Dual Nature of Radiation and Matter — Page 16 | JEE Physics

Q151

In an experiment with photoelectric effect, the stopping potential,

A is

\left(\dfrac{1}{e}\right)

times the maximum kinetic energy of the emitted photoelectrons

B increases with increase in the intensity of the incident light

C decreases with increase in the intensity of the incident light

D increases with increase in the wavelength of the incident light

Correct Answer

Option A

Solution

We know,

e{V_0} = hv - \phi

.... (i)

\Rightarrow e{V_0} = K{E_{\max }}

(where $\phi$ = work function of the metal and $v_0$ = stopping potential)

\Rightarrow {V_0} = \left( {{1 \over e}} \right)K{E_{\max }}

Stopping potential depends solely on the frequency of the incident light $+(v)$ & the work function of the metal, not the intensity.

From eq. (i), we can see,

v \uparrow \Rightarrow {v_0} \uparrow

\Rightarrow {e \over {\lambda \downarrow }} \Rightarrow {v_0} \uparrow

(as

v = {c \over \lambda }

) So, stopping potential increases with decrease in the wavelength of the incident light. Hence, option 1 is correct.

Q152

Given below are two statements: Statement I : Out of microwaves, infrared rays and ultraviolet rays, ultraviolet rays are the most effective for the emission of electrons from a metallic surface. Statement II : Above the threshold frequency, the maximum kinetic energy of photoelectrons is inversely proportional to the frequency of the incident light. In the light of above statements, choose the correct answer form the options given below

A Both Statement I and Statement II are true

B Statement I is true but statement II is false

C Statement I is false but statement II is true

D Both Statement I and Statement II are false

Correct Answer

Option B

Solution

Now let's analyze both statements: Statement I is correct.

According to the photoelectric effect, the ability to emit electrons from a metallic surface depends on the energy of the incident light.

The energy of a photon is given by

E = h\nu

, where

h

is Planck's constant and $\nu$ is the frequency of the light.

Ultraviolet rays have higher frequencies and thus higher energies compared to microwaves and infrared rays, making them more effective for the emission of electrons from a metallic surface.

Statement II is incorrect.

The maximum kinetic energy of photoelectrons is given by

K_{max} = h\nu - h\nu_0

, where

\nu_0

is the threshold frequency.

This equation shows that the maximum kinetic energy of photoelectrons is directly proportional to the frequency of the incident light (above the threshold frequency), not inversely proportional.

Based on the analysis, the correct answer is: Statement I is true, and Statement II is false.

Q153

Two particles move at right angle to each other. Their de-Broglie wavelengths are

\lambda _1

and

\lambda _2

respectively. The particles suffer perfectly inelastic collision. The de-Broglie wavelength

\lambda _2

of the final particle, is given by :

A

\lambda = {{{\lambda _1} + {\lambda _2}} \over 2}

B

{1 \over {{\lambda ^2}}} = {1 \over {\lambda _1^2}} + {1 \over {\lambda _2^2}}

C

\lambda = \sqrt {{\lambda _1}{\lambda _2}}

D

{2 \over \lambda } = {1 \over {{\lambda _1}}} + {1 \over {{\lambda _2}}}

Correct Answer

Option B

Solution

Let the two particles be moving along x-direction and y-direction. So, the net momentum initially is

\sqrt {{{{h^2}} \over {\lambda _1^2}} + {{{h^2}} \over {\lambda _2^2}}}

and final momentum will be

{h \over \lambda }

. Applying momentum conservation,

{h \over \lambda } = \sqrt {{{{h^2}} \over {\lambda _1^2}} + {{{h^2}} \over {\lambda _2^2}}}

$\Rightarrow$

{1 \over {{\lambda ^2}}} = {1 \over {\lambda _1^2}} + {1 \over {\lambda _2^2}}

Q154

In photoelectric effect an em-wave is incident on a metal surface and electrons are ejected from the surface. If the work function of the metal is 2.14 eV and stopping potential is 2 V , what is the wavelength of the em-wave? (Given

\mathrm{hc}=1242 \mathrm{eVnm}

where h is the Planck's constant and c is the speed of light in vaccum.)

A 400 nm

B 600 nm

C 300 nm

D 200 nm

Correct Answer

Option C

Solution

Let's break down the problem step by step: In the photoelectric effect, the energy of a photon is given by:

h\nu = \phi + K_e

where: $\phi$ is the work function of the metal,

K_e

is the kinetic energy of the ejected electron.

The kinetic energy of the ejected electrons is provided by the stopping potential:

K_e = eV

Given that the stopping potential is

V = 2 \text{ V}

, we have:

K_e = 2 \text{ eV}

Now, we can calculate the photon energy:

h\nu = \phi + K_e = 2.14 \text{ eV} + 2 \text{ eV} = 4.14 \text{ eV}

The relation between the photon's energy and its wavelength is:

h\nu = \frac{hc}{\lambda}

Rearranging for $\lambda$ :

\lambda = \frac{hc}{h\nu}

Given that

hc = 1242 \text{ eV·nm}

, substitute the values:

\lambda = \frac{1242 \text{ eV·nm}}{4.14 \text{ eV}}

Calculate the wavelength:

\lambda \approx \frac{1242}{4.14} \approx 300 \text{ nm}

Thus, the wavelength of the incident electromagnetic wave is approximately 300 nm.

The correct answer is Option C: 300 nm.

Q155

Two identical photocathodes receive the light of frequencies f1 and f2 respectively. If the velocities of the photo-electrons coming out are v1 and v2 respectively, then

A

{v_1} - {v_2} = {\left[ {{{2h} \over m}({f_1} - {f_2})} \right]^{{1 \over 2}}}

B

v_1^2 + v_2^2 = {{2h} \over m}[{f_1} + {f_2}]

C

{v_1} + {v_2} = {\left[ {{{2h} \over m}({f_1} + {f_2})} \right]^{{1 \over 2}}}

D

v_1^2 - v_2^2 = {{2h} \over m}[{f_1} - {f_2}]

Correct Answer

Option D

Solution

{1 \over 2}mv_1^2 = h{f_1} - \phi

___________(1)

{1 \over 2}mv_2^2 = h{f_2} - \phi

___________(2) Subtracting equation (1) by equation (2)

{1 \over 2}mv_1^2 - {1 \over 2}mv_2^2 = h{f_1} - h{f_2}

v_1^2 - v_2^2 = {{2h} \over m}({f_1} - {f_2})

Q156

Both the nucleus and the atom of some element arein their respective first excited states. They get de-excted by emitting photons of wavelengths

\lambda

N,

\lambda

A respectively. The ratio

{{{}^\lambda N} \over {{}^\lambda A}}

is closest to :

A 10

-

6

B 10

C 10

-

10

D 10

-

1

Correct Answer

Option A

Solution

We know that

E = {{hc} \over \lambda }

So, for atom

{E_A} = {{hc} \over {{\lambda _A}}}

And for neutron

{E_N} = {{hc} \over {{\lambda _N}}}

Then,

{{{E_A}} \over {{E_N}}} = {{hc} \over {{\lambda _A}}} \times {{{\lambda _N}} \over {hc}} \Rightarrow {{{\lambda _N}} \over {{\lambda _A}}}

Here, EA is order of eV and EN is order of MeV. Therefore,

{{{\lambda _N}} \over {{\lambda _A}}} = {{{E_A}} \over {{E_N}}} = {{eV} \over {MeV}} = {{1eV} \over {{{10}^6}eV}}

{\lambda _N} = {10^{ - 6}}\,{\lambda _A}

Q157

Two sources of light emit X-rays of wavelength 1 nm and visible light of wavelength 500 nm, respectively. Both the sources emit light of the same power 200 W. The ratio of the number density of photons of X-rays to the number density of photons of the visible light of the given wavelengths is :

A

{1 \over {500}}

B 500

C 250

D

{1 \over {250}}

Correct Answer

Option A

Solution

Given, wavelength of x ray ( $\lambda$ 1) = 1 nm And wavelength of visible light ( $\lambda$ 2) = 500 nm we know, Power

(P) = {{nhc} \over \lambda }

As P = constant and h, c also constant So,

{n \over \lambda } = constant

$\Rightarrow$

{{{n_1}} \over {{n_2}}} = {{{\lambda _1}} \over {{\lambda _2}}} = {{1nm} \over {500nm}} = {1 \over {500}}

Q158

An electron (mass

\mathrm{m}

) with an initial velocity

\vec{v}=v_{0} \hat{i}\left(v_{0}>0\right)

is moving in an electric field

\vec{E}=-E_{0} \hat{i}\left(E_{0}>0\right)

where

E_{0}

is constant. If at

\mathrm{t}=0

de Broglie wavelength is

\lambda_{0}=\dfrac{h}{m v_{0}}

, then its de Broglie wavelength after time t is given by

A

\lambda_{0}

B

\lambda_{0}\left(1+\dfrac{e E_{0} t}{m v_{0}}\right)

C

\lambda_{0} t

D

\dfrac{\lambda_{0}}{\left(1+\dfrac{e E_{0} t}{m v_{0}}\right)}

Correct Answer

Option D

Solution

\text{ At } t=0, \lambda_0=\frac{h}{m v_0}

Since $\vec{v}=v_0 \hat{i}$ and $\overrightarrow{\mathrm{E}}=\mathrm{E}_0 \hat{i}$ its velocity $v$ at any time $t$ is given by

v=v_o+\frac{\varepsilon \mathrm{E}_{\mathrm{o}}}{m} t

De Broglie wavelength $\lambda$ at any time $t$ is given by

\begin{aligned} \lambda & =\frac{h}{m v}=\frac{h}{m\left(v_0+\frac{e \mathrm{E}_0}{m} t\right)} \\\\ & =\frac{h}{m v_0\left(1+\frac{e \mathrm{E}_0}{m v_0} t\right)} \\\\ & =\frac{\lambda_0}{1+\frac{e \mathrm{E}_0}{m v_0} t} \end{aligned}

Q159

A source of monochromatic light liberates 9

\times

1020 photon per second with wavelength 600 nm when operated at 400 W. The number of photons emitted per second with wavelength of 800 nm by the source of monochromatic light operating at same power will be :

A 12

\times

1020

B 6

\times

1020

C 9

\times

1020

D 24

\times

1020

Correct Answer

Option A

Solution

As we know

\begin{aligned} &I=\frac{E}{A t}=\frac{n h v}{A t} \Rightarrow \frac{n}{t}=\frac{I A \lambda}{h C} \Rightarrow \frac{n}{t}=\frac{\rho \lambda}{h c} \Rightarrow \frac{n}{t}=\rho \lambda \\\\ &\Rightarrow\left(\frac{n}{t}\right)_{2}=\left(\frac{n}{t}\right)_{1} \times \frac{p_{2} \lambda_{2}}{p_{1} \lambda 1}=9 \times 10^{20} \times \frac{P}{P} \times \frac{800}{600}=12 \times 10^{20} \end{aligned}

Q160

An electron with mass ' m ' with an initial velocity

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

enters a magnetic field

\overrightarrow{\mathrm{B}}=\mathrm{B}_0 \hat{j}

. If the initial de-Broglie wavelength at

\mathrm{t}=0

is

\lambda_0

then its value after time ' t ' would be :

A

\dfrac{\lambda_0}{\sqrt{1-\dfrac{\mathrm{e}^2 \mathrm{~B}_0^2 \mathrm{t}^2}{\mathrm{~m}^2}}}

B

\lambda_0

C

\lambda_0 \sqrt{1+\dfrac{\mathrm{e}^2 \mathrm{~B}_0^2 \mathrm{t}^2}{\mathrm{~m}^2}}

D

\dfrac{\lambda_0}{\sqrt{1+\dfrac{\mathrm{e}^2 \mathrm{~B}_0^2 \mathrm{t}^2}{\mathrm{~m}^2}}}

Correct Answer

Option B

Solution

Magnetic field does not work $\therefore$ Speed will not charge, so De-Broglie wavelength remains same.