Physics Ncert Exemplar 2019 Solutions for Class 12 Science Physics Chapter 11 - Dual Nature Of Radiation And Matter

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Page No 68:

Question 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^1/2
(c) H⁰
(d) H ^–1/2

Answer:

de-Broglie wavelength, $λ = \frac{h}{P}$ (here, h = plank's constant and p = mv = momentum)
$\Rightarrow λ = \frac{h}{m ν} . . . . . (1)$

Velocity of a body falling from a height 'H' is given by, $v = \sqrt{2 g H}$

Putting this value of 'v' in equation (1)

$\Rightarrow λ = \frac{h}{m \sqrt{2 g H}} \Rightarrow λ = \frac{h}{m \sqrt{2 g}} \cdot \frac{1}{\sqrt{H}} \Rightarrow λ = K \cdot \frac{1}{\sqrt{H}} . . . . . (2)$
Where 'K' is constant with value $\frac{h}{m \sqrt{2 g}} .$
From equation (2)
$\Rightarrow λ \propto \frac{1}{\sqrt{H}} or λ \propto H^{- \frac{1}{2}}$

Hence, the correct answer is option (d).

Page No 68:

Question 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:

Energy of photon, E = 1 MeV = 1.6 × 10^–13J

But, E = hf (h = plank's constant and f = frequency)

where, $f = \frac{c}{λ}$

$\Rightarrow E = \frac{h c}{λ} \Rightarrow λ = \frac{h c}{E} = \frac{6.6 \times 10^{- 34} \times 3 \times 10^{8}}{1.6 \times 10^{- 13}} \Rightarrow λ = 1.2 \times 10^{- 3} nm$

Hence, the correct answer is option (b).

Page No 68:

Question 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:

When a beam of electrons (each electron with energy E₀) â€‹is incident on a metal surface kept in an evacuated chamber, electrons can be emitted with maximum energy E_â€‹0 â€‹ (due to elastic collision). But when part of incident energy of electrons is used in liberating the electrons from the surface of metal then the energy of emitted electron would be less than E₀ .
Hence, the correct answer is option (d).

Page No 69:

Question 4:

Consider Fig. 11.7 in the NCERT text book 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:

In Davisson - Germer experiment, the de-Broglie wavelength associated with electron is:

$λ = \frac{12.27}{\sqrt{v}} Å (where, v = stopping potential) . . . (1)$

If here is a maxima of the diffracted electrons at angle θ, then

2d sinθ = λ ...(2)

From equation(1), $λ \propto \frac{1}{\sqrt{v}}$
From equation(2), λ ∝ θ

Thus, when the voltage applied to A is increased. The diffracted beam will have the maximum value at an angle 'θ' less then the earlier value.

Hence, the correct answer is option (c).

Page No 69:

Question 5:

A proton, a neutron, an electron and an α-particle have same energy. Then their de Broglie wavelengths compare as
(a) λ_p = λ_n > λ_e > λ_α
(b) λ_α < λ_p = λ_n > λ_e
(c) λ_e < λ_p = λ_n > λ_α
(d) λ_e = λ_p = λ_n = λ_α

Answer:

de-Broglie wavelength, $λ = \frac{h}{p} . . . . . (1)$
But, momentum, $p = \sqrt{2 mk} . . . . . (2)$
where, m = mass
K = Kinetic energy
from equation (1) and (2)
$\Rightarrow λ = \frac{h}{\sqrt{2 mK}} \Rightarrow λ \propto \frac{1}{\sqrt{m}} \{∴ \frac{h}{\sqrt{2 K}} = constant\}$

Since, m_P = m_n, hence λ_P = λ_n
As, m_∝ > m_P, therefore λ_p > λα
As, m_e < m_n, therefore λ_e > λ_p
Hence, λ_α < λ_p = λ_n < λ_e

Disclaimer: None of the given options are correct.

Page No 69:

Question 6:

An electron is moving with an initial velocity v = v₀ $\hat{i}$ and is in a magnetic field B = B₀ $\hat{j}$ . Then it’s de Broglie wavelength
(a) remains constant.
(b) increases with time.
(c) decreases with time.
(d) increases and decreases periodically.

Answer:

$\vec{v} = v_{0} \hat{i} \vec{B} = B_{0} \hat{i}$

Force on a charge moving in a magnetic field.

$\vec{F} = q (\vec{v} \times \vec{B})$

For an electron (e^–),
$\vec{F} = - e (\vec{v} \times \vec{B})$

$\Rightarrow \vec{F} = - e (v_{0} \hat{i} \times B_{0} \hat{j}) \Rightarrow \vec{F} = - e v_{0} B_{0} \hat{k}$
This force is perpendicular to both $\vec{v}$ and $\vec{B} .$
So the magnitude of $\vec{v}$ will not change.
If the magnitude of velocity is constant, then magnitude of momentum (p = mv) will also remain constant.
de-Broglie wavelength, $λ = \frac{h}{m v}$
Here, p = mv will not change
So, λ = remains constant

Hence, the correct answer is option (a).

Page No 69:

Question 7:

An electron (mass m ) with an initial velocity v = v₀ $\hat{i}$ $(v_{0} > 0)$ is in an electric field $E = - E_{0} \hat{i}$ (E₀ = constant > 0). It’s de Broglie wavelength at time t is given by

$(a) \frac{λ_{0}}{(1 + \frac{e E_{0}}{m} \frac{t}{v_{0}})} (b) λ_{0} (1 + \frac{e E_{0} t}{m v_{0}}) (c) λ_{0} (d) λ_{0} t$

Answer:

Initial velocity, $\vec{v} = v_{0} \hat{i}$
Electric field, $\vec{E} = - E_{0} \hat{i}$
Consider the initial de - Broglie wavelength is 'λ₀'
Where, $λ_{0} = \frac{h}{m v_{0}}$
Force on electron is electric field, $\vec{F} = q \vec{E} = - e \vec{E}$

$\Rightarrow \vec{F} = - e (- E_{0} \hat{i}) ..... (1)$

From Newton's II law, $\vec{F} = m \vec{a} ..... (2)$

From equation (1) and (2)

$\begin{array}{l} m \vec{a} = + e E_{0} \hat{i} \\ \vec{a} = + \frac{e E_{0} \hat{i}}{m} \end{array}$
Equation of motion, v = u + at
consider the new velocity of electron = v′
$\begin{array}{l} \Rightarrow v^{'} = v_{0} \hat{i} + (\frac{e E_{0}}{m} \cdot t) \hat{i} \\ \Rightarrow v^{'} = (v_{0} + \frac{e E_{0}}{m} \cdot t) \hat{i} = v_{0} (1 + \frac{e E_{0}}{m v_{0}} \cdot t) \hat{i} \end{array}$

Consider the new wavelength $= λ = \frac{h}{m v^{'}}$
$\begin{array}{l} = \frac{h}{m v_{0} (1 + \frac{e E_{0}}{m v_{0}} \cdot t)} \end{array} = \frac{λ_{ο}}{(1 + \frac{e E_{0}}{m v_{0}} \cdot t)} (∵ λ_{ο} = \frac{h}{m v_{0}})$

Hence, the correct answer is option (a).

Page No 70:

Question 8:

An electron (mass m) with an initial velocity $v = v_{0} \hat{i}$ is in an electric field $E = E_{0} \hat{j}$ . If λ₀ = h/mv₀, it’s de Breoglie wavelength at time t is given by

$(a) λ_{0} (b) λ_{0} \sqrt{1 + \frac{e^{2} E_{0}^{2} t^{2}}{m^{2} {v^{2}}_{0}}} (c) \frac{λ_{0}}{\sqrt{1 + \frac{e^{2} E_{0}^{2} t^{2}}{m^{2} {v^{2}}_{0}}}} (d) \frac{λ_{0}}{(1 + \frac{e^{2} E_{0}^{2} t^{2}}{m^{2} {v^{2}}_{0}})}$

Answer:

Initial velocity, $\vec{v} = v_{0} \hat{i}$

Electric field, $\vec{E} = + {\vec{E}}_{0} \hat{j}$

Initial de Broglie wavelength, $λ = \frac{h}{m v_{0}}$

Force on electron,
$\vec{F} = - e \vec{E} = - e E_{0} \hat{j}$

Acceleration on electron,
$a = \frac{\vec{F}}{m} = \frac{- e E_{0}}{m} \hat{j}$

It is clear that acceleration is acting along negative y-axis.

Initial velocity of electron along x-axis, ${\vec{v}}_{x 0} = v_{o} \hat{i}$

Initial velocity of electron along y-axis, ${\vec{v}}_{y 0} = 0$

Velocity of electron along x-axis at time 't',
${\vec{v}}_{x} = v_{0} \hat{i} + 0 \times t = v_{0} \hat{i}$
Velocity of electron along y-axis, ${\vec{v}}_{y} = 0 + (\frac{- e E_{0}}{m} \hat{j}) t = \frac{- e E_{0}}{m} t \hat{j}$

Resultant velocity at time 't',

$\begin{array}{l} v = \sqrt{v_{x}^{2} + v_{y}^{2}} = \sqrt{v_{0}^{2} + {(\frac{- e E_{0}}{m} t)}^{2}} \\ \Rightarrow v = v_{0} \sqrt{1 + \frac{e^{2} E_{0}^{2} t^{2}}{m^{2} v_{0}^{2}}} \end{array}$
Now, de - Broglie wavelength at time t,
$λ' = \frac{h}{m v}$
$\Rightarrow λ' = \frac{h}{m v_{o} \sqrt{1 + \frac{e^{2} E_{0}^{2} t^{2}}{m^{2} v_{0}^{2}}}}$

$λ' = \frac{λ_{0}}{\sqrt{1 + \frac{e^{2} E_{0}^{2} t^{2}}{m^{2} v_{0}^{2}}}} (as, λ_{0} = \frac{h}{m v_{o}})$

Hence, the correct answer is option (c).

Page No 70:

Question 9:

Relativistic corrections become neccssary when the expression for the kinetic energy $\frac{1}{2} m v^{2}$ , 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 =10 nm
(b) λ =10^–1nm
(c) λ =10^–4nm
(d) λ =10^–6nm

Answer:

de - Broglie wavelength, $λ = \frac{h}{m v}$
So, velocity,
$\Rightarrow v = \frac{h}{m λ}$

In option (a)
λ₁ = 10 nm = 10^–8 m

$\Rightarrow v_{1} = \frac{h}{m λ_{1}}$

⇒ v₁ ~ 10⁵ m/s

(b) λ₂ = 10^–1 nm = 10^–10 m

$\Rightarrow v_{2} = \frac{h}{m λ_{2}}$

⇒ v₂ ~ 10^–1 m/s

(c) λ₃ = 10^–13 m

$\Rightarrow v_{3} = \frac{h}{m λ_{3}}$

⇒ v₃ ~ 10¹⁰ m/s

(d) λ₄ = 10^–15 m

$\Rightarrow v_{4} = \frac{h}{m λ_{4}}$

⇒ v₄ ~ 10¹² m/s

Only v₃ and v₄ are greater than 3 × 10⁸ m/s

Hence, the correct options are (c) and (d).

Page No 70:

Question 10:

Two particles A₁ sand 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:

de - Broglie wavelength, $λ = \frac{h}{m v} = \frac{h}{p}$
$\Rightarrow p = \frac{h}{λ}$
So, $p \propto \frac{1}{λ} \Rightarrow \frac{p_{1}}{p_{2}} = \frac{λ_{2}}{λ_{1}}$

But λ₁ = λ₂ = λ

Then, p₁ = p₃ .....(1)

Thus, their momenta is same.

Also, kinetic energy,
K.E = $\frac{P^{2}}{2 m}$

From equation (1), 'p' is constant

$K . E \propto \frac{1}{m} ∴ \frac{K . E_{1}}{K . E_{2}} = \frac{m_{2}}{m_{1}} < 1 \Rightarrow K . E_{1} < K . E_{2}$

Hence, the correct options are (a) and (c).

Page No 70:

Question 11:

The de Broglie wavelength of a photon is twice the de Broglie wavelength of an electron. The speed of the electron is v_e = $\frac{c}{100}$ Then
$(a) \frac{E_{e}}{E_{p}} = 10^{- 4} (b) \frac{E_{e}}{E_{p}} = 10^{- 2} (c) \frac{p_{e}}{m_{e} c} = 10^{- 2} (d) \frac{p_{e}}{m_{e} c} = 10^{- 4}$

Answer:

For electron: $λ_{e} = \frac{h}{m_{e} v_{e}} = \frac{h}{m_{e} (\frac{c}{100})} = \frac{100 h}{m_{e} c} ..... (1)$
Also, $λ_{e} = \frac{h}{\sqrt{2 m_{e} E_{e}}}$

$\Rightarrow Kinetic energy, E_{e} = \frac{h^{2}}{2 λ_{e}^{2} m_{e}} ..... (2)$

For photon,

$E_{P} = \frac{h c}{λ_{p}} = \frac{h c}{2 λ_{e}} (∵ λ_{p} = 2 λ_{e}) . . . . . (3)$

From equation (2) and (3)

$\Rightarrow \frac{E_{e}}{E_{p}} = \frac{h}{c λ_{e} m_{e}} = \frac{h}{c m_{e} (\frac{100 h}{m_{e} c})} = \frac{1}{100} \Rightarrow \frac{E_{e}}{E_{p}} = 10^{- 2}$

For electron, momentum $p_{e} = m_{e} v_{e} = m_{e} \times \frac{c}{100}$
So, $\frac{p_{e}}{m_{e} c} = \frac{1}{100} = 10^{- 2}$

Hence, the correct options are (b) and (c).

Page No 71:

Question 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:

Energy spent to convert ice into water = mL

= mass × Latent heat

= (1000 g) × (80 cal/g)

= 80000 cal.

Energy of photons used = nTE = nT × hv

So, nThv = mL

$\Rightarrow T = \frac{m L}{n h v}$

$∴ T \propto \frac{1}{n},$ when v is constant

$T \propto \frac{1}{v},$ when n is constant

$\Rightarrow T \propto \frac{1}{n^{v}}$

Thus, T is constant, if nv is constant.

Hence, the correct answer are option A, B and C.

Page No 71:

Question 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 wave length is λ₁,the particle is nearer the origin than when its value is λ₂.
(d) When the de Broglic wavelength is λ₂, the particle is nearer the origin than when its value is λ₁.

Answer:

Consider the particle moving in the ellipticle path as shown below:

Let v₁, v₂ be the speed of particle at positions A and P respectively and origin is at focus O. If λ₁ and λ₂ are the de - Broglie wavelengths associated with particle while moving at points A and P respectively.
Then,

$λ_{1} = \frac{h}{m v_{1}}$

and $λ_{2} = \frac{h}{m v_{2}}$

$∴ \frac{λ_{1}}{λ_{2}} = \frac{v_{2}}{v_{1}}$
By law of conservation of angular momentum, the particle moves faster when it is closer to focus).
The origin O is closed to P than A.
v₂ > v₁
So, λ₁ > λ₂
Hence, we can say that when the de Broglic wavelength is λ₂, the particle is nearer the origin than when its value is λ₁and the particle could be moving in an elliptic orbit with origin as its focus as shown in the diagram.

Hence, the correct options are (b) and (d).

Page No 71:

Question 14:

A proton and an α-particle are accelerated, using the same potential difference. How are the deBroglie wavelengths λ_p and λ_α related to each other?

Answer:

de - Broglie wavelength, $λ = \frac{h}{p} = \frac{h}{\sqrt{2 m K . E}}$
Kinetic energy, K.E = qV

$\Rightarrow λ = \frac{h}{\sqrt{2 m q V}} ∴ λ \propto \frac{1}{\sqrt{m q}} \frac{λ_{p}}{λ_{α}} = \frac{\sqrt{m_{α} q_{α}}}{\sqrt{m_{p} q_{p}}} = \frac{\sqrt{4 m_{p} \times 2_{e}}}{\sqrt{m_{p} \times e}} = \sqrt{8} ∴ λ_{p} = \sqrt{8} λ_{α}$
i.e., the deBroglie wavelength of proton is $\sqrt{8}$ times the wavelength of α-particle.

Page No 71:

Question 15:

(i) In the explanation of photo electric effect, we asssume one photon of frequency ν collides with an electron and transfers its energy. This leads to the equation for the maximum energy
E_max of the emitted electron as
E_max = hν – $ϕ_{0}$
where $ϕ_{0}$ 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) It is given that the electron absorbs two photons each of frequency v then
The energy absorbed by the electron = hv+hv = 2hv

the, maximum energy for emitted electrons is
E_max = 2hv – Ï•₀ (Work function Ï•₀ will remain unchanged)

(ii) The probability of absorbing 2 photon by the same electron is very low. Hence, such emission will be negligible.

Page No 72:

Question 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:

According to first statement, when the materials which absorb photons of shorter wavelength has the energy of the incident photon on the material is high and the energy of emitted photon in low when it has a higher wavelength.
But, in second statement, the energy of the incident photon is low for the substances which has to absorbs photons of larger wavelength and energy of emitted photon is high to emit light of shorter wavelength. This means in this statement material has to supply the energy for the emission of photons.
But this is not possible for a stable substance.

Page No 72:

Question 17:

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

Answer:

When photon incident on the metal surface, the most of the electrons get scattered into the metal by absorbing a photon.
Thus, all the electrons that absorbs a photon does not come out of the surface of the metal.

Page No 72:

Question 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 number of photons of X-rays to the photons of visible light of the given wavelength?

Answer:

Let the wavelength of X-Rays is 'λ₁' and the wavelength of visible light is 'λ₂'
Given,
P = 100 W
λ₁ = 1nm
λ₂ = 500 nm
n₁ = Number of photons of X-Rays emitted per second
n₂ = Number of photons of visible emitted per second
So power,
$P = \frac{E}{t} = n_{1} \frac{h c}{λ_{1}} = n_{2} \frac{h c}{λ_{2}}$

$\Rightarrow \frac{n_{1}}{λ_{1}} = \frac{n_{2}}{λ_{2}} \Rightarrow \frac{n_{1}}{n_{2}} = \frac{λ_{1}}{λ_{2}} = \frac{1}{500}$

Page No 72:

Question 19:

Consider Fig.11.1 for photoemission.
How would you reconcile with momentum-conservation? Note light (photons) have momentum in a different direction than the emitted electrons.

Answer:

When a photon is incident on a metal surface. The momentum of photon gets transferred to the electron.
In result of which the electron gets excited and if the frequency of incident radiation is greater than threshold frequency, then the excited electron is emitted. Therefore, due to the conservation of momentum, the momentum of incident photon transferred to the nucleus and electrons.

Page No 72:

Question 20:

Consider a metal exposed to light of wavelength 600 nm. The maximum energy of the electron doubles when light of wavelength 400 nm is used. Find the work function in eV.

Answer:

Case I
Wavelength of light, λ = 600 nm

Case II
Wavelength of light, λ′ = 400 nm

It is given, maximum Kinetic energy for the second condition is equal to the twice of the Kinetic energy in first condition.

i.e., K′_max = 2K_max

Here, $K'_{\max} = \frac{h c}{λ'} - ϕ_{0}$

$\Rightarrow 2 K_{\max} = \frac{h c}{λ'} - ϕ_{0} \Rightarrow 2 (\frac{h c}{λ} - ϕ_{0}) = \frac{h c}{λ'} - ϕ_{0} \Rightarrow ϕ = \frac{1230}{1200} = 1.02 eV$

Page No 72:

Question 21:

Assuming an electron is confined to a 1nm wide region , find the uncertainty in momentum using Heisenberg Uncertainty principle (Ref Eq 11.12 of NCERT Textbook). You can assume the uncertainty in position âˆ†x as 1nm. Assuming pâ‰ƒâˆ†p, find the energy of the electron in electron volts.

Answer:

Given, âˆ†x = 1nm = 10^–9m

âˆ†P = ?

$∆ P \times ∆ x ≃ \frac{h}{2 π} \Rightarrow ∆ P ≃ \frac{h}{2 π ∆ x} = \frac{6.6 \times 10^{- 34}}{2 \times 3.14 \times 10^{- 9}}$

So, âˆ†P = 1.05 × 10^–25 kg ms^–1

Now energy of electron,
$K = \frac{P^{2}}{2 m} = \frac{{(∆ P)}^{2}}{2 m} = \frac{{(1.05 \times 10^{- 25})}^{2}}{2 \times 9.8 \times 10^{- 31}} J [∵ P ≃ ∆ P]$

⇒ K = 3.8 × 10^–2 eV

Page No 72:

Question 22:

Two monochromatic beams A and B of equal intensity I, hit a screen. The number of photons hitting the screen by beam A is twice that by beam B. Then what inference can you make about their frequencies?

Answer:

According to the question,

n_A = 2n_B
It is given,
Intensity of A = intensity of B

$n_{A} ν_{A} = n_{B} ν_{B} (As, frequency = ν) \Rightarrow \frac{v_{A}}{v_{B}} = \frac{n_{B}}{n_{A}} = \frac{1}{2} \Rightarrow v_{B} = 2 v_{A}$

Thus, the frequency of beam B is twice of beam A.

Page No 73:

Question 23:

Two particles A and B of de Broglie wavelengths λ₁ and λ₂ combine to form a particle C. The process conserves momentum. Find the de Broglie wavelength of the particle C. (The motion is one dimensional).

Answer:

Conservation of momentum,

|P_A| + |P_B| = |P_C| ....(1)

From de - Broglie hypothesis, $λ = \frac{h}{P}$

$\Rightarrow P = \frac{h}{λ}$

From equation(1)

$\Rightarrow \frac{h}{λ_{A}} + \frac{h}{λ_{B}} = \frac{h}{λ_{C}} \Rightarrow λ_{C} = \frac{λ_{A} λ_{B}}{λ_{A} + λ_{B}}$

Case I : Both P_A and P_B are positive.

then, $λ_{C} = \frac{λ_{A} λ_{B}}{λ_{A} + λ_{B}}$

Case II : P_A is positive and P_B is negative

then, $λ_{C} = \frac{λ_{A} λ_{B}}{λ_{B} - λ_{A}}$

Case III : P_A is negative and P_B is positive

$λ_{C} = \frac{λ_{A} λ_{B}}{λ_{A} - λ_{B}}$

Case IV : Both P_A and P_B are negative

$λ_{C} = \frac{λ_{A} λ_{B}}{λ_{A} + λ_{B}}$

Page No 73:

Question 24:

A neutron beam of energy E scatters from atoms on a surface with a spacing d = 0.1nm. The first maximum of intensity in the reflected beam occurs at θ = 30°. What is the kinetic energy E of the beam in eV?

Answer:

Given, d = 0.l nm

θ = 30°, when n = 1

⇒ 2dsinθ = nλ

⇒ λ = 10^–10 m

Now, $λ = \frac{h}{m v} = \frac{h}{P}$

$\Rightarrow P = \frac{h}{λ} = 6.62 \times 10^{- 24} kg {ms}^{- 1} \Rightarrow K . E = \frac{P^{2}}{2 m} \Rightarrow K . E = 0.21 eV$

Page No 73:

Question 25:

Consider a thin target (10^–2m square, 10^–3m thickness) of sodium, which produces a photocurrent of 100 μA when a light of intensity 100 W/m² (λ = 660 nm) falls on it. Find the probability that a photoelectron is produced when a photons strikes a sodium atom. [Take density of Na = 0.97 kg/m³].

Answer:

Given, A = 10^–4 m², d = 10^–3 m, i = 10^–4A
So, volume of the target, V = Ad =10^-7 m³

Intensity, I = 100 W/m²

Wavelength, λ = 660 × 10^–9m

density of Na = ρ_Na= 0.97 Kg/m³

Mass of 1 mole $(6.022 \times 10^{23})$ Na atoms = 23 g = 0.023 kg

Volume occupied by one Na atom $= \frac{0.023}{0.97 \times 6.022 \times 10^{23}} = 3.95 \times 10^{- 26} m^{3}$

Number of Na atoms in the target $= \frac{10^{- 7}}{3.95 \times 10^{- 26}} = 2.53 \times 10^{18}$ (volume of the target, V = 10^-7 m³)

Let 'n' be the number of photons falling per second on the target.

Total energy falling on the target, $\frac{n h c}{λ} = I A$

$∴ n = \frac{I A λ}{n c}$

⇒ n = 3.3 × 10¹⁶

Let 'p' be the probability of emission per atom per photon.

So, the number of photoelectrons emitted per second
N $= P \times 3.1 \times 10^{16} \times 2.53 \times 10^{18}$

Now, according to question,

i = 100 µA = 10^–4A

Current, i = Ne

⇒10^–4= P × 3.3 × 10¹⁶ × 2.53 × 10¹⁸ × 1.6 × 10^–19

⇒ P $\approx$ 8 × 10^–21

Thus, the probability of emission by a single photon on a single atom is very much less than 1. It is due to this reason, the absorption of two photons by an atom is negligible.

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Question 26:

Consider an electron in front of metallic surface at a distance d (treated as an infinite plane surface). Assume the force of attraction by the plate is given as $\frac{1}{4} \frac{q^{2}}{4 π \in_{0} d^{2}}$
Calculate work in taking the charge to an infinite distance from the plate. Taking d = 0.1nm, find the work done in electron volts. [Such a force law is not valid for d < 0.1nm].

Answer:

d = 0.1 nm = 10^–10m

$F = \frac{q^{2}}{4 \times 4 π ε_{0} d^{2}}$

Let the electron be at distance x from metallic surface.

Then force of attraction, $F_{x} = \frac{q^{2}}{4 \times 4 π ε_{0} x^{2}}$

Work done by external agency in taking charge from distance 'd' to '∞'.

$W = \int_{d}^{\infty} F d x = \int_{d}^{\infty} (\frac{q^{2}}{4 \times 4 π ε_{0}}) \times \frac{d x}{x^{2}} \Rightarrow W = \frac{q^{2}}{4 \times 4 π ε_{0}} \times \frac{1}{d} = \frac{{(1.6 \times 10^{- 19})}^{2} \times 9 \times 10^{9}}{4 \times (10^{- 10})} J \Rightarrow W = \frac{{(1.6 \times 10^{- 19})}^{2} \times 9 \times 10^{9}}{4 \times (10^{- 10}) \times 1.6 \times 10^{- 19}} eV \Rightarrow W = 3.6 eV$

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Question 27:

A student performs an experiment on photoelectric effect, using two materials A and B. A plot of V_stopvs ν is given in Fig. 11.2.

(i) Which material A or B has a higher work function?
(ii) Given the electric charge of an electron = 1.6 × 10^–19 C, find the value of h obtained from the experiment for both A and B.
Comment on whether it is consistent with
Einstein’s theory:

Answer:

(i) Threshold frequency for metal A,
v_oA = 5 × 10¹⁴Hz.
For B,
v_oB = 10¹⁵Hz

Work function, Ï• = hv₀ or Ï•₀ = v₀

So, $\frac{ϕ_{O A}}{ϕ_{O B}} = \frac{5 \times 10^{14}}{10 \times 10^{4}} < 1$
Stopping potential = 0 at a higher frequency for B. Hence it has a higher work function.

(ii) For metal A, Slope $= \frac{h}{e} = \frac{2}{(10 - 5) \times 10^{14}}$

⇒ h = 6.4 × 10^–34 Js

For metal B, slope $= \frac{h}{e} = \frac{2.5}{(15 - 10) \times 10^{14}}$

⇒ h = 8 × 10^–34 Js.

Value of 'h' are different. Hence, the experiment is not consistent with theory

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Question 28:

A particle A with a mass m_A is moving with a velocity v and hits a particle B (mass m_B) at rest (one dimensional motion). Find the change in the de Broglie wavelength of the particle A. Treat the collision as elastic.

Answer:

The collision is elastic.

According to law of conservation of momentum.

$m_{A} v_{A} + m_{B} \times 0 = m_{A} v_{1} + m_{B} v_{2} (v_{A} = v and the body B is at rest initially)$

According to law of conservation of kinetic energy

$\frac{1}{2} m_{A} v^{2} = \frac{1}{2} m_{A} v_{1}^{2} + \frac{1}{2} m_{B} v_{2}^{2} . . . . . (1) \Rightarrow m_{A} (v - v_{1}^{2}) = m_{B} v_{2}^{2} . . . . . (2)$

Solving the above equations we get:

⇒v + v₁ = v₂
or, v = v₂ – v₁ .....(3)

From (1) and (3)

$v_{1} = (\frac{m_{A} - m_{B}}{m_{A} + m_{B}}) v and v_{2} = (\frac{2 m_{A}}{m_{A} + m_{B}}) v So, inatially de - Broglie wavelength of A, λ_{i} = \frac{h}{m_{A} v} and finally de - Broglie wavelength of A, λ_{f} = \frac{h}{m_{A} v_{1}} = \frac{h (m_{A} + m_{B})}{m_{A} (m_{A} - m_{B}) v} ∆ λ = λ_{f} - λ_{i} = \frac{h}{m_{A} v} [\frac{m_{A} + m_{B}}{m_{A} - m_{B}} - 1]$

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Question 29:

Consider a 20 W bulb emitting light of wavelength 5000 Å and shining on a metal surface kept at a distance 2 m. Assume that the metal surface has work function of 2 eV and that each atom on the metal surface can be treated as a circular disk of radius 1.5 Å.

(i) Estimate no. of photons emitted by the bulb per second. [Assume no other losses]
(ii) Will there be photoelectric emission?
(iii) How much time would be required by the atomic disk to receive energy equal to work function (2 eV)?
(iv) How many photons would atomic disk receive within time duration calculated in (iii) above?
(v) Can you explain how photoelectric effect was observed instantaneously?
[Hint: Time calculated in part (iii) is from classical consideration and you may further take the target of surface area say 1cm² and estimate what would happen?]

Answer:

Given, p = 20 W,
λ = 5000 Å = 5000 × 10^–10 m
d = 2 m,
Ï•₀ = 2 eV,
r = 1.5 × 10^–10 m

(i) Number of photon emitted by bulb per second is $n' = \frac{d n}{d t} = \frac{p}{(\frac{h c}{λ})} = \frac{p λ}{h c}$
⇒ n′ = 5 × 10¹⁹ s^–1.

(ii) Energy of incident radiation $= \frac{h c}{λ} = 2.48 eV$
As this energy is greater than the work function of the surface, i.e, 2 eV, hence the photoelectric emission will take place.

(iii) Let 'âˆ†t' be the time spent in getting the energy Ï• = work function of metal from the figure.

$\frac{p}{4 π d^{2}} \times π r^{2} ∆ t = ϕ_{0} \Rightarrow ∆ t = \frac{4 ϕ_{0} d^{2}}{p r^{2}} Now, putting the required values we get : ∆ t ≃ 28.4 s$

(iv) Number of photoelectrons received by disc in time 'âˆ†t'

$N = \frac{n' \times π r^{2}}{4 π d^{2}} ∆ t ≃ 2$

(v) As time of emission of electrons is 11.04 s.
Hence, the photoelectric emission is not instantaneous in this problem.

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