Dual Nature of Radiation and Matter — Page 13 | JEE Physics

Q121

The threshold frequency of a metal is

f_{0}

. When the light of frequency

2 f_{0}

is incident on the metal plate, the maximum velocity of photoelectrons is

v_{1}

. When the frequency of incident radiation is increased to

5 \mathrm{f}_{0}

, the maximum velocity of photoelectrons emitted is

v_{2}

. The ratio of

v_{1}

to

v_{2}

is :

A

\dfrac{v_{1}}{v_{2}}=\dfrac{1}{2}

B

\dfrac{v_{1}}{v_{2}}=\dfrac{1}{16}

C

\dfrac{v_{1}}{v_{2}}=\dfrac{1}{4}

D

\dfrac{v_{1}}{v_{2}}=\dfrac{1}{8}

Correct Answer

Option A

Solution

\begin{aligned} & \frac{1}{2} m v^2=h f-h f_0 \\\\ & \Rightarrow \frac{1}{2} m v_1^2=2 h f_0-h f_0=h f_0 \\\\ & \text{ also, } \frac{1}{2} m v_2^2=5 h f_0-h f_0=4 h f_0 \end{aligned}

taking ratio,

\frac{v_1^2}{v_2^2}=\frac{1}{4} \Rightarrow \frac{v_1}{v_2}=\frac{1}{2}

Q122

An electron accelerated through a potential difference

V_{1}

has a de-Broglie wavelength of

\lambda

. When the potential is changed to

V_{2}

, its de-Broglie wavelength increases by

50 \%

. The value of

\left(\dfrac{V_{1}}{V_{2}}\right)

is equal to

A

\dfrac{3}{2}

B 4

C 3

D

\dfrac{9}{4}

Correct Answer

Option D

Solution

P = \sqrt {2\,eVm}

\lambda = \left( {{h \over {{P_1}}}} \right)

..... (i)

{{3\lambda } \over 2} = {h \over {{P_2}}}

..... (ii) Dividing (i) by (ii)

\Rightarrow {2 \over 3} = \left( {{{{P_2}} \over {{P_1}}}} \right) = \sqrt {{{{v_2}} \over {{v_1}}}}

\Rightarrow {4 \over 9} = \left( {{{{v_2}} \over {{v_1}}}} \right)

{{{v_1}} \over {{v_2}}} = \left( {{9 \over 4}} \right)

Q123

The kinetic energy of emitted electron is E when the light incident on the metal has wavelength

\lambda

. To double the kinetic energy, the incident light must have wavelength:

A

\dfrac{\mathrm{hc}}{\mathrm{E} \lambda-\mathrm{hc}}

B

\dfrac{\mathrm{hc} \lambda}{\mathrm{E} \lambda+\mathrm{hc}}

C

\dfrac{\mathrm{h} \lambda}{\mathrm{E} \lambda+\mathrm{hc}}

D

\dfrac{\text{ hc } \lambda}{\mathrm{E} \lambda-\mathrm{hc}}

Correct Answer

Option B

Solution

k = {{hc} \over \lambda } - \phi = E

and,

2k = {{hc} \over {{\lambda _2}}} - \phi = 2E

\Rightarrow {{hc} \over \lambda } - E = {{hc} \over {{\lambda _2}}} - 2E

\Rightarrow {{hc} \over {{\lambda _2}}} = {{hc} \over \lambda } + E

\Rightarrow {\lambda _2} = {{hc\lambda } \over {hc + \lambda E}}

Q124

From the photoelectric effect experiment, following observations are made. Identify which of these are correct. A. The stopping potential depends only on the work function of the metal. B. The saturation current increases as the intensity of incident light increases. C. The maximum kinetic energy of a photo electron depends on the intensity of the incident light. D. Photoelectric effect can be explained using wave theory of light. Choose the correct answer from the options given below :

A A, B, D only

B A, C, D only

C B, C only

D B only

Correct Answer

Option D

Solution

(A) From Einstein's equation

K_{\max }=e V_{s}=h v-\phi

Form the stopping potential $\left(\mathrm{V}_{\mathrm{s}}\right)$ depends on $\phi$ $\& \,v$ .

(B) Saturation current is proportional to intensity, i.e., number of incident photons.

(C) $K_{\max }$ only depends on nature of photon and $\phi$ .

(D) Einstein used particle behaviour of photon to explain photon electric effect.

Only B is correct.

Q125

If the total energy transferred to a surface in time

\mathrm{t}

is

6.48 \times 10^5 \mathrm{~J}

, then the magnitude of the total momentum delivered to this surface for complete absorption will be:

A

2.16 \times 10^{-3} \mathrm{~kg} \mathrm{~m} / \mathrm{s}

B

2.46 \times 10^{-3} \mathrm{~kg} \mathrm{~m} / \mathrm{s}

C

1.58 \times 10^{-3} \mathrm{~kg} \mathrm{~m} / \mathrm{s}

D

4.32 \times 10^{-3} \mathrm{~kg} \mathrm{~m} / \mathrm{s}

Correct Answer

Option A

Solution

\mathrm{p=\frac{E}{C}=\frac{6.48 \times 10^5}{3 \times 10^8}=2.16 \times 10^{-3}}

Q126

Given below are two statements : Statement I : Stopping potential in photoelectric effect does not depend on the power of the light source. Statement II : For a given metal, the maximum kinetic energy of the photoelectron depends on the wavelength of the incident light. In the light of above statements, choose the most appropriate answer from the options given below

A Both Statement I and Statement II are incorrect

B Statement I is correct but Statement II is incorrect

C Both Statement I and Statement II are correct

D Statement I is incorrect but Statement II is correct

Correct Answer

Option C

Solution

Statement I is correct as stopping potential is independent of power of light used.

Statement II is correct as maximum kinetic energy of photoelectron depends on wavelength of light.

Q127

In photo electric effect A. The photocurrent is proportional to the intensity of the incident radiation B. Maximum Kinetic energy with which photoelectrons are emitted depends on the intensity of incident light. C. Max. K.E with which photoelectrons are emitted depends on the frequency of incident light. D. The emission of photoelectrons require a minimum threshold intensity of incident radiation. E. Max. K.E of the photoelectrons is independent of the frequency of the incident light. Choose the correct answer from the options given below:

A A and E only

B A and B only

C B and C only

D A and C only

Correct Answer

Option D

Solution

The photoelectric effect is the phenomenon of emission of electrons (or photoelectrons) from the surface of a metal when it is illuminated by light of sufficient energy.

The observations from the photoelectric effect led to the development of quantum theory.

According to the principles of the photoelectric effect: A.

The photocurrent (number of photoelectrons ejected per unit time) is indeed proportional to the intensity of the incident radiation.

More intense light means more photons hitting the surface and thus more electrons being ejected.

B.

The maximum kinetic energy of the photoelectrons does not depend on the intensity of the incident light but rather on its frequency.

Increasing the intensity of light increases the number of photoelectrons (current) but does not increase their maximum kinetic energy.

C.

The maximum kinetic energy of the photoelectrons does indeed depend on the frequency of the incident light.

If the frequency of the incident light is below a certain threshold frequency specific to the metal, no photoelectrons are emitted regardless of the intensity of the light.

Above this threshold, the maximum kinetic energy of the photoelectrons increases linearly with the frequency of the light.

D.

The emission of photoelectrons does not require a minimum threshold intensity of incident radiation, but rather a minimum threshold frequency.

E.

The maximum kinetic energy of the photoelectrons is not independent of the frequency of the incident light, but rather depends on it.

Therefore, only statements A and C are correct.

Q128

The de-Broglie wavelength of an electron is the same as that of a photon. If velocity of electron is

25 \%

of the velocity of light, then the ratio of K.E. of electron and K.E. of photon will be:

A

\dfrac{1}{4}

B

\dfrac{8}{1}

C

\dfrac{1}{8}

D

\dfrac{1}{1}

Correct Answer

Option C

Solution

We know that the de-Broglie wavelength $\lambda$ of a particle is given by:

\lambda = \frac{h}{p}

where:

h

is Planck's constant,

p

is the momentum of the particle. For a photon (which has zero rest mass), its energy

E

and momentum

p

are related by the equation:

E = cp

and its de-Broglie wavelength $\lambda$ is given by:

\lambda = \frac{h}{E} \times \frac{1}{c}

Now, since the photon and electron are said to have the same de-Broglie wavelength:

\lambda_{electron} = \lambda_{photon}

\frac{h}{p_{electron}} = \frac{h}{E_{photon}} \times \frac{1}{c}

For the electron, its momentum

p_{electron}

is given by:

p_{electron} = m_{e}v_{electron}

where:

m_{e}

is the electron's rest mass,

v_{electron}

is the electron's velocity. The kinetic energy

K.E.

of the electron is:

K.E._{electron} = \frac{1}{2}m_{e}v_{electron}^2

The question states that

v_{electron}

is

25\%

(

0.25c

) of the speed of light

c

. So we write:

v_{electron} = 0.25c

Plugging this into the kinetic energy formula, we get:

K.E._{electron} = \frac{1}{2}m_{e}(0.25c)^2 = \frac{1}{2}m_{e} \times \frac{1}{16}c^2

K.E._{electron} = \frac{1}{32}m_{e}c^2

For a photon,

p_{photon} = \frac{E_{photon}}{c}

and hence its kinetic energy (which, for a photon, is simply its energy) is:

K.E._{photon} = E_{photon} = cp_{photon}

Now, comparing the kinetic energies:

\frac{K.E._{electron}}{K.E._{photon}} = \frac{\frac{1}{32}m_{e}c^2}{cp_{photon}}

Since

p_{electron} = p_{photon}

(from the de-Broglie relation), we can replace

p_{photon}

with

p_{electron}

which is

m_{e}v_{electron}

:

\frac{K.E._{electron}}{K.E._{photon}} = \frac{\frac{1}{32}m_{e}c^2}{m_{e}v_{electron}c}

\frac{K.E._{electron}}{K.E._{photon}} = \frac{\frac{1}{32}c}{0.25c}

\frac{K.E._{electron}}{K.E._{photon}} = \frac{1}{32} \times \frac{1}{0.25}

\frac{K.E._{electron}}{K.E._{photon}} = \frac{1}{32} \times 4

\frac{K.E._{electron}}{K.E._{photon}} = \frac{1}{8}

So the correct answer is Option C

\frac{1}{8}

.

Q129

Two sources of light emit with a power of

200 \mathrm{~W}

. The ratio of number of photons of visible light emitted by each source having wavelengths

300 \mathrm{~nm}

and

500 \mathrm{~nm}

respectively, will be :

A

5: 3

B

3: 5

C

1: 5

D

1: 3

Correct Answer

Option B

Solution

\begin{aligned} & \mathrm{n}_1 \times \frac{\mathrm{hc}}{\lambda_1}=200 \\ & \mathrm{n}_2 \times \frac{\mathrm{hc}}{\lambda_2}=200 \\ & \frac{\mathrm{n}_1}{\mathrm{n}_2}=\frac{\lambda_1}{\lambda_2}=\frac{300}{500} \\ & \frac{\mathrm{n}_1}{\mathrm{n}_2}=\frac{3}{5} \end{aligned}

Q130

Given below are two statements: one is labelled as Assertion

\mathbf{A}

and the other is labelled as Reason R. Assertion A: Number of photons increases with increase in frequency of light. Reason R: Maximum kinetic energy of emitted electrons increases with the frequency of incident radiation. In the light of the above statements, choose the most appropriate answer from the options given below:

A

\mathbf{A}

is not correct but

\mathbf{R}

is correct.

B

\mathbf{A}

is correct but

\mathbf{R}

is not correct.

C Both

\mathbf{A}

and

\mathbf{R}

are correct and

\mathbf{R}

is the correct explanation of

\mathbf{A}

.

D Both

\mathbf{A}

and

\mathbf{R}

are correct and

\mathbf{R}

is NOT the correct explanation of

\mathbf{A}

.

Correct Answer

Option A

Solution

In order to determine the most appropriate answer to the question, let's analyze the given Assertion A and Reason R in detail.

Assertion A: Number of photons increases with increase in frequency of light.

This statement is not correct.

The number of photons is determined by the intensity (or power) of the light and is given by the formula:

N = \frac{P}{hf}

where

N

is the number of photons per second,

P

is the power (intensity) of the light,

h

is Planck's constant, and

f

is the frequency of the light.

As the frequency increases, the energy per photon increases, but it does not necessarily mean that the number of photons increases unless the power of the light also increases proportionally.

Reason R: Maximum kinetic energy of emitted electrons increases with the frequency of incident radiation.

This statement is correct.

According to the photoelectric effect, the maximum kinetic energy of emitted electrons is given by:

K.E_{\text{max}} = hf - \phi

where

K.E_{\text{max}}

is the maximum kinetic energy of the emitted electrons,

h

is Planck's constant,

f

is the frequency of the incident radiation, and $\phi$ is the work function of the material.

As the frequency of the incident light increases, the kinetic energy of the emitted electrons increases.

Now, let's match the statements with the options given: Option A:

\mathbf{A}

is not correct but

\mathbf{R}

is correct. This option is correct because Assertion A is incorrect while Reason R is correct. Option B:

\mathbf{A}

is correct but

\mathbf{R}

is not correct. This option is incorrect because Assertion A is not correct. Option C: Both

\mathbf{A}

and

\mathbf{R}

are correct and

\mathbf{R}

is the correct explanation of

\mathbf{A}

. This option is incorrect because Assertion A is not correct. Option D: Both

\mathbf{A}

and

\mathbf{R}

are correct and

\mathbf{R}

is NOT the correct explanation of

\mathbf{A}

. This option is incorrect because Assertion A is not correct. Therefore, the most appropriate answer is: Option A:

\mathbf{A}

is not correct but

\mathbf{R}

is correct.