NCERT Exemplar for Class 12 Physics Chapter 11

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Access the NCERT Exemplar Class 12 Physics Chapter 11 Dual Nature of Radiation and Matter

Multiple Choice Questions I

11.1. A particle is dropped from a height H. The de Broglie wavelength of the particle as a function of height is proportional to

(a) H

(b) H ½

(c) H0

(d) H–1/2

Answer:

The correct answer is d) H–1/2

11.2. The wavelength of a photon needed to remove a proton from a nucleus which is bound to the nucleus with 1 MeV energy is nearly

(a) 1.2 nm

(b) 1.2 × 10–3 nm

(c) 1.2 × 10–6 nm

(d) 1.2 × 101 nm

Answer:

The correct answer is b) 1.2 × 10–3 nm

11.3. Consider a beam of electrons (each electron with energy E0) incident on a metal surface kept in an evacuated chamber. Then

(a) no electrons will be emitted as only photons can emit electrons

(b) electrons can be emitted but all with an energy, E0

(c) electrons can be emitted with any energy, with a maximum of E0 – φ (φ is the work function)

(d) electrons can be emitted with any energy, with a maximum of E0

Answer:

The correct answer is d) electrons can be emitted with any energy, with a maximum of E0

11.4. Consider Fig. 11.7 in the NCERT textbook of physics for Class XII. Suppose the voltage applied to A is increased. The diffracted beam will have the maximum at a value of θ that

(a) will be larger than the earlier value

(b) will be the same as the earlier value

(c) will be less than the earlier value

(d) will depend on the target

Answer:

The correct answer is c) will be less than the earlier value

11.5. A proton, a neutron, an electron and an α-particle have the same energy. Then their de Broglie wavelengths compare as

(a) λp = λn > λe > λα

(b) λα < λp = λn < λe

(c) λe < λp = λn > λα

(d) λe = λp = λn = λα

Answer:

The correct answer is b) λα < λp = λn < λe

11.6. An electron is moving with an initial velocity (\vec{v} = v_0 \hat{i}) and is in a magnetic field (\vec{B} = B_0 \hat{j}). What happens to its de Broglie wavelength? \begin{enumerate} \item remains constant \item increases with time \item decreases with time \item increases and decreases periodically \end{enumerate} \section*

{Solution} The correct option is \textbf{A: remains constant}. The magnetic force on a moving electron is given by: [ \vec{F} = -e (\vec{v} \times \vec{B}) ] Substituting the given vectors: [ \vec{F} = -e (v_0 \hat{i} \times B_0 \hat{j}) = -e v_0 B_0 \hat{k} ] Since the force is perpendicular to both (\vec{v}) and (\vec{B}), the magnitude of (\vec{v}) will not change, which means the momentum of the electron remains constant. The de Broglie wavelength is given by: [ \lambda = \frac{h}{p} ] where ( p = mv ) is the momentum. Since the momentum ( p ) remains constant, the de Broglie wavelength (\lambda) also remains constant.

11.7. An electron of mass ( m ) with an initial velocity (\vec{V} = V_0 \hat{i} ) (where ( V_0 > 0 )) enters an electric field (\vec{E} = -E_0 \hat{i} ) (where ( E_0 = \text{constant} > 0 )) at ( t = 0 ). If (\lambda_0) is its de-Broglie wavelength initially, then its de-Broglie wavelength at time ( t ) is: \begin{enumerate} \item (\lambda_0 \left( 1 + \frac{e E_0 t}{m V_0} \right)) \item (\lambda_0 t) \item (\lambda_0 \left( 1 + \frac{e E_0 t}{m V_0} \right)) \item (\lambda_0) \end{enumerate} \section*

{Solution} The correct option is {A:} (\lambda_0 \left( 1 + \frac{e E_0 t}{m V_0} \right)). \begin{itemize} \item Initial velocity: ( V_0 ) \item Since the acceleration is constant, the final velocity at time ( t ) is given by: [ v = V_0 + \frac{e E_0}{m} t ] \item At ( t = 0 ), the de-Broglie wavelength is (\lambda_0). Therefore: [ \lambda_0 = \frac{h}{m V_0} ] \item Let the de-Broglie wavelength at time ( t ) be (\lambda). Then: [ \lambda = \frac{h}{m v} = \frac{h}{m \left( V_0 + \frac{e E_0}{m} t \right)} ] \item Substitute ( \lambda_0 = \frac{h}{m V_0} ): [ \lambda = \frac{\frac{h}{m V_0}}{1 + \frac{e E_0 t}{m V_0}} = \frac{\lambda_0}{1 + \frac{e E_0 t}{m V_0}} ] \item Rearranging: [ \lambda = \lambda_0 \left( 1 + \frac{e E_0 t}{m V_0} \right) ]

11.8. An electron (mass ( m )) with an initial velocity (\vec{v} = v_0 \hat{i}) is in an electric field (\vec{E} = E_0 \hat{j}). If (\lambda_0 = \frac{h}{m v_0}), its de Broglie wavelength at time ( t ) is given by: \begin{enumerate} \item (\lambda_0) \item (\lambda_0 \sqrt{1 + \frac{e^2 E_0^2 t^2}{m^2 v_0^2}}) \item (\lambda_0 \sqrt{1 + \frac{e^2 E_0^2 t^2}{m^2 v_0^2}}) \item (\lambda_0 \left(1 + \frac{e^2 E_0^2 t^2}{m^2 v_0^2}\right)) \end{enumerate} \section*

{Solution} The correct option is {C:} (\lambda_0 \sqrt{1 + \frac{e^2 E_0^2 t^2}{m^2 v_0^2}}). \begin{itemize} \item Initial de Broglie wavelength of the electron: [ \lambda_0 = \frac{h}{m v_0} ] \item Force on the electron in the electric field: [ \vec{F} = -e \vec{E} = -e E_0 \hat{j} ] \item Acceleration of the electron: [ \vec{a} = \frac{\vec{F}}{m} = \frac{e E_0}{m} \hat{j} ] This acceleration is along the negative ( y )-axis. \item Initial velocity components: [ \vec{v}{x0} = v_0 \hat{i}, \quad \vec{v}{y0} = 0 ] \item Velocity of the electron after time ( t ): [ \vec{v}_x = v_0 \hat{i} ] [ \vec{v}_y = 0 + \left(\frac{e E_0}{m}\right) t = \frac{e E_0 t}{m} \hat{j} ] \item Magnitude of the velocity after time ( t ): [ |\vec{v}| = \sqrt{v_x^2 + v_y^2} = \sqrt{v_0^2 + \left(\frac{e E_0 t}{m}\right)^2} ] [ |\vec{v}| = v_0 \sqrt{1 + \frac{e^2 E_0^2 t^2}{m^2 v_0^2}} ] \item De Broglie wavelength at time ( t ): [ \lambda = \frac{h}{m v} = \frac{h}{m \left(v_0 \sqrt{1 + \frac{e^2 E_0^2 t^2}{m^2 v_0^2}}\right)} ] [ \lambda = \frac{\lambda_0}{\sqrt{1 + \frac{e^2 E_0^2 t^2}{m^2 v_0^2}}} ] \item Simplify: [ \lambda = \lambda_0 \sqrt{1 + \frac{e^2 E_0^2 t^2}{m^2 v_0^2}} ] \end{itemize}

Multiple Choice Questions II

11.9. Relativistic corrections become necessary when the expression for the kinetic energy 1/2 mv2 , becomes comparable with mc2, where m is the mass of the particle. At what de Broglie wavelength will relativistic corrections become important for an electron?

(a) λ =10nm

(b) λ =10–1nm

(c) λ =10–4nm

(d) λ =10–6nm

Answer:

The correct answers are

c) λ =10–4nm

d) λ =10–6nm

11.10. Two particles A1 and A2 of masses m1, m2 (m1 > m2) have the same de Broglie wavelength. Then

(a) their momenta are the same

(b) their energies are the same

(c) energy of A1 is less than the energy of A2

(d) energy of A1 is more than the energy of A2

Answer:

The correct answers are

a) their momenta are the same

c) energy of A1 is less than the energy of A2

11.11. The de Broglie wavelength of a photon is twice the de Broglie wavelength of an electron. The speed of the electron is ve = c/100. Then

(a) Ee/Ep = 10-4

(b) Ee/Ep = 10-2

(c) Pe/mec = 10-2

(d) Pe/mec = 10-4

Answer:

The correct answers are

b) Ee/Ep = 10-2

c) Pe/mec = 10-2

11.12. Photons absorbed in matter are converted to heat. A source emitting n photon/sec of frequency ν is used to convert 1kg of ice at 0°C to water at 0°C. Then, the time T taken for the conversion

(a) decreases with increasing n, with ν fixed

(b) decreases with n fixed, ν increasing

(c) remains constant with n and ν changing such that nν = constant

(d) increases when the product nν increases

Answer:

The correct answers are

a) decreases with increasing n, with ν fixed

b) decreases with n fixed, ν increasing

c) remains constant with n and ν changing such that nν = constant

11.13. A particle moves in a closed orbit around the origin, due to a force which is directed towards the origin. The de Broglie wavelength of the particle varies cyclically between two values λ1, λ2 with λ1>λ2. Which of the following statement are true?

(a) The particle could be moving in a circular orbit with origin as centre

(b) The particle could be moving in an elliptic orbit with origin as its focus

(c) When the de Broglie wavelength is λ1, the particle is nearer the origin than when its value is λ2

(d) When the de Broglie wavelength is λ2, the particle is nearer the origin than when its value is λ1

Answer:

The correct answers are

b) The particle could be moving in an elliptic orbit with origin as its focus

d) When the de Broglie wavelength is λ2, the particle is nearer the origin than when its value is λ1

Very Short Answers

11.14. A proton and an α-particle are accelerated, using the same potential difference. How are the de Broglie wavelengths λp and λa related to each other?

Answer:

The proton and α-particle are accelerated at the same potential difference.

λ = h/√2mqv

λ is proportional to 1/√mq

λp/ λa = √maqa/mpqp = √8

Therefore, the wavelength of the proton is √8 times the wavelength of α-particle.

11.15. (i) In the explanation of the photoelectric effect, we assume one photon of frequency ν collides with an electron and transfers its energy. This leads to the equation for the maximum energy Emax of the emitted electron as Emax = hν – φ0 where φ0 is the work function of the metal. If an electron absorbs 2 photons (each of frequency ν ) what will be the maximum energy for the emitted electron?

(ii) Why is this fact (two-photon absorption) not taken into consideration in our discussion of the stopping potential?

Answer:

i) From the question, the energy absorbed by the electron is two protons with frequency v and v’= 2v where v’ is the frequency from the emitted electron.

Emax = hv – ϕ0

ii) The emission is negligible as the chances of the same proton absorbing the energy is very low.

11.16. There are materials which absorb photons of shorter wavelength and emit photons of longer wavelength. Can there be stable substances which absorb photons of larger wavelength and emit light of shorter wavelength?

Answer:

The wavelength of the photon increases when the frequency decreases. There are two possible cases and they are; case one in which the photons have a smaller wavelength and the energy is consumed against work function. In case two, the photons might have a longer wavelength and emit photons with shorter and lesser energy.

11.17. Do all the electrons that absorb a photon come out as photoelectrons?

Answer:

No, not all electrons that are absorbed as a photon come out as a photoelectron.

11.18. There are two sources of light, each emitting with a power of 100 W. One emits X-rays of wavelength 1nm and the other visible light at 500 nm. Find the ratio of a number of photons of X-rays to the photons of visible light of the given wavelength?

Answer:

Ex = hvx

Ev = hvv

Let nx and nv be the no.of photons from x-rays and visible light are of equal energies and they emit 100 W power.

nxhvx = nvhvv

nx/nv = vv/vx = λx/λv

nx/nv = 1 nm/500 nm

nx:nv = 1:500

NCERT Exemplar For Class 12 Science

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