Dual Nature of Radiation and Matter — Page 15 | JEE Physics

Q141

The magnetic field associated with a light wave is given, at the origin, by B = B0 [sin(3.14

\times

107)ct + sin(6.28

\times

107)ct]. If this light falls on a silver plate having a work function of 4.7 eV, what will be the maximum kinetic energy of the photo electrons ? (Take c = 3

\times

108 ms

-

1, h = 6.6

\times

10

-

34J-s)

A 6.82 eV

B 12.5 eV

C 8.52 eV

D 7.72 eV

Correct Answer

Option D

Solution

Given that, B = B0[sin (3.14 $\times$ 107) ct + sin(6.28 $\times$ 107) ct] This light wave is non-monochromotic wave as here is two different frequency in the magnetic field.

Given work function ( $\phi$ ) = 4.7 eV Question says to find the maximum kinetic energy, so we have to use maximum frequency among the available frequency.

As, Kmax = Emax $-$ $\phi$ Emax = hF = h $\times$

{\omega \over {2\pi }}

= 6.6 $\times$ 10 $-$ 34 $\times$

{{6.28 \times {{10}^7} \times 3 \times {{10}^8}} \over {2\pi }}

= 6.6 $\times$ 3 $\times$ 10 $-$ 19 J =

{{6.6 \times 3 \times {{10}^{ - 19}}} \over {1.6 \times {{10}^{ - 19}}}}eV

= 12.375

eV

$\therefore$ Kmax = 12.375 $-$ 4.7 = 7.675

eV

\simeq

7.7

eV

Q142

Given below are two statements : Statement I : Davisson-Germer experiment establishes the wave nature of electrons. Statement II : If electrons have wave nature, they can interfere and show diffraction. In the light of the above statements choose the correct answer from the option given below :

A Both Statement I and Statement II are true.

B Both Statement I and Statement II are false.

C Statement I is true but Statement II is false.

D Statement I is false but Statement II is true.

Correct Answer

Option A

Solution

Davisson-Germer experiment is done and establishes the wave nature of electrons.

Interference and diffraction establishes wave nature.

Q143

An electron and proton are separated by a large distance. The electron starts approaching the proton with energy 3 eV. The proton captures the electron and forms a hydrogen atom in second excited state. The resulting photon is incident on a photosensitive metal of threshold wavelength 4000

\mathop A\limits^o

. What is the maximum kinetic energy of the emitted photoelectron?

A 7.61 eV

B 1.41 eV

C 3.3 eV

D No photoelectron would be emitted

Correct Answer

Option B

Solution

initially, energy of electron = + 3eV finally, in 2nd excited state, energy of electron =

- {{(13.6eV)} \over {{3^2}}}

= - 1.51eV

Loss in energy is emitted as photon, So, photon energy

{{hc} \over \lambda } = 4.51eV

Now, photoelectric effect equation

K{E_{\max }} = {{hc} \over \lambda } - \phi = 4.51 - \left( {{{hc} \over {{\lambda _{th}}}}} \right)

= 4.51eV - {{12400eV\mathop A\limits^o } \over {4000\mathop A\limits^o }}

= 1.41eV

Q144

The de Broglie wavelength of an electron having kinetic energy

\mathrm{E}

is

\lambda

. If the kinetic energy of electron becomes

\dfrac{E}{4}

, then its de-Broglie wavelength will be :

A

\sqrt{2} \lambda

B

2 \lambda

C

\dfrac{\lambda}{2}

D

\dfrac{\lambda}{\sqrt{2}}

Correct Answer

Option B

Solution

\lambda = \frac{h}{\sqrt{2mE}}

where $h$ is Planck's constant, $m$ is the mass of the particle, and $E$ is its kinetic energy.

We are given that the de Broglie wavelength of an electron with kinetic energy $E$ is $\lambda$ , and we want to find the de Broglie wavelength of the same electron when its kinetic energy becomes $\dfrac{E}{4}$ .

To do this, we can use the formula for the de Broglie wavelength again, but with the new kinetic energy $\dfrac{E}{4}$ :

\lambda' = \frac{h}{\sqrt{2m\left(\frac{E}{4}\right)}} = \frac{2h}{\sqrt{2mE}} = 2\lambda

where we have used the fact that $\sqrt{\dfrac{1}{4}} = \dfrac{1}{\sqrt{4}} = \dfrac{1}{2}$ to simplify the expression.

Therefore, the de Broglie wavelength of the electron when its kinetic energy becomes $\dfrac{E}{4}$ is twice its original value, or $\boxed{2\lambda}$ .

Q145

A proton and an

\alpha

-particle are accelerated from rest by

2 \mathrm{~V}

and

4 \mathrm{~V}

potentials, respectively. The ratio of their de-Broglie wavelength is :

A 4 : 1

B 2 : 1

C 8 : 1

D 16 : 1

Correct Answer

Option A

Solution

The de-Broglie wavelength of a particle is given by:

\lambda = \frac{h}{p}

where

h

is the Planck's constant and

p

is the momentum of the particle. The momentum of a particle of mass

m

and charge

q

accelerated through a potential difference

V

is given by:

p = \sqrt{2 m q V}

For a proton,

m = 1.67 \times 10^{-27} \mathrm{~kg}

and

q = 1.6 \times 10^{-19} \mathrm{~C}

, and for an $\alpha$ -particle,

m = 6.64 \times 10^{-27} \mathrm{~kg}

and

q = 2 \times 1.6 \times 10^{-19} \mathrm{~C}

. Therefore, the ratio of their de-Broglie wavelengths is:

\frac{\lambda_p}{\lambda_\alpha} = \frac{p_\alpha}{p_p} = \sqrt{\frac{m_\alpha}{m_p} \cdot \frac{q_\alpha V_\alpha}{q_p V_p}} = \sqrt{\frac{6.64 \times 10^{-27}}{1.67 \times 10^{-27}} \cdot \frac{2 \times 1.6 \times 10^{-19} \times 4}{1.6 \times 10^{-19} \times 2}} = \sqrt{16} = 4

Therefore, the ratio of their de-Broglie wavelengths is 4 : 1.

Q146

The work functions of cesium (Cs) and lithium (Li) metals are 1.9 eV and 2.5 eV , respectively. If we incident a light of wavelength 550 nm on these two metal surfaces, then photo-electric effect is possible for the case of

A Both Cs and Li

B Neither Cs nor Li

C Li only

D Cs only

Correct Answer

Option D

Solution

For photo-electric effect, energy of the photon must be greater than the work function of the metal. We know,

E = {{1240} \over \lambda } = {{1240} \over {550}}

(as $\lambda=550$ nm given) $\Rightarrow E=2.25$ eV here,

E > {\phi _{cs}}

So, $C_S$ only.

Q147

An

\alpha

-particle, a proton and an electron have the same kinetic energy. Which one of the following is correct in case of their de-Broglie wavelength:

A

{\lambda _\alpha } > {\lambda _p} < {\lambda _e}

B

{\lambda _\alpha } > {\lambda _p} > {\lambda _e}

C

{\lambda _\alpha } = {\lambda _p} = {\lambda _e}

D

{\lambda _\alpha } < {\lambda _p} < {\lambda _e}

Correct Answer

Option D

Solution

$\lambda=\dfrac{h}{m v}=\dfrac{h}{\sqrt{2 m k}}$ So, $\lambda \propto \dfrac{1}{\sqrt{m}}$ So, $\lambda_{e}>\lambda_{p}>\lambda_{\alpha}$

Q148

If the kinetic energy of a free electron doubles, it's deBroglie wavelength changes by the factor

A

2

B

{1 \over 2}

C

{\sqrt 2 }

D

{1 \over {\sqrt 2 }}

Correct Answer

Option D

Solution

de-Broglie wavelength,

\lambda = {h \over p} = {h \over {\sqrt {2.m,\left( {K.E} \right)} }}

$\therefore$

\lambda \propto {1 \over {\sqrt {K.E} }}

If

K.E

is doubled, wavelength becomes

{\lambda \over {\sqrt 2 }}

Q149

Given below are two statements: one is labelled as Assertion (A) and the other is labelled as Reason (R).Assertion (A) : Emission of electrons in photoelectric effect can be suppressed by applying a sufficiently negative electron potential to the photoemissive substance.Reason (R) : A negative electric potential, which stops the emission of electrons from the surface of a photoemissive substance, varies linearly with frequency of incident radiation.In the light of the above statements, choose the most appropriate answer from the options given below :

A Both (A) and (R) are true and (R) is the correct explanation of (A).

B Both (A) and (R) are true but (R) is not the correct explanation of (A).

C (A) is false but (R) is true

D (A) is true but (R) is false

Correct Answer

Option B

Solution

Assertion (A) is true because a sufficiently negative potential can repel and stop the emitted electrons from leaving the surface.

Reason (R) is also true because the stopping potential $V_{\text{stop}}$ varies linearly with the frequency $\nu$ of the incident light ( $V_{\text{stop}} = \dfrac{h}{e}\,\nu - \dfrac{\phi}{e}$ ).

However, (R) does not directly explain why a negative potential suppresses electron emission; it only shows how much potential is needed for different frequencies.

Therefore, both (A) and (R) are true, but (R) is not the correct explanation of (A).

Q150

The electric field of light wave is given as

\overrightarrow E = {10^{ - 3}}\cos \left( {{{2\pi x} \over {5 \times {{10}^{ - 7}}}} - 2\pi \times 6 \times {{10}^{14}}t} \right)\mathop x\limits^ \wedge {{\rm N} \over C}

This light falls on a metal plate of work function 2eV. The stopping potential of the photoelectrons is : Given, E (in eV) = 12375/

\lambda

$(inÅ)

A 2.48 V

B 0.48 V

C 0.72 V

D 2.0 V

Correct Answer

Option B

Solution

\omega = 6 \times {10^{14}} \times 2\pi

f = 6 × 1014 C = f $\lambda$

\lambda = {C \over f} = {{3 \times {{10}^8}} \over {6 \times {{10}^{14}}}} = 5000

Å Energy of photon

\Rightarrow {{12375} \over {5000}} = 2.475\,eV

From Einstein’s equation KEmax = E – $\phi$ eVs = E – $\phi$ eVs = 2.475 – 2 eVo = 0.475 eV Vo = 0.48 V