg, h = 6.63 × 10^-34 J∙s
KE = \(\frac{1}{2}\) mv² = 5 × 10^-21 J
∴ v = \(\sqrt{\frac{2 \mathrm{KE}}{m}}=\sqrt{\frac{(2)\left(5 \times 10^{-21}\right)}{1.675 \times 10^{-27}}}\)
= 2.443 × 10³ m/s
This is the speed of the neutrons. The de Broglie wavelength associated with the neutron,
λ = \(\frac{h}{m v}=\frac{6.63 \times 10^{-34}}{\left(1.675 \times 10^{-27}\right)\left(2.443 \times 10^{3}\right)}\)
= 1.620 × 10^-10 m = 1.620 Å

Question 18.
Find the ratio of the de Broglie wavelengths of an electron and a proton when both are moving with the (a) same speed, (b) same energy and (c) same momentum? State which of the two will have the longer wavelength in each case?
Answer:
Data: m_p = 1836 m_e

12th Physics Digest Chapter 14 Dual Nature of Radiation and Matter Intext Questions and Answers

Remember This (Textbook Page No. 316)

Question 1.
Is a solar cell a photocell?
Answer:
Yes

Remember This (Textbook Page No. 317)

Question 1.
Can you estimate the de Broglie wavelength of the Earth?
Answer:
Taking the mass of the Earth as (about) 6 × 10²⁴ kg, and the linear speed of the earth around the Sun as (about) 3 × 10⁴ m/s, we have, the de Brogue wave length of the Earth as
λ = \(\frac{h}{p}=\frac{h}{M v}=\frac{6.63 \times 10^{-34} \mathrm{~J} \cdot \mathrm{s}}{\left(6 \times 10^{24} \mathrm{~kg}\right)\left(3 \times 10^{4} \mathrm{~m} / \mathrm{s}\right)}\)
= 3.683 × 10^-63 m (extremely small)

Question 2.
The expression p = E/c defines the momentum of a photon. Can this expression be used for the momentum of an electron or proton?
Answer:
No

Remember This (Textbook Page No. 319)

Diffraction results described above can be produced in the laboratory using an electron diffraction tube as shown in the figure. It has a filament which on heating produces electrons. This filament acts as a cathode. Electrons are accelerated to quite high speeds by creating large potential difference between the cathode and a positive electrode. On its way, the beam of electrons comes across a thin sheet of

graphite. The electrons are diffracted by the atomic layers in the graphite and form diffraction rings on the phosphor screen. By changing the voltage between the cathode and anode, the energy, and therefore the speed, of the electrons can be changed. This will change the wavelength of the electrons and a change will be seen in the diffraction pattern. By increasing the voltage, the radius of the diffraction rings will decrease. Try to explain why?
Answer:
When the accelerating voltage is increased, the kinetic energy and hence the momentum of the electron increases. This decreases the de Brogue wavelength of the electron. Hence, the radius of the diffraction ring decreases.

Remember This (Textbook Page No. 320)

Question 1.
On what scale or under which circumstances are the wave nature of matter apparent?
Answer:
When the de Brogue wavelength of a particle such as an electron, atom, or molecule is comparable to the interatomic spacing in a crystal, the wave nature of matter is revealed in diffraction/interference.

12th Std Physics Questions And Answers:

• Current Electricity Class 12 Physics Questions And Answers
• Magnetic Fields due to Electric Current Class 12 Physics Questions And Answers
• Magnetic Materials Class 12 Physics Questions And Answers
• Electromagnetic Induction Class 12 Physics Questions And Answers
• AC Circuits Class 12 Physics Questions And Answers
• Dual Nature of Radiation and Matter Class 12 Physics Questions And Answers
• Structure of Atoms and Nuclei Class 12 Physics Questions And Answers
• Semiconductor Devices Class 12 Physics Questions And Answers

Balbharti Maharashtra State Board 12th Physics Textbook Solutions Chapter 13 AC Circuits Textbook Exercise Questions and Answers.

AC Circuits Class 12 Exercise Question Answers Solutions Maharashtra Board

Class 12 Physics Chapter 13 Exercise Solutions Maharashtra Board

Physics Class 12 Chapter 13 Exercise Solutions

1. Choose the correct option.

i) If the RMS current in a 50 Hz AC circuit is 5A, the value of the current \(\frac{1}{300}\) seconds after its value becomes zero is
(A) 5\(\sqrt {2}\) A
(B) 5\(\sqrt{\frac{3}{2}}\) A
(C) \(\frac{5}{6}\) A
(D) \(\frac{5}{\sqrt{2}}\) A
Answer:
(B) 5\(\sqrt{\frac{3}{2}}\) A

ii) A resistor of 500 Ω and an inductance of 0.5 H are in series with an AC source which is given by V = 100 \(\sqrt {2}\) sin (1000t). The power factor of the combination
(A) \(\frac{1}{\sqrt{2}}\)
(B) \(\frac{1}{\sqrt{3}}\)
(C) 0.5
(D) 0.6
Answer:
(A) \(\frac{1}{\sqrt{2}}\)

iii) In a circuit L, C & R are connected in series with an alternating voltage of frequency f. the current leads the voltage by 450. The value of C is

Answer:
(B) \(\frac{1}{2 \pi f(2 \pi f L-R)}\)

iv) In an AC circuit, e and i are given by e = 150 sin (150t) V and i = 150 sin (150 t + \(\frac{\pi}{3}\)) A. the power dissipated in the circuit is
(A) 106W
(B) 150W
(C) 5625W
(D) Zero
Answer:
(C) 5625W

v) In a series LCR circuit the phase difference between the voltage and the current is 45°. Then the power factor will be
(A) 0.607
(B) 0.707
(C) 0.808
(D) 1
Answer:
(B) 0.707

2. Answer in brief.

i) An electric lamp is connected in series with a capacitor and an AC source is glowing with a certain brightness. How does the brightness of the lamp change on increasing the capacitance ?
Answer:
Impedance, Z = \(\sqrt{R^{2}+\frac{1}{\omega^{2} C^{2}}}\), where R is the resistance of the lamp, w is the angular frequency of AC and C is the capacitance of the capacitor connected in series with the AC source and the lamp. When C is increased, \(\) decreases. Hence, Z increases.
Power factor, cos Φ = \(\frac{R}{Z}\)
As Z increases, the power factor decreases.
Now, the average power over one cycle,
P_av = v_rms i_rms cos Φ
= V_rms \(\left(\frac{V_{\mathrm{rms}}}{Z}\right)\) cos Φ
= \(\frac{V_{\mathrm{rms}}^{2}}{\mathrm{Z}} \cos \phi\)
∴ P_av decreases as Z increases and cos Φ decreases.
As the current through the lamp \(\left(\frac{V_{\mathrm{rms}}}{Z}\right)\) decreases, the brightness of the lamp will decrease when C is increased.

ii) The total impedance of a circuit decreases when a capacitor is added in series with L and R. Explain why ?
Answer:
For an LR circuit, the impedance,
Z_LR = \(\sqrt{R^{2}+X_{\mathrm{L}}^{2}}\), where X_L is the reactance of the inductor.
When a capacitor of capacitance C is added in series with L and R, the impedance,
Z_LCR = \(\sqrt{R^{2}+\left(X_{\mathrm{L}}-X_{\mathrm{C}}\right)^{2}}\) because in the case of an inductor the current lags behind the voltage by a phase angle of \(\frac{\pi}{2}\) rad while in the case of a capacitor the current leads the voltage by a phase angle of \(\frac{\pi}{2}\) rad. The decrease in net reactance decreases the total impedance (Z_LCR < Z_LR).

iii) For very high frequency AC supply, a capacitor behaves like a pure conductor. Why ?
Answer:
The reactance of a capacitor is X_C = \(\frac{1}{2 \pi f C}\), where f is the frequency of the AC supply and C is the capacitance of the capacitor. For very high frequency, f, X_C is very small. Hence, for very high frequency AC supply, a capacitor behaves like a pure conductor.

iv) What is wattless current ?
Answer:
The current that does not lead to energy consumption, hence zero power consumption, is called wattless current.
In the case of a purely inductive circuit or a purely capacitive circuit, average power consumed over a complete cycle is zero and hence the corresponding alternating current in the circuit is called wattless current.
[Note : In this case, the power factor is zero.]

v) What is the natural frequency of L C circuit ? What is the reactance of this circuit at this frequency
Answer:
The natural frequency of LC circuit is \(\frac{1}{2 \pi \sqrt{L C}}\) ,
where L is the inductance and C is the capacitance. The reactance of this circuit at this frequency is

Question 3.
In a series LR circuit X_L = R and power factor of the circuit is P₁. When capacitor with capacitance C such that X_L = X_C is put in series, the power factor becomes P₂. Calculate P₁ / P₂ .
Answer:
For a series LR circuit, power factor,

Question 4.
When an AC source is connected to an ideal inductor show that the average power supplied by the source over a complete cycle is zero.
Answer:
In an AC circuit containing only an ideal inductor, the current i lags behind the emf e by a phase angle of \(\frac{\pi}{2}\) rad. Here, for e = e₀ sin ωt, we have, i = i₀ sin(ωt – \(\frac{\pi}{2}\))
Instantaneous power, P = ei
= (e₀ sin ωt) [i₀ (sin ωt cos \(\frac{\pi}{2}\) – cos ωt sin \(\frac{\pi}{2}\))]
= – e₀i₀ sin ωt cos ωt as cos \(\frac{\pi}{2}\) = 0 and sin \(\frac{\pi}{2}\) = 1.
Average power over one cycle,


= e_rms i_rms cos Φ = e_rms i_rms (\(\frac{R}{Z}\)), where the impedance Z = \(\sqrt{R^{2}+\left(X_{\mathrm{L}}-X_{\mathrm{C}}\right)^{2}}\).
∴ P_av = 0, i.e., the circuit does not dissipate power.

Question 5.
Prove that an ideal capacitor in an AC circuit does not dissipate power
Answer:
In an AC circuit containing only an ideal inductor, the current i lags behind the emf e by a phase angle of \(\frac{\pi}{2}\) rad. Here, for e = e₀ sin ωt, we have, i = i₀ sin(ωt – \(\frac{\pi}{2}\))
Instantaneous power, P = ei
= (e₀ sin ωt) [i₀ (sin ωt cos \(\frac{\pi}{2}\) + cos ωt sin \(\frac{\pi}{2}\))]
= – e₀i₀ sin ωt cos ωt as cos \(\frac{\pi}{2}\) = 0 and sin \(\frac{\pi}{2}\) = 1.
Average power over one cycle, P_av


= e_rms i_rms cos Φ = e_rms i_rms (\(\frac{R}{Z}\)), where the impedance Z = \(\sqrt{R^{2}+\left(X_{\mathrm{L}}-X_{\mathrm{C}}\right)^{2}}\).
∴ P_av = 0, i.e., the circuit does not dissipate power.

Question 6.
(a) An emf e = e₀ sin ωt applied to a series L – C – R circuit derives a current I = I₀ sinωt in the circuit. Deduce the expression for the average power dissipated in the circuit.
(b) For circuits used for transporting electric power, a low power factor implies large power loss in transmission. Explain.
Answer:
(a) Instantaneous power,
P = ei
= (e₀ sin ωt) [i₀ (sin ωt ± Φ)]
= e₀i₀ sin ωt(sin ωt cos Φ ± cos ωt sin Φ)
= e₀i₀ sin² ωt ± e₀i₀ sin Φ sin ωt cos ωt
Average power over one cycle,


= e_rms i_rms cos Φ = e_rms i_rms (\(\frac{R}{Z}\)), where the impedance Z = \(\sqrt{R^{2}+\left(X_{\mathrm{L}}-X_{\mathrm{C}}\right)^{2}}\).

(b) P_av = e_rms i_rms cos Φ
The factor cos Φ is called as power factor. For circuits used for transporting electric power, a low power factor means the power available on transportation is much less than e_rms i_rms It means there is significant loss of power during transportation.

Question 7.
A device Y is connected across an AC source of emf e = e₀ sinωt. The current through Y is given as i = i₀ sin(ωt + π/2)
a) Identify the device Y and write the expression for its reactance.
b) Draw graphs showing variation of emf and current with time over one cycle of AC for Y.
c) How does the reactance of the device Y vary with the frequency of the AC ? Show graphically
d) Draw the phasor diagram for the device Y.
Answer:
(a) The device Y is a capacitor. Its reactance is X_c = \(\frac{1}{\omega C}\),
where ω is the angular frequency of the applied emf and C is the capacitance of the capacitor.

(b)

(c) X_C = \(\frac{1}{\omega C}=\frac{1}{2 \pi f C}\). Thus X_C ∝ \(\frac{1}{f}\), where f is the frequency of AC. Suppose C = \(\left(\frac{1000}{2 \pi}\right)\) pF
For f= 100 Hz, X_C = 1 × 10⁷Ω = 10MΩ;
for f = 200 Hz, X_C = 5 MΩ;
for f = 300 Hz, X_C = \(\frac{10}{3}\) MΩ;
for f = 400 Hz, X_C = 2.5 MΩ
for f = 500 Hz, X_C = 2 MΩ and so on

(d)

The phasor representing the peak emf (e₀) makes an angle (ωt) in an anticlockwise direction with respect to the horizontal axis. As the current leads the voltage by 90°, the phasor representing the peak current (i₀) is turned 90° anticlockwise with respect to the phasor representing emf e₀. The projections of these phasors on the vertical axis give instantaneous values of e and i.

Question 8.
Derive an expression for the impedance of an LCR circuit connected to an AC power supply.
Answer:
Figure shows an inductor of inductance L, capacitor of capacitance C, resistor of resistance R, key K and source (power supply) of alternating emf (e) connected to form a closed series circuit.

We assume the inductor, capacitor and resistor to be ideal. As these are connected in series, at any instant, they carry the same current i = i₀ sin ωt. The voltage across the resistor, e_R = Ri, is in phase with the current. The voltage across the inductor, e_L = X_Li, leads the current by \(\frac{\pi}{2}\) rad and that across the capacitor, e_C = X_Ci, lags behind the current by \(\frac{\pi}{2}\) rad. This is shown in the phasor diagram.

is the effective resistance of the circuit. It is called the impedance.

Question 9.
Compare resistance and reactance.
Answer:
(1) Resistance is opposition to flow of charges (current) and appears in a DC circuit as well as in an AC circuit.
The term reactance appears only in an AC circuit. It occurs when an inductor and/or a capacitor is used.

(2) In a purely resistive circuit, current and voltage are always in phase.
When reactance is not zero, there is nonzero phase difference between current and voltage.

(3) Resistance does not depend on the frequency of AC.
Reactance depends on the frequency of AC. In case of an inductor, reactance increases linearly with frequency. In case of a capacitor, reactance decreases as frequency of AC increases; it is inversely proportional to frequency.

(4) Resistance gives rise to production of Joule heat in a component.
In a circuit with pure reactance, there is no production of heat.

Question 10.
Show that in an AC circuit containing a pure inductor, the voltage is ahead of current by π/2 in phase.
Answer:
Figure 13.8 shows an AC source, generating a voltage e = e₀ sin ωt, connected to a key K and a pure inductor of inductance L to form a closed circuit.

On closing the key K, an emf is induced in the inductor as the magnetic flux linked with it changes with time. This emf opposes the applied emf and according to the laws of electromagnetic induction by Faraday and Lenz, we have,
e’ = -L\(\frac{d i}{d t}\) ………………. (1)
where e’ is the induced emf and i is the current through the inductor. To maintain the current; e and e’ must be equal in magnitude and opposite in direction.

According to Kirchhoff’s voltage law, as the resistance of the inductor is assumed to be zero, we

where C is the constant of integration. C must be time independent and have the dimension of current. As e oscillates about zero, i also oscillates about zero and hence there cannot be any time independent component of current.
∴ C = 0. ∴ i = –\(\frac{e_{0}}{\omega L}\)cos ωt = – \(\frac{e_{0}}{\omega L}\)sin(\(\frac{\pi}{2}\) – ωt)
∴ i = \(\frac{e_{0}}{\omega L}\) sin(ωt – \(\frac{\pi}{2}\)) ……………. (3)
as sin (-θ) = – sin θ
From Eq. (3), i_peak = i₀ = \(\frac{e_{0}}{\omega L}\)
∴ i = i₀ sin(ωt – \(\frac{\pi}{2}\)) ………………. (4)
Comparison of this equation with e = e₀ sin ωt shows that e leads i by \(\frac{\pi}{2}\) rad, i.e., the voltage is ahead of current by \(\frac{\pi}{2}\) rad in phase.

Question 11.
An AC source generating a voltage e = e₀ sinωt is connected to a capacitor of capacitance C. Find the expression for the current i flowing through it. Plot a graph of e and i versus ωt.
Answer:
Figure 13.12 shows an AC source, generating a voltage e = e₀ sin ωt, connected to a capacitor of capacitance C. The plates of the capacitor get charged due to the applied voltage. As the alternating voltage is reversed in each half cycle, the

capacitor is alternately charged and discharged. If q is the charge on the capacitor, the corresponding potential difference across the plates of the capacitor is V = \(\frac{q}{C}\) ∴ q = CV. q and V are functions of time, with V = e = e₀ sin ωt. The instantaneous current in the circuit is i = \(\frac{d q}{d t}=\frac{d}{d t}\)(CV) = C \(\frac{d v}{d t}\) = C \(\frac{d}{d t}\) (e₀ sin ωt) = ωC e₀ cos ωt
∴ i = \(\frac{e_{0}}{(1 / \omega C)} \sin \left(\omega t+\frac{\pi}{2}\right)=i_{0} \sin \left(\omega t+\frac{\pi}{2}\right)\)
where i₀ = \(\frac{e_{0}}{(1 / \omega C)}\) is the peak value of the current.

Table gives the values of e and i for different values of cot and Fig shows graphs of e and i versus ωt. i leads e by phase angle of \(\frac{\pi}{2}\) rad.

Question 12.
If the effective current in a 50 cycle AC circuit is 5 A, what is the peak value of current? What is the current 1/600 sec.after if was zero ?
Answer:
Data : f = 50 Hz, i_rms = 5 A, t = \(\frac{1}{600}\) s
The peak value of the current,
i₀ = i_rms\(\sqrt {2}\) = (5)(1.414) = 7.07 A
= i₀sin (2πft)
= 7.07 sin [2π(5o) (\(\frac{1}{600}\))]
= 7.07 sin (\(\frac{\pi}{6}\)) = (7.07)(0.5)
= 3.535 A
This is the required current.

Question 13.
A light bulb is rated 100W for 220 V AC supply of 50 Hz. Calculate (a) resistance of the bulb. (b) the rms current through the bulb.
Answer:
Data: Power (V_rms i_rms) = 100 W, V_rms = 220V,
f = 50 Hz
The rms current through the bulb,
i_rms = \(\frac{\text { power }}{V_{\mathrm{rms}}}=\frac{100}{220}\) = 0.4545 A
The resistance of the bulb,
R = \(\frac{V_{\mathrm{rms}}}{i_{\mathrm{rms}}}=\frac{220}{(100 / 220)}\) = (22) (22) = 484 Ω

Question 14.
A 15.0 µF capacitor is connected to a 220 V, 50 Hz source. Find the capacitive reactance and the current (rms and peak)
in the circuit. If the frequency is doubled, what will happen to the capacitive reactance and the current.
Answer:
Data : C = 15 µF = 15 × 10-6 F, V_rms = 220V, f = 50 Hz,
The capacitive reactance = \(\frac{1}{2 \pi f C}\)

If the frequency is doubled, the capacitive reactance will be halved and the current will be doubled.

Question 15.
An AC circuit consists of only an inductor of inductance 2 H. If the current is represented by a sine wave of amplitude 0.25 A and frequency 60 Hz, calculate the effective potential difference across the inductor (π = 3.142)
Answer:
Data : L = 2H, i₀ = 0.25 A, f = 60 Hz, π = 3.142
ωL = 2πfL = 2(3.142)(60)(2) = 754.1 Ω
The effective potential difference across the inductor = ωLi_rms = ωL \(\frac{i_{0}}{\sqrt{2}}\)
= \(\frac{(754.1)(0.25)}{1.414}\) = 133.3 V

Question 16.
Alternating emf of e = 220 sin 100 πt is applied to a circuit containing an inductance of (1/π) henry. Write an equation for instantaneous current through the circuit. What will be the reading of the AC galvanometer connected in the circuit?
Answer:
Data: e = 220 sin 100 πt, L = (\(\frac{1}{\pi}\))H
Comparing e = 220 sin 100 πt with
e = e₀ sin ωt, we get
ω = 100 π ∴ ωL = (100 π) (\(\frac{1}{\pi}\)) = 100 Ω
∴ The instantaneous current through the circuit
= i = \(\frac{e_{0}}{\omega L}\) sin(100 πt – \(\frac{\pi}{2}\))
= \(\frac{220}{100}\) sin (100 πt – \(\frac{\pi}{2}\)) = 2.2 sin (100 πt – \(\frac{\pi}{2}\)) in ampere [assuming that e is in volt.]
i_rms = \(\frac{i_{0}}{\sqrt{2}}=\frac{2.2}{1.414}\) = 1.556 A is the reading of the AC galvanometer connected in the circuit.

Question 17.
A 25 µF capacitor, a 0.10 H inductor and a 25Ω resistor are connected in series with an AC source whose emf is given by e = 310 sin 314 t (volt). What is the frequency, reactance, impedance, current and phase angle of the circuit?
Answer:
Data: C = 25 µF = 25 × 10^-6F, L = 0.10H, R = 25 Ω ,
e = 310 sin (314 t) [volt]
Comparing e = 310 sin (314 t) with
e = e₀ sin (2πft), we get,
the frequency of the alternating emf as

cos Φ = \(\frac{R}{Z}=\frac{25}{99.2}\) = 0.2520
∴ The phase angle, Φ = cos^-1(0.2520) = 75.40° = 1.316 rad

Question 18.
A capacitor of 100 µF, a coil of resistance 50Ω and an inductance 0.5 H are connected in series with a 110 V-50Hz source. Calculate the rms value of current in the circuit.
Answer:
Data : C = 100 µF = 100 × 10^-6 F = 10^-4 F,
R = 50 Ω, L = 0.5H, f = 50 Hz, V_rms = 110 V
∴ ωL = 2πfL = 2 (3.142)(50)(0.5) = 157.1 Ω

2500 + 15700 = 18200 Ω²
∴ Impedance, Z = \(\sqrt {18200}\) Ω = 134.9 Ω
The rms value of the current in the circuit,
i_rms = \(\frac{V_{\mathrm{rms}}}{\mathrm{Z}}=\frac{110}{134.9} \mathrm{~A}\)
= 0.8154 A

Question 19.
Find the capacity of a capacitor which when put in series with a 10Ω resistor makes the power factor equal to 0.5. Assume an 80V-100Hz AC supply.
Answer:
Data : R = 10 Ω, power factor = 0.5, f = 100 Hz
Power factor = \(\frac{1}{2 \pi f C R}\)
∴ 0.5 = \(\frac{1}{2(3.142)(100) C(10)}\)
∴ C = \(\frac{1}{3.142 \times 10^{3}}\)
= \(\frac{10 \times 10^{-4}}{3.142}\)
= 3.182 × 10^-4 F
This is the capacity of the capacitor.

Question 20.
Find the time required for a 50 Hz alternating current to change its value from zero to the rms value.
Answer:
Data : f = 50 Hz, i = \(\frac{i_{0}}{\sqrt{2}}\) ∴ \(\frac{i}{i_{0}}=\frac{1}{\sqrt{2}}\)
i = i₀ sinωt
∴ sinωt = \(\frac{i}{i_{0}}=\frac{1}{\sqrt{2}}\)
∴ ωt = \(\frac{\pi}{4}\) rad
∴ 2πft = \(\frac{\pi}{4}\)
∴ t = \(\frac{1}{8 f}=\frac{1}{8(50)}=\frac{1}{400}\)
= \(\frac{1000 \times 10^{-3}}{400}\) = 2.5 × 10^-3 s
This is the required time.

Question 21.
Calculate the value of capacitance in picofarad, which will make 101.4 micro henry inductance to oscillate with frequency of one megahertz.
Answer:
Data : f_r = 10⁶ Hz, L = 101.4 × 10^-6 H

= \(\frac{10000 \times 10^{-10}}{4(3.142)^{2}(101.4)}\) = 2.497 × 10^-10 F
= 249.7 × 10^-12 F = 249.7 picofarad
This is the value of the capacity.

Question 22.
A 10 µF capacitor is charged to a 25 volt of potential. The battery is disconnected and a pure 100 m H coil is connected across the capacitor so that LC oscillations are set up. Calculate the maximum current in the coil.
Answer:
Data: C = 10 µF = 10 × 10^-6F = 10^-5F,
L = 100mH = 100 × 10^-3 H = 10^-1 H, V = 25V
For reference, see the solved example (8) above.
\(\frac{1}{2}\)CV² = \(\frac{1}{2}\)Li²
∴i² = \(\frac{C}{L} V^{2}=\frac{10^{-5}}{10^{-1}}(25)^{2}\)
∴i = 25 × 10^-2 A = 0.25 A
This is the maximum current in the coil.

Question 23.
A 100 µF capacitor is charged with a 50 V source supply. Then source supply is removed and the capacitor is connected across an inductance, as a result of which 5A current flows through the inductance. Calculate the value of the inductance.
Answer:
Data: C = 100 µF = 100 × 10^-6 F = 10^-4 F,
V = 50V, i = 5A
The energy stored in the electric field in the capacitor
= \(\frac{1}{2}\)CV²
The energy stored in the magnetic field in the inductor = \(\frac{1}{2}\)Li²
Here, \(\frac{1}{2}\)CV² = \(\frac{1}{2}\)Li²
∴ L = C\(\frac{V^{2}}{i^{2}}\)
∴ L = C\(\left(\frac{V}{i}\right)^{2}=10^{-4}\left(\frac{50}{5}\right)^{2}\) = 10^-4 × 10²
= 10^-2H
This is the value of the inductance.

12th Std Physics Questions And Answers:

• Current Electricity Class 12 Physics Questions And Answers
• Magnetic Fields due to Electric Current Class 12 Physics Questions And Answers
• Magnetic Materials Class 12 Physics Questions And Answers
• Electromagnetic Induction Class 12 Physics Questions And Answers
• AC Circuits Class 12 Physics Questions And Answers
• Dual Nature of Radiation and Matter Class 12 Physics Questions And Answers
• Structure of Atoms and Nuclei Class 12 Physics Questions And Answers
• Semiconductor Devices Class 12 Physics Questions And Answers

Balbharti Maharashtra State Board 12th Physics Textbook Solutions Chapter 12 Electromagnetic Induction Textbook Exercise Questions and Answers.

Electromagnetic Induction Class 12 Exercise Question Answers Solutions Maharashtra Board

Class 12 Physics Chapter 12 Exercise Solutions Maharashtra Board

Physics Class 12 Chapter 12 Exercise Solutions

1. Choose the correct option.

i) A circular coil of 100 turns with a cross-sectional area (A) of 1 m² is kept with its plane perpendicular to the magnetic field (B) of 1 T. What is the magnetic flux linkage with the coil?
(A) 1 Wb
(B) 100 Wb
(C) 50 Wb
(D) 200 Wb
Answer:
(B) 100 Wb

ii) A conductor rod of length (l) is moving with velocity (v) in a direction normal to a uniform magnetic field (B). What will be the magnitude of induced emf produced between the ends of the moving conductor?
(A) BLv
(B) BLv²
(C) \(\frac{1}{2}\)Blv
(D) \(\frac{2 B l}{\mathrm{v}}\)
Answer:
(A) BLv

iii) Two inductor coils with inductance 10 mH and 20 mH are connected in series. What is the resultant inductance of the combination of the two coils?
(A) 20 mH
(B) 30 mH
(C) 10 mH
(D) \(\frac{20}{3}\) mH
Answer:
(A) 20 mH

iv) A current through a coil of self inductance 10 mH increases from 0 to 1 A in 0.1 s. What is the induced emf in the coil?
(A) 0.1 V
(B) 1 V
(C) 10 V
(D) 0.01 V
Answer:
(A) 0.1 V

v) What is the energy required to build up a current of 1 A in an inductor of 20 mH?
(A) 10 mJ
(B) 20 mJ
(C) 20 J
(D) 10 J
Answer:
(A) 10 mJ

2. Answer in brief.

i) What do you mean by electromagnetic induction? State Faraday’s law of induction.
Answer:
The phenomenon of production of emf in a conductor or circuit by a changing magnetic flux through the circuit is called electromagnetic induction.

Faraday’s laws of electromagnetic induction :
(1) First law ; Whenever there is a change in the magnetic flux associated with a circuit, an emf is induced in the circuit.
(2) Second law : The magnitude of the induced emf is directly proportional to the time rate of change of magnetic flux through the circuit.
[Note : The phenomenon was discovered in 1830 by Joseph Henry (1797-1878), US physicist, and independently in 1832 by Michael Faraday (1791 -1867), British chemchemist and physicist.]

ii) State and explain Lenz’s law in the light of principle of conservation of energy.
Answer:
Lenz’s law : The direction of the induced current is such as to oppose the change that produces it.

The change that induces a current may be (i) the motion of a conductor in a magnetic field or (ii) the change of the magnetic flux through a stationary circuit.

Explanation : Consider Faraday’s magnet-and- coil experiment. If the bar magnet is moved towards the coil with its N-pole facing the coil, as in Fig., the number of magnetic lines of induction (pointing to the left) through the coil increases. The induced current in the coil sets up a magnetic field of its own pointing to the right (as given by Amperes right-hand rule) to oppose the growing flux due to the magnet. Hence, to move the magnet towards the coil against this repulsive flux of the induced current, we must do work. The work done shows up as electric energy in the coil.

When the magnet is withdrawn, with its N-pole still facing the coil, the number of magnetic lines of induction (pointing left) through the coil decreases. The induced current reverses its direction to supplement the decreasing flux with its own, as shown in Fig.. Facing the coil along the magnet, the induced current is in the clockwise sense. The electric energy in the coil comes from the work done to withdraw the magnet, now against an attractive force. Thus, we see that Lenz’s law is a consequence of the law of conservation of energy.

[Note : The above law was discovered by Heinrich Friedrich Emil Lenz (1804-65), Russian physicist.]

iii) What are eddy currents? State applications of eddy currents.
Answer:
Whenever a conductor or a part of it is moved in a magnetic field “cutting” magnetic field lines, or placed in a changing magnetic field, the free electrons in the bulk of the metal start circulating in closed paths equivalent to current-carrying loops. These loop currents resemble eddies in a fluid stream and are hence called eddy or Foucault currents [after Jean Bernard Leon Foucault (1819-68), French physicist, who first detected them].

Applications :
(1) Dead-beat galvanometer : A pivoted moving-coil galvanometer used for measuring current has the coil wound on a light aluminium frame. The rotation of the metal frame in magnetic field produces eddy currents in the frame which opposes the rotation and the coil is brought to rest quickly. This makes the galvanometer dead-beat.

(2) Electric brakes : When a conducting plate is pushed into a magnetic field, or pulled out, very quickly, the interaction between the eddy currents in the moving conductor and the field retards the motion. This property of eddy currents is used as a method of braking in vehicles.

iv) If the copper disc of a pendulum swings between the poles of a magnet, the pendulum comes to rest very quickly. Explain the reason. What happens to the mechanical energy of the pendulum?
Answer:
As the copper disc enters and leaves the magnetic field, the changing magnetic flux through it induces eddy current in the disc. In both cases, Fleming’s right hand rule shows that opposing magnetic force damps the motion. After a few swings, the mechanical energy becomes zero and the motion comes to a stop.

Joule heating due to the eddy current warms up the disc. Thus, the mechanical energy of the pendulum is transformed into thermal energy.

v) Explain why the inductance of two coils connected in parallel is less than the inductance of either coil.
Answer:
Assuming that their mutual inductance can be ignored, the equivalent inductance of a parallel combination of two coils is given by
\(\frac{1}{L_{\text {parallel }}}=\frac{1}{L_{1}}+\frac{1}{L_{2}}\) or L_parallel = \(\frac{L_{1} L_{2}}{L_{1}+L_{2}}\)
Hence, the equivalent inductance is less than the inductance of either coil.

Question 3.
In a Faraday disc dynamo, a metal disc of radius R rotates with an angular velocity ω about an axis perpendicular to the plane of the disc and passing through its centre. The disc is placed in a magnetic field B acting perpendicular to the plane of the disc. Determine the induced emf between the rim and the axis of the disc.
Answer:
Suppose a thin conducting disc of radius R is rotated anticlockwise, about its axis, in a plane perpendicular to a uniform magnetic field of induction \(\vec{B}\) (see the figure in the above Note for reference). \(\vec{B}\) points downwards. Let the constant angular speed of the disc be ω.

Consider an infinitesimal element of radial thickness dr at a distance r from the rotation axis. In one rotation, the area traced by the element is dA = 2πrdr. Therefore, the time rate at which the element traces out the area is
\(\frac{d A}{d t}\) = frequency of rotation x dA = fdA
where f = \(\frac{\omega}{2 \pi}\) is the frequency of rotation.
.’. \(\frac{d A}{d t}=\frac{\omega}{2 \pi}\) (2 πr dr) = ωr dr
The total emf induced between the axle and the rim of the rotating disc is
\(|e|=\int B \frac{d A}{d t}=\int_{0}^{R} B \omega r d r=B \omega \int_{0}^{R} r d r=B \omega \frac{R^{2}}{2}\)
For anticlockwise rotation in \(\vec{B}\) pointing down, the axle is at a higher potential.

Question 4.
A horizontal wire 20 m long extending from east to west is falling with a velocity of 10 m/s normal to the Earth’s magnetic field of 0.5 × 10^-4 T. What is the value of induced emf in the wire?
Answer:
Data : l = 20 m, v = 10 m/s. B = 5 × 10^-5 T
The magnitude of the induced emf,
|e| = Blv = (5 × 10^-5)(20)(10) = 10^-2V = 10 mV

Question 5.
A metal disc is made to spin at 20 revolutions per second about an axis passing through its centre and normal to its plane. The disc has a radius of 30 cm and spins in a uniform magnetic field of 0.20 T, which is parallel to the axis of rotation. Calculate
(a) The area swept out per second by the radius of the disc,
(b) The flux cut per second by a radius of the disc,
(c) The induced emf in the disc.
Answer:
Data: R = 0.3m, f = 20 rps, B = 0.2T
(a) The area swept out per unit time by a given radius = (the frequency of rotations) × (the area swept out per rotation) = f(πr²)
= (20)(3.142 × 0.09) = 5.656 m²

(b) The time rate at which a given radius cuts magnetic flux
= \(\frac{d \Phi_{\mathrm{m}}}{d t}\) = B f(πr²)
= (0.2)(5.656) = 1.131 Wb/s

(c) The magnitude of the induced emf,
|e| = \(\frac{d \Phi_{\mathrm{m}}}{d t}\) = 1.131 V

Question 6.
A pair of adjacent coils has a mutual inductance of 1.5 H. If the current in one coil changes from 0 to 10 A in 0.2 s, what is the change of flux linkage with the other coil?
Answer:
Data: M = 1.5 H, I_1i = 0, I_1f = 10A, ∆f = 0.2s
The flux linked per unit turn with the second coil due to current I₁ in the first coil is
Φ₂₁ = MI₁
Therefore, the change in the flux due to change in I₁ is
∆₂₁ =M(∆I₁) = M(I_1f – I_1i) = 1.5 (10 – 0)
= 15 Wb
[Note: The rate of change of flux linkage is M(∆I₁/∆t) = 15/0.2 = 75 Wb/s] .

Question 7.
A long solenoid has 1500 turns/m. A coil C having cross sectional area 25 cm2 and 150 turns (N_c) is wound tightly around the centre of the solenoid. If a current of 3.0A flows through the solenoid, calculate :
(a) the magnetic flux density at the centre of the solenoid,
(b) the flux linkage in the coil C,
(c) the average emf induced in coil C if the direction of the current in the solenoid is reversed in a time of 0.5 s. (µ₀ = 4π × 10^-7 H/m)
Answer:
Data: n = 1.5 × 10³ m , A = 25 × 10^-4 m²,
N_c = 150, I = 3A, ∆t = 0.5s,
µ₀ = 4π × 10^-7 H/m
(a) Magnetic flux density inside the solenoid,
B = u₀ nI = (4π × 10^-7)(1500)(3)
= 5.656 × 10^-3 T = 5.656 mT

(b) Flux per unit turn through the coils of the solenoid, Φ_m = BA
Since the coil C is wound tightly over the solenoid, the flux linkage of C is
N_CΦ_m = N_CBA = (150)(5.656 × 10^-3)(25 × 10^-4)
= 2.121 × 10^-3 Wb = 2.121 mWb

(c) Initial flux through coil C,
Φ_i = N_CΦ_m = 2.121 × 10^-3 Wb
Reversing the current in the solenoid reverses the flux through coil C, the magnitude remaining the same. But since the flux enters through the other face of the coil, the final flux through C is Φ_f = -2.121 × 10^-3 Wb
Therefore, the average emf induced in coil C,
e = \(-\frac{\Phi_{\mathrm{f}}-\Phi_{1}}{\Delta t}=-\frac{(-2.121-2.121) \times 10^{-3}}{0.5}\)
= 2 × 4.242 × 10^-3 = 8.484 × 10^-3 V = 8.484 mV

Question 8.
A search coil having 2000 turns with area 1.5 cm² is placed in a magnetic field of 0.60T. The coil is moved rapidly out of the field in a time of 0.2 second. Calculate the induced emf across the search coil.
Answer:
Data: N = 2000, A_i = 1.5 × 10^-4 m², A_f = 0,
B = 0.6T, ∆t = 0.2s
Initial flux, NΦ_f = NBA_i = 2000(0.6)(1.5 × 10^-4)
= 0.18 Wb
Final flux, NΦ_f = 0, since the coil is withdrawn out of the field.
Induced emf,e = \(-N \frac{\Delta \Phi_{\mathrm{m}}}{\Delta t}=-N \frac{\Phi_{\mathrm{f}}-\Phi_{1}}{\Delta t}\)
∴ e = \(-\frac{0-0.18}{0.2}\) = 0.9V

Question 9.
An aircraft of wing span of 50 m flies horizontally in earth’s magnetic field of 6 × 10^-5 T at a speed of 400 m/s. Calculate the emf generated between the tips of the wings of the aircraft.
Answer:
Data : l = 50 m, B = 6 × 10^-5T, v = 400 m/s
The magnitude of the induced emf,
|e| = Blv = (6 × 10^-5)(400)(50) = 1.2V

Question 10.
A stiff semi-circular wire of radius R is rotated in a uniform magnetic field B about an axis passing through its ends. If the frequency of rotation of the wire is f, calculate the amplitude of the alternating emf induced in the wire.
Answer:
In one rotation, the wire traces out a circle of radius R, i.e., an area A = πR².
Therefore, the rate at which the wire traces out the area is
\(\frac{d A}{d t}\) = frequency or rotation × A = fA
If the angle between the uniform magnetic field \(\vec{B}\) and the rotation axis is θ, the magnitude of the induced emf is
|e|= B\(\frac{d A}{d t}\) cosθ = BfA cosθ = Bf(πR²)cosθ
so that the required amplitude is equal to Bf(πR²).

Question 11.
Calculate the value of induced emf between the ends of an axle of a railway carriage 1.75 m long traveling on level ground with a uniform velocity of 50 km per hour. The vertical component of Earth’s magnetic field (B_v) is given to be 5 × 10^-5T.
Answer:
Data: l = 1.75 m, v = 50 km/h = 50 × \(\frac{5}{18}\) m/s.
B_v = 5 × 10^-5 T
The area swept out by the wing per unit time = 1v.
∴ The magnetic flux cut by the wing per unit time
= \(\frac{d \Phi_{\mathrm{m}}}{d t}\) = B_v(lv)
=(5 × 10^-5)(1.75)(50 × \(\frac{5}{18}\))= 121.5 × 10^-5
= 1.215 mWb/s
Therefore, the magnitude of the induced emf,
|e| =1.215 mV
[Note: In the northern hemisphere, the vertical com ponent of the Earth’s magnetic induction is downwards. Using Fleming’s right hand rule, the port (left) wing-tip would be positive.]

Question 12.
The value of mutual inductance of two coils is 10 mH. If the current in one of the coil changes from 5A to 1A in 0.2 s, calculate the value of emf induced in the other coil.
Answer:
Data: M = 10 mH = 10^-2 H, I_1i = 5A, I_1f = 1 A,
∆t = 0.2s
The mutually induced emf in coil 2 due to the changing current in coil 1,
e₂₁ = \(-M \frac{\Delta I_{1}}{\Delta t}=-M \frac{I_{1 \mathrm{f}}-I_{1 \mathrm{i}}}{\Delta t}\)
= -(10^-2) \(\left(\frac{1-5}{0.2}\right)\) = 0.2 V

Question 13.
An emf of 96.0 mV is induced in the windings of a coil when the current in a nearby coil is increasing at the rate of 1.20 A/s. What is the mutual inductance (M) of the two coils?
Answer:
Data: |e₂| = 9.6 × 10^-2 V, dI₁/dt = 1.2 A/s
|e₂| = M\(\frac{d I_{1}}{d t}\)
Mutual inductance,
M = \(\frac{\left|e_{2}\right|}{d I_{1} / d t}=\frac{9.6 \times 10^{-2}}{1.2}\)
= 8 × 10^-2 H
= 80 mH

Question 14.
A long solenoid of length l, cross-sectional area A and having N₁ turns (primary coil) has a small coil of N₂ turns (secondary coil) wound about its centre. Determine the Mutual inductance (M) of the two coils.
Answer:
We assume the solenoid to be ideal and that all the flux from the solenoid passes through the outer coil C. For a steady current Is through the solenoid, the uniform magnetic field inside the solenoid is
B = µ₀nI_s ……………….. (1)
Then, the magnetic flux through each turn of the coil due to the current in the solenoid is
Φ_CS = BA = (µ₀nI_s)(πR²) ……………… (2)
Thus, their mutual inductance is
M = \(\frac{N \Phi_{\mathrm{CS}}}{I_{\mathrm{S}}}\) = µ₀πR²nN ……………. (3)
Equation (2) is true as long as the magnetic field of the solenoid is entirely contained within the cross section of the coil C. Hence, M does not depend on the shape, size, or possible lack of close packing of the coil.
replacing n with N₁/l and N with N₂. M = µ₀A = \(\frac{N_{1} N_{2}}{l}\)
[Note: The answer given in the textbook misses out the factor of 1.] .

Question 15.
The primary and secondary coil of a transformer each have an inductance of 200 × 10^-6H. The mutual inductance (M) between the windings is 4 × 10^-6H. What percentage of the flux from one coil reaches the other?
Answer:
Data: L_P = L_S = 2 × 10^-4 H, M = 4 × 10^-6 H
M = K\(\sqrt{L_{\mathrm{P}} L_{\mathrm{S}}}\)
The coupling coefficient is
K = \(\frac{M}{\sqrt{L_{\mathrm{P}} L_{\mathrm{S}}}}=\frac{4 \times 10^{-6}}{\sqrt{\left(2 \times 10^{-4}\right)^{2}}}=\frac{4 \times 10^{-6}}{2 \times 10^{-4}}\)
= 2 × 10^-2
Therefore, the percentage of flux of the primary reaching the secondary is
0.02 × 100% = 2%

Question 16.
A toroidal ring, having 100 turns per cm of a thin wire is wound on a nonmagnetic metal rod of length 1 m and diameter 1 cm. If the permeability of bar is equal to that of free space (µ₀), calculate the magnetic field inside the bar (B) when the current (i) circulating through the turns is 1 A. Also determine the self-inductance (L) of the coil.
Answer:
Data : l = 1 m, d = 1 cm, n = 100 cm^-1 = 10⁴ m^-1,
I = 100 A, µ₀ = 4π × 10^-7 H/m
The radius of cross section, r = \(\frac{d}{2}\) = 0.5 cm
= 5 × 10^-3 m
(a) Magnetic field inside the toroid,
B = µ₀nI = (4π × 10^-7)(10⁴)(100)
= 0.4 × 3.142 = 1.257 T

(b) Self inductance of the toroid,
L = µ₀2πRn²A = µ₀n²lA = µ₀n²l(πr²)
= (4π × 10^-7)(10⁴)²(1) [π(5 × 10^-3)²]
= π² × 10^-3 = 9.87 × 10^-3 H = 9.87 mH

Question 17.
A uniform magnetic field B(t), pointing upward fills a circular region of radius, s in horizontal plane. If B is changing with time, find the induced electric field.
[Hint : Part of Maxwell’s equation, applied to a time varying magnetic flux, leads us to the equation \(\oint \vec{E} \cdot \overline{\mathrm{d} l}=\frac{-d \phi_{m}}{d t}\), where \(\vec{E}\) is the electric field induced when the magnetic flux changes at the rate of \(\frac{d \phi_{m}}{d t}\)]
Answer:
The area of the region, A = πs², remains constant while B = B(f) is a function of time. Therefore, the induced emf,
e = \(-\frac{d \Phi_{\mathrm{m}}}{d t}=-\frac{d}{d t}(B A)=-A \frac{d B(t)}{d t}=-\pi s^{2} \frac{d B(t)}{d t}\)
[Note : Emf and electric field are different physical quantities, whose respective SI units are the volt and the volt per metre. The question has accordingly been corrected.]

12th Physics Digest Chapter 12 Electromagnetic induction Intext Questions and Answers

Do you know (Textbook Page No. 274)

Question 1.
If a wire without any current is kept in a magnetic field, then it experiences no force as shown in figure (a). But when the wire is carrying a current into the plane of the paper in the magnetic field, a force will be exerted on the wire towards the left as shown in the figure (b). The field will be strengthened on the right side of the wire where the lines of force are in the same direction as that of the magnetic field and weakened on the left side where the field lines are in opposite direction to that of the applied magnetic field. For a wire carrying a current out of the plane of the paper, the force will act to the right as shown in figure (c).

Answer:
Force on a current-carrying conductor in a magnetic field, \(\vec{F}=I \vec{L} \times \vec{B}\) (Refer unit 10.5). The field due to acurrent-carrying straight conductor is given by right- hand grip rule. As shown in the figure below, the combined field due to a permanent magnet and a current-carrying conductor force the conductor out of the field. The field is strengthened where the two fields are in the same direction and add constructively while the field is weakened where the two fields are opposite in direction.

Use your brain power (Textbook Page No. 282)

Question 1.
It can be shown that the mutual potential energy of two circuits is W = MI₁I₂. Therefore, the mutual inductance (M) may also be defined as the mutual potential energy (W) of two circuits corresponding to unit current flowing in each circuit.
M = \(\frac{W}{I_{1} I_{2}}\)
M = W[I₁ = I₂ = 1]
Answer:
Mutual inductance of two magnetically linked coils equals the potential energy for unit currents in the coils.
1 H = 1 T∙m²/A (= 1 V∙s/A = 1 Ω ∙ s = 1 J/A²)

Use your brain power (Textbook Page No. 284)

Question 1.
Prove that the inductance of parallel wires of length l in the same circuit is given by L = \(\left(\frac{\mu_{0} l}{\pi}\right)\) ln (d / a), where a is the radius of wire and d is separation between wire axes.
Answer:
If l is the current in each wire, from Ampe’re’s law the magnitude of the magnetic field outside each wire is
B = \(\frac{\mu_{0} I}{2 \pi r}\)
By right hand grip rule, the direction of the magnetic field due to both the wires are in the same direction at the point shown. Hence, by the symmetry of the setup, the total magnetic flux through an area dA = l dr shown is two times that due to one wire.

Do you know (Textbook Page No. 285)

Question 1.
The flux rule is the terminology that Feynman used to refer to the law relating magnetic flux to emf. (RP Feynman, Feynman Lectures on Physics, Vol II)
Answer:
Modern applications of Faraday’s law of induction :

• Electric generators and motors
• Dynamos in vehicles
• Transformers
• Induction furnaces (industrial), induction cooking stoves (domestic)
• Radio communication
• Magnetic flow meters and energy meters
• Metal detectors at security checks
• Magnetic hard disk and tape, storage and retrieval
• Graphics tablets
• ATM Credit/debit cards, ATM and point-of-sale (POS) machines
• Pacemakers

Faraday’s second law of electromagnetic induction is referred to by some as the “flux rule”.

12th Std Physics Questions And Answers:

• Current Electricity Class 12 Physics Questions And Answers
• Magnetic Fields due to Electric Current Class 12 Physics Questions And Answers
• Magnetic Materials Class 12 Physics Questions And Answers
• Electromagnetic Induction Class 12 Physics Questions And Answers
• AC Circuits Class 12 Physics Questions And Answers
• Dual Nature of Radiation and Matter Class 12 Physics Questions And Answers
• Structure of Atoms and Nuclei Class 12 Physics Questions And Answers
• Semiconductor Devices Class 12 Physics Questions And Answers