Dual Nature of Radiation and Matter — Page 14 | JEE Physics

Q131

A proton and an electron have the same de Broglie wavelength. If

\mathrm{K}_{\mathrm{p}}

and

\mathrm{K}_{\mathrm{e}}

be the kinetic energies of proton and electron respectively, then choose the correct relation :

A

\mathrm{K_p>K_e}

B

\mathrm{K_p=K_e}

C

\mathrm{K}_{\mathrm{p}}<\mathrm{K}_{\mathrm{e}}

D

\mathrm{K}_{\mathrm{p}}=\mathrm{K}_{\mathrm{e}}{ }^2

Correct Answer

Option C

Solution

To determine the correct relation between the kinetic energies of a proton (

\mathrm{K}_{\mathrm{p}}

) and an electron (

\mathrm{K}_{\mathrm{e}}

) when they have the same de Broglie wavelength, we need to use the de Broglie wavelength formula:

\lambda = \frac{h}{p}

where $\lambda$ is the de Broglie wavelength,

h

is Planck's constant, and

p

is the momentum of the particle. The momentum

p

of a particle is given by:

p = \sqrt{2mK}

where

m

is the mass of the particle and

K

is its kinetic energy. For a proton and an electron with the same de Broglie wavelength:

\lambda_{\mathrm{p}} = \lambda_{\mathrm{e}}

This implies the momenta should be the same:

p_{\mathrm{p}} = p_{\mathrm{e}}

Thus, we can write:

\sqrt{2 m_{\mathrm{p}} K_{\mathrm{p}}} = \sqrt{2 m_{\mathrm{e}} K_{\mathrm{e}}}

Squaring both sides to eliminate the square roots:

2 m_{\mathrm{p}} K_{\mathrm{p}} = 2 m_{\mathrm{e}} K_{\mathrm{e}}

m_{\mathrm{p}} K_{\mathrm{p}} = m_{\mathrm{e}} K_{\mathrm{e}}

Rearranging to solve for

K_{\mathrm{p}}

in terms of

K_{\mathrm{e}}

:

K_{\mathrm{p}} = \frac{m_{\mathrm{e}}}{m_{\mathrm{p}}} K_{\mathrm{e}}

Since the mass of a proton

m_{\mathrm{p}}

is much greater than the mass of an electron

m_{\mathrm{e}}

:

m_{\mathrm{p}} \gg m_{\mathrm{e}}

This means:

\frac{m_{\mathrm{e}}}{m_{\mathrm{p}}} \ll 1

Therefore, it implies:

K_{\mathrm{p}} So, the correct option is: Option C:

\mathrm{K}_{\mathrm{p}}

Q132

An electron in the ground state of the hydrogen atom has the orbital radius of

5.3 \times 10^{-11} \mathrm{~m}

while that for the electron in third excited state is

8.48 \times 10^{-10} \mathrm{~m}

. The ratio of the de Broglie wavelengths of electron in the ground state to that in the excited state is

A 4

B 3

C 9

D 16

Correct Answer

Option A

Solution

We know,

\lambda = {h \over {mv}}

and

mvr = {{nh} \over {2\pi }}

\Rightarrow mv = {{nh} \over {2\pi r}}

So,

\lambda = {h \over {nh}}2\pi r = 2\pi {r \over n}

\Rightarrow \lambda \propto {r \over n}

We can write,

{{{\lambda _g}} \over {{\lambda _e}}} = \left( {{{{r_g}} \over {{r_e}}}} \right)\left( {{{{n_e}} \over {{n_g}}}} \right)

\Rightarrow {{{\lambda _g}} \over {{\lambda _e}}} = \left( {{{5.3 \times {{10}^{ - 11}}} \over {84.8 \times {{10}^{ - 11}}}}} \right)\left( {{4 \over 1}} \right) = {1 \over 4}

\Rightarrow {{{\lambda _e}} \over {{\lambda _g}}} = 4

Q133

A light source of wavelength

\lambda

illuminates a metal surface and electrons are ejected with maximum kinetic energy of 2 eV . If the same surface is illuminated by a light source of wavelength

\dfrac{\lambda}{2}

, then the maximum kinetic energy of ejected electrons will be (The work function of metal is 1 eV )

A 5 eV

B 3 eV

C 2 eV

D 6 eV

Correct Answer

Option A

Solution

Let's break down the problem step by step. The photoelectric equation is given by:

K_{\text{max}} = \frac{hc}{\lambda} - \phi

where:

K_{\text{max}}

is the maximum kinetic energy of the electrons.

\frac{hc}{\lambda}

is the energy of the incident photon. $\phi$ is the work function of the metal.

For the initial light source of wavelength $\lambda$ , we know:

2 \text{ eV} = \frac{hc}{\lambda} - 1 \text{ eV}

Solving for

\frac{hc}{\lambda}

gives:

\frac{hc}{\lambda} = 2 \text{ eV} + 1 \text{ eV} = 3 \text{ eV}

Now, if the wavelength is halved to

\frac{\lambda}{2}

, the photon energy becomes:

\frac{hc}{\lambda/2} = \frac{2hc}{\lambda} = 2 \times 3 \text{ eV} = 6 \text{ eV}

The maximum kinetic energy for the new wavelength is then:

K'_{\text{max}} = 6 \text{ eV} - 1 \text{ eV} = 5 \text{ eV}

Therefore, the maximum kinetic energy of the ejected electrons when the surface is illuminated by a light source of wavelength

\frac{\lambda}{2}

is 5 eV. The correct answer is Option A.

Q134

In photoelectric effect, the stopping potential

\left(\mathrm{V}_0\right) \mathrm{v} / \mathrm{s}

frequency

(v)

curve is plotted. ( h is the Planck's constant and

\phi_0

is work function of metal ) (A)

\mathrm{V}_0 \mathrm{v} / \mathrm{s} v

is linear. (B) The slope of

\mathrm{V}_0 \mathrm{v} / \mathrm{s} v

curve

=\dfrac{\phi_0}{\mathrm{~h}}

(C) h constant is related to the slope of

\mathrm{V}_0 \mathrm{v} / \mathrm{s} v

line. (D) The value of electric charge of electron is not required to determine h using the

\mathrm{V}_0 \mathrm{v} / \mathrm{s} v

curve. (E) The work function can be estimated without knowing the value of

h

. Choose the correct answer from the options given below :

A (A), (C) and (E) only

B (C) and (D) only

C (A), (B) and (C) only

D (D) and (E) only

Correct Answer

Option A

Solution

\begin{aligned} & \mathrm{hv}=\phi+\mathrm{KE}_{\max } \\ & \mathrm{KE}_{\max }=\mathrm{eV}_0 \\ & \mathrm{~V}_0=\frac{\mathrm{hv}-\phi}{\mathrm{e}} \end{aligned}

(A) $\mathrm{V}_0 \mathrm{v} / \mathrm{s} \mathrm{V}$ is linear correct (B) Slope

\mathrm{v}_0=\left(\frac{\mathrm{h}}{\mathrm{e}}\right) \mathrm{v}-\frac{\phi}{\mathrm{e}} \text{ Wrong }

Slope $\dfrac{\mathrm{h}}{\mathrm{e}}$ (C) Correct (D) Incorrect (E) Correct

Q135

An electron of mass ' m ' with an initial velocity

\overrightarrow{\mathrm{v}}=\mathrm{v}_0 \hat{i}\left(\mathrm{v}_0>0\right)

enters an electric field

\overrightarrow{\mathrm{E}}=-\mathrm{E}_{\mathrm{o}} \hat{\mathrm{k}}

. If the initial de Broglie wavelength is

\lambda_0

, the value after time t would be

A

\dfrac{\lambda_o}{\sqrt{1-\dfrac{\mathrm{e}^2 \mathrm{E}_{\mathrm{o}}^2 \mathrm{t}^2}{\mathrm{~m}^2 \mathrm{v}_{\mathrm{o}}^2}}}

B

\lambda_0

C

\dfrac{\lambda_o}{\sqrt{1+\dfrac{\mathrm{e}^2 \mathrm{E}_{\mathrm{o}}^2 \mathrm{t}^2}{\mathrm{~m}^2 v_o^2}}}

D

\lambda_{\mathrm{o}} \sqrt{1+\dfrac{\mathrm{e}^2 \mathrm{E}_{\mathrm{o}}^2 \mathrm{t}^2}{\mathrm{~m}^2 \mathrm{v}_{\mathrm{o}}^2}}

Correct Answer

Option C

Solution

Derivation of the de Broglie wavelength after time

t

The de Broglie wavelength is given by

\lambda = \frac{h}{p},

where

h

is Planck's constant and

p

is the momentum of the particle. Initial Conditions: Initially, the electron has a velocity

\vec{v} = v_0\,\hat{i},

so its momentum is

p_0 = m v_0.

The initial de Broglie wavelength is therefore

\lambda_0 = \frac{h}{m v_0}.

Effect of the Electric Field: The electron enters an electric field

\vec{E} = -E_0\,\hat{k}.

For an electron with charge

-e

, the force is given by

\vec{F} = -e\,\vec{E} = -e\left(-E_0\,\hat{k}\right) = e E_0\,\hat{k}.

This force causes an acceleration in the

\hat{k}

direction:

a_z = \frac{F_z}{m} = \frac{e E_0}{m}.

Momentum Components After Time

t

: In the

\hat{i}

direction, there is no force, so the momentum remains constant:

p_x = m v_0.

In the

\hat{k}

direction, starting from rest, the momentum becomes:

p_z = m v_z = m \left(a_z t\right) = e E_0 t.

Total Momentum of the Electron: The total momentum is the vector sum of the components:

p = \sqrt{p_x^2 + p_z^2} = \sqrt{(m v_0)^2 + (e E_0 t)^2}.

New de Broglie Wavelength: Substituting the expression for the total momentum into the de Broglie relation gives:

\lambda = \frac{h}{\sqrt{m^2 v_0^2 + e^2 E_0^2 t^2}}.

Expressing this in terms of the initial de Broglie wavelength

\lambda_0 = \frac{h}{m v_0}

:

\lambda = \frac{\lambda_0}{\sqrt{1+\frac{e^2 E_0^2 t^2}{m^2 v_0^2}}}.

This final expression matches the option:

\frac{\lambda_0}{\sqrt{1+\frac{e^2 E_0^2 t^2}{m^2 v_0^2}}}.

Q136

A photoemissive substance is illuminated with a radiation of wavelength

\lambda_i

so that it releases electrons with de-Broglie wavelength

\lambda_e

. The longest wavelength of radiation that can emit photoelectron is

\lambda_o

. Expression for de-Broglie wavelength is given by: (m: mass of the electron, h: Planck's constant and c: speed of light)

A

\lambda_e = \dfrac{\sqrt{h \lambda_i}}{\sqrt{2mc}}

B

\lambda_e = \dfrac{h}{\sqrt{2mc \left( \dfrac{1}{\lambda_i} - \dfrac{1}{\lambda_o} \right)}}

C

\lambda_{\mathrm{e}}=\sqrt{\dfrac{\mathrm{h}}{2 \mathrm{mc}\left(\dfrac{1}{\lambda_i}-\dfrac{1}{\lambda_{\mathrm{o}}}\right)}}

D

\lambda_e=\sqrt{\dfrac{h \lambda_0}{2 m c}}

Correct Answer

Option C

Solution

Kinetic Energy (K.E.): The kinetic energy of the emitted electron is given by subtracting the work function (W) of the material from the energy of the incident radiation (E).

$\text{K.E.} = E - W$ Expressions for Energy: The energy of the incoming radiation $E$ is calculated using the formula: $E = \dfrac{hc}{\lambda_i}$ The work function $W$ , which is the minimum energy needed to release an electron, when the radiation wavelength is at its longest $\lambda_0$ , is: $W = \dfrac{hc}{\lambda_0}$ De-Broglie Wavelength of the Electron: The de-Broglie wavelength $\lambda_e$ of the emitted electron is given by: $\lambda_e = \dfrac{h}{\sqrt{2m \cdot \text{K.E.}}}$ Substituting the Values: The kinetic energy of the emitted electron can be expressed in terms of energies as: $\dfrac{h^2}{2m \lambda_e^2} = \dfrac{hc}{\lambda_i} - \dfrac{hc}{\lambda_0}$ Final Expression for $\lambda_e$ : Solving for $\lambda_e$ , the de-Broglie wavelength of the electron, we have: $\lambda_e = \sqrt{\dfrac{h}{2mc\left(\dfrac{1}{\lambda_i} - \dfrac{1}{\lambda_0}\right)}}$ This expression determines the de-Broglie wavelength of the emitted electron based on the wavelengths of the incident and threshold radiation.

Q137

The radiation pressure exerted by a 450 W light source on a perfectly reflecting surface placed at 2 m away from it, is

A

3 \times 10^{-8}

Pascals

B 0

C

1.5 \times 10^{-8}

Pascals

D

6 \times 10^{-8}

Pascals

Correct Answer

Option D

Solution

First, the intensity

I

at a distance

r

from a point light source spreading its energy isotropically is given by

I = \frac{P}{4\pi r^2},

where

P = 450 \, \text{W}

r = 2 \, \text{m}.

So,

I = \frac{450}{4\pi (2)^2} = \frac{450}{16\pi} \approx \frac{450}{50.27} \approx 8.96 \, \text{W/m}^2.

For a perfectly reflecting surface, the radiation pressure

p

is given by

p = \frac{2I}{c},

where

c \approx 3 \times 10^8 \, \text{m/s}

is the speed of light. Substituting the value of

I

:

p = \frac{2 \times 8.96}{3 \times 10^8} \approx \frac{17.92}{3 \times 10^8} \, \text{Pa}.

Calculating the final value:

p \approx 5.97 \times 10^{-8} \, \text{Pa},

which is approximately

6 \times 10^{-8} \, \text{Pa}.

Thus, the correct answer is: Option D:

6 \times 10^{-8}

Pascals.

Q138

Given below are two statements: one is labelled as Assertion A and the other is labelled as Reason

\mathbf{R}

Assertion A : In photoelectric effect, on increasing the intensity of incident light the stopping potential increases. Reason R : Increase in intensity of light increases the rate of photoelectrons emitted, provided the frequency of incident light is greater than threshold frequency. In the light of the above statements, choose the correct answer from the options given below

A Both

\mathbf{A}

and

\mathbf{R}

are true and

\mathbf{R}

is the correct explanation of

\mathbf{A}

B

\mathbf{A}

is false but

\mathbf{R}

is true

C Both

\mathbf{A}

and

\mathbf{R}

are true but

\mathbf{R}

is NOT the correct explanation of

\mathbf{A}

D

\mathbf{A}

is true but

\mathbf{R}

is false

Correct Answer

Option B

Solution

In the photoelectric effect, the stopping potential ( $V_s$ ) is determined by the equation: $V_s = \dfrac{hv - \phi}{e}$ where $h$ is Planck's constant, $v$ is the frequency of the incident light, $\phi$ is the work function of the material, and $e$ is the charge of the electron.

This shows that the stopping potential depends on the frequency of the incident light and the work function, not on the intensity of the light.

The intensity of light ( $I$ ) affects the number of photons hitting the surface per second, given by: $I = \dfrac{\eta hv}{A}$ where $\eta$ is the efficiency of photoelectron emission and $A$ is the area.

Increasing the intensity increases the number of photons per second ( $n$ ), which in turn increases the number of emitted electrons.

However, the stopping potential remains unaffected by changes in intensity; it is solely influenced by the frequency of the incident light relative to the work function of the material.

Q139

A photoelectric surface is illuminated successively by monochromatic light of wavelengths

\lambda

and

{\lambda \over 2}.

If the maximum kinetic energy of the emitted photoelectrons in the second case is 3 times that in the first case, the work function of the surface is :

A

{{hc} \over {3\lambda }}

B

{{hc} \over {2\lambda }}

C

{{hc} \over {\lambda }}

D

{3\,{hc} \over {\lambda }}

Correct Answer

Option B

Solution

We know, Einstein's photo electric equation, (KE)max =

{{hc} \over \lambda }

$-$ $\phi$ 0 In first case, K =

{{hc} \over \lambda }

$-$ $\phi$ 0 . . .(1) In second case, 3K =

{{2hc} \over \lambda }

$-$ $\phi$ 0 . . .(2) $\Rightarrow$

3\left( {{{hc} \over \lambda } - {\phi _0}} \right)

=

{{2hc} \over \lambda } - {\phi _0}

$\Rightarrow$

{{3hc} \over \lambda }

$-$ 3 $\phi$ 0 =

{{2hc} \over \lambda }

$-$ $\phi$ 0 $\Rightarrow$

{{hc} \over \lambda }

= 2 $\phi$ 0 $\Rightarrow$ $\phi$ =

{{hc} \over {2\lambda }}

Q140

When photons of wavelength

{\lambda _1}

are incident on an isolated sphere, the corresponding stopping potential is found to be V. When photons of wavelength

{\lambda _2}

are used, the corresponding stopping potential was thrice that of the above value. If light of wavelength

{\lambda _3}

is used then find the stopping potential for this case :

A

{{hc} \over e}\left[ {{1 \over {{\lambda _3}}} - {1 \over {{\lambda _2}}} - {1 \over {{\lambda _1}}}} \right]

B

{{hc} \over e}\left[ {{1 \over {{\lambda _3}}} + {1 \over {{\lambda _2}}} - {1 \over {{\lambda _1}}}} \right]

C

{{hc} \over e}\left[ {{1 \over {{\lambda _3}}} + {1 \over {2{\lambda _2}}} - {3 \over {2{\lambda _1}}}} \right]

D

{{hc} \over e}\left[ {{1 \over {{\lambda _3}}} + {1 \over {2{\lambda _2}}} - {1 \over {{\lambda _1}}}} \right]

Correct Answer

Option C

Solution

We know, Einstein's photoelectric equation,

eV = {{hc} \over \lambda } - {\phi _0}

and

{\phi _0}

,

{{hc} \over {{\lambda _0}}}

, where

{{\lambda _0}}

is the threashhold wavelength. $\therefore$ In first case, eV

= {{hc} \over {{\lambda _1}}} - {{hc} \over {{\lambda _0}}}

. . .(1) and in second case, 3 eV

= {{hc} \over {{\lambda _2}}} - {{hc} \over {{\lambda _0}}}

. . .(2) Now, let slopping potential = V1 when light of wavelength $\lambda$ 3 is used then, eV1

= {{hc} \over {{\lambda _3}}} - {{hc} \over {{\lambda _0}}}

. . .(3) From (1) and (2) get,

3\left( {{{hc} \over {{\lambda _1}}} - {{hc} \over {{\lambda _0}}}} \right) = {{hc} \over {{\lambda _2}}} - {{hc} \over {{\lambda _0}}}

$\Rightarrow$

{{3hc} \over {{\lambda _1}}}

$-$

{{hc} \over {{\lambda _2}}}

$=$

{{2hc} \over {{\lambda _0}}}

$\Rightarrow$

{{hc} \over {{\lambda _0}}}

$=$

{{3hc} \over {2{\lambda _1}}}

$-$

{{hc} \over {2{\lambda _2}}}

Putting this value of

{{hc} \over {{\lambda _0}}}

in equation 3, eV1

= {{hc} \over {{\lambda _3}}} - {{3hc} \over {2{\lambda _1}}} + {{hc} \over {2{\lambda _2}}}

$\Rightarrow$ V1

= {{hc} \over e}

\left[ {{1 \over {{\lambda _3}}} - {3 \over {2{\lambda _1}}} + {1 \over {2{\lambda _2}}}} \right]