Dual Nature of Radiation and Matter — Page 8 | JEE Physics

Q71

A monochromatic neon lamp with wavelength of 670.5 nm illuminates a photo-sensitive material which has a stopping voltage of 0.48 V. What will be the stopping voltage if the source light is changed with another source of wavelength of 474.6 nm?

A 0.96 V

B 1.25 V

C 0.24 V

D 1.5 V

Correct Answer

Option B

Solution

k{E_{\max }} = {{hc} \over {{\lambda _i}}} + \phi

or

e{V_o} = {{hc} \over {{\lambda _i}}} + \phi

when $\lambda$ i = 670.5 nm ; Vo = 0.48 when $\lambda$ i = 474.6 nm ; Vo = ? So,

e(0.48) = {{1240} \over {670.5}} + \phi

..... (1)

e({V_o}) = {{1240} \over {474.6}} + \phi

.....(2) (2) $-$ (1)

e({V_o} - 0.48) = 1240\left( {{1 \over {474.6}} - {1 \over {670.5}}} \right)eV

{V_o} = 0.48 + 1240\left( {{{670.5 - 474.6} \over {474.6 \times 670.5}}} \right)

Volts Vo = 0.48 + 0.76 Vo = 1.24 V

\simeq

1.25 V

Q72

An electron of mass m and a photon have same energy E. The ratio of wavelength of electron to that of photon is : (c being the velocity of light)

A

{1 \over c}{\left( {{{2m} \over E}} \right)^{1/2}}

B

{1 \over c}{\left( {{E \over {2m}}} \right)^{1/2}}

C

{\left( {{E \over {2m}}} \right)^{1/2}}

D

c{(2mE)^{1/2}}

Correct Answer

Option B

Solution

For photon, E =

{{hc} \over \lambda }

{\lambda _p} = {{hc} \over E}

.... (i) For electron,

{\lambda _e} = {{hc} \over {\sqrt {2mE} }}

.... (ii)

{{{\lambda _e}} \over {{\lambda _p}}} = {{{{hc} \over {\sqrt {2mE} }}} \over {{{hc} \over E}}} = \sqrt {{E \over {2m{c^2}}}} = {1 \over c}{\left( {{E \over {2m}}} \right)^{1/2}}

Q73

The radiation corresponding to 3

\to

2 transition of a hydrogen atom falls on a gold surface to generate photoelectrons. The electrons are passed through a magnetic field of 5

\times

10

-

4 T. Assume that the radius of the largest circular path followed by these electrons is 7 mm, the work function of the metal is : (Mass of electron = 9.1

\times

10

-

31 kg)

A 1.36 eV

B 1.88 eV

C 0.82 eV

D 0.16 eV

Correct Answer

Option C

Solution

Energy of photon can be given as

{E_p} = 13.6\left[ {{1 \over {n_1^2}} - {1 \over {n_2^2}}} \right]

eV where, n1 = lower energy level and n2 = higher energy level. As per question, n1 = 2, n2 = 3 $\therefore$

{E_p} = 13.6\left[ {{1 \over {{{(2)}^2}}} - {1 \over {{{(3)}^2}}}} \right]

= 13.6\left[ {{1 \over 4} - {1 \over 9}} \right] = 13.6\left[ {{{9 - 4} \over {36}}} \right] = 1.89

eV We know that work-function is the minimum energy required to eject photoelectrons from metal surface.

For gold plate, it will be $\phi$ = EP $-$ KEmax .... (i) [Given, B = 5 $\times$ 10 $-$ 4 T, r = 7 mm = 7 $\times$ 10 $-$ 3 m, q = 1.6 $\times$ 10 $-$ 19 C and m = 9.1 $\times$ 10 $-$ 31 kg] Therefore, velocity of photoelectrons will be

v = {{Bqr} \over m} = {{5 \times {{10}^{ - 4}} \times 1.6 \times {{10}^{ - 19}} \times 7 \times {{10}^{ - 3}}} \over {9.1 \times {{10}^{ - 31}}}} = 6.15 \times {10^5}

ms $-$ 1 Kinetic energy will be $\therefore$

KE = {1 \over 2}m{v^2} = {{1 \times 9.1 \times {{10}^{ - 31}} \times {{(6.15 \times {{10}^5})}^2}} \over {2 \times 1.6 \times {{10}^{ - 19}}}}

eV = 1.075 eV Now, putting the values, in Eq. (i), we get $\phi$ = (1.89 $-$ 1.075) eV = 0.82 eV

Q74

Which of the following phenomena cannot be explained by wave theory of light?

A Refraction of light

B Reflection of light

C Diffraction of light

D Compton effect

Correct Answer

Option D

Solution

Solution Explanation Among the four listed phenomena: Reflection of light Wave theory (Huygens’s principle, for example) adequately explains reflection.

Refraction of light Wave theory also explains refraction via the change of wave speed in different media.

Diffraction of light Wave theory (again, Huygens–Fresnel principle) explains the bending of waves around obstacles.

Compton effect This involves inelastic scattering of photons by electrons (where the photon loses energy and is scattered with a longer wavelength).

Wave theory alone cannot explain the discrete energy exchange and wavelength shift; it requires the particle (photon) concept from quantum theory.

Hence, the phenomenon not explained by the (classical) wave theory is the Compton effect.

Answer: (D) Compton effect.

Q75

When a certain photosensitive surface is illuminated with monochromatic light of frequency v, the stopping potential for the current is –V0/2. When the surface is illuminated by monochromatic light of frequency v/2, the stopping potential is – V0. The threshold frequency for photoelectric emission is :

A 2

v

B

{4 \over 3}v

C

{{3v} \over 2}

D

{{5v} \over 3}

Correct Answer

Option C

Solution

Einstein’s photoelectric equation in the two cases is given by

{{e{V_0}} \over {\Delta E}} = h\upsilon - h{\upsilon _0}

......(i) and

e{V_0} = {{h\upsilon } \over 2} - h{\upsilon _0}

.....(ii) From eqn. (i) and (ii),

{1 \over 2} = {{h\upsilon - h{\upsilon _0}} \over {{{h\upsilon } \over 2} - h{\upsilon _0}}}

$\Rightarrow$

{\upsilon _0} = {3 \over 2}\upsilon

Q76

A free electron of 2.6 eV energy collides with a H+ ion. This results in the formation of a hydrogen atom in the first excited state and a photon is released. Find the frequency of the emitted photon. (h = 6.6

\times

10

-

34 Js)

A 1.45

\times

1016 MHz

B 0.19

\times

1015 MHz

C 1.45

\times

109 MHz

D 9.0

\times

1027 MHz

Correct Answer

Option C

Solution

For every large distance P.E. = 0 & total energy = 2.6 + 0 = 2.6 eV Finally in first excited state of H atom total energy = $-$ 3.4 eV Loss in total energy = 2.6 $-$ ( $-$ 3.4) = 6 eV It is emitted as photon

\lambda = {{1240} \over 6} = 206

nm

f = {{3 \times {{10}^8}} \over {206 \times {{10}^{ - 9}}}}

= 1.45 $\times$ 1015 Hz = 1.45 $\times$ 109 Hz

Q77

An electron moving with speed v and a photon moving with speed c, have same D-Broglie wavelength. The ratio of kinetic energy of electron to that of photon is :

A

{{3c} \over v}

B

{v \over {3c}}

C

{v \over {2c}}

D

{{2c} \over v}

Correct Answer

Option C

Solution

{\lambda _e} = {\lambda _{Ph}}

{h \over {{p_e}}} = {h \over {{p_{ph}}}}

\sqrt {2m{k_e}} = {{{E_{ph}}} \over c}

2m{k_e} = {{{{({E_{ph}})}^2}} \over {{c^2}}}

{{{k_e}} \over {{E_{ph}}}} = {{{E_{ph}}} \over {{c^2}}}\left( {{1 \over {2m}}} \right)

= {{{p_{ph}}} \over c}\left( {{1 \over {2m}}} \right)

= {{{p_e}} \over c}\left( {{1 \over {2m}}} \right)

= {{mv} \over c}{1 \over {2m}}

= {v \over {2c}}

Q78

The de-Broglie wavelength (

\lambda

B) associated with the electron orbiting in the second excited state of hydrogen atom is related to that in the ground state (

\lambda

G) by :

A

\lambda

B = 2

\lambda

G

B

\lambda

B = 3

\lambda

G

C

\lambda

B =

\lambda

G/2

D

\lambda

B =

\lambda

G/3

Correct Answer

Option B

Solution

We know that,

\lambda = {h \over {mv}}

From third Bohr's postulate, we have

mvr = n{h \over {2\pi }}

{h \over {mv}} = {{2\pi r} \over n} \Rightarrow \lambda = {{2\pi r} \over n}

Since,

r = {a_0}{{{n^2}} \over Z}

, where a0 is radius of Bohr's orbit having value (0.53) 12

\mathop A\limits^o

= 0.53

\mathop A\limits^o

, therefore,

\lambda = {{2\pi {a_0}{n^2}} \over {n\,.\,Z}} = {{2\pi {a_0}} \over Z}\,.\,n

For Hydrogen Z = 1. Therefore,

{\lambda _G} = {{2\pi {a_0} \times 1} \over 1} = 2\pi {a_0}

and

{\lambda _B} = {{2\pi {a_0} \times 3} \over 1} = 6\pi {a_0}

Then,

{\lambda _B} = 3{\lambda _G}

Q79

Two identical photo-cathodes receive light of frequencies

{f_1}

and

{f_2}

. If the velocities of the photo electrons (of mass

m

) coming out are respectively

{v_1}

and

{v_2},

then

A

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

B

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

C

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

D

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

Correct Answer

Option A

Solution

For one photo cathode

h{f_1} - W = {1 \over 2}mv_1^2\,\,\,\,\,\,\,\,\,\,\,\,\,...\left( i \right)

For another photo cathode

h{f_2} - W = {1 \over 2}mv_2^2\,\,\,\,\,\,\,\,\,\,\,\,\,...\left( {ii} \right)

Subtracting

(ii)

from

(i)

we get

\left( {h{f_1} - W} \right) - \left( {h{f_2} - W} \right) = {1 \over 2}mv_1^2 - {1 \over 2}mv_2^2

$\therefore$

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

$\therefore$

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

Q80

A proton of mass '

m_P

' has same energy as that of a photon of wavelength '

\lambda

'. If the proton is moving at non-relativistic speed, then ratio of its de Broglie wavelength to the wavelength of photon is.

A

\dfrac{1}{c} \sqrt{\dfrac{E}{m_p}}

B

\dfrac{1}{\mathrm{c}} \sqrt{\dfrac{2 \mathrm{E}}{\mathrm{m}_{\mathrm{p}}}}

C

\dfrac{1}{\mathrm{2c}} \sqrt{\dfrac{ \mathrm{E}}{\mathrm{m}_{\mathrm{p}}}}

D

\dfrac{1}{\mathrm{c}} \sqrt{\dfrac{\mathrm{E}}{2 \mathrm{~m}_{\mathrm{p}}}}

Correct Answer

Option D

Solution

To find the ratio of the de Broglie wavelength of a proton to the wavelength of a photon when they have the same energy, we follow these steps: Energy Equivalence: Both the proton and the photon have the same energy, denoted by E.

Energy of the Photon: The energy of a photon with wavelength λ is given by: $E = \dfrac{hc}{\lambda}$ Kinetic Energy of the Proton: Since the proton moves at a non-relativistic speed, its kinetic energy can be expressed as: $\text{KE}_{\text{proton}} = \dfrac{1}{2}m_P v^2 = E$ Momentum of the Proton: The momentum p of the proton is derived from its kinetic energy: $p = \sqrt{2m_P E}$ de Broglie Wavelength of the Proton: The de Broglie wavelength λ_{proton} is: $\lambda_{\text{proton}} = \dfrac{h}{p} = \dfrac{h}{\sqrt{2m_P E}}$ Ratio of Wavelengths: The ratio of the de Broglie wavelength of the proton to the wavelength of the photon is: $\dfrac{\lambda_{\text{proton}}}{\lambda_{\text{photon}}} = \left( \dfrac{h}{\sqrt{2m_P E}} \right) \times \left( \dfrac{E}{hc} \right)$ Simplifying, we get: $\dfrac{\lambda_{\text{proton}}}{\lambda_{\text{photon}}} = \dfrac{1}{c} \sqrt{\dfrac{E}{2m_P}}$ Thus, the ratio of the de Broglie wavelength of the proton to the wavelength of the photon is $\dfrac{1}{c} \sqrt{\dfrac{E}{2m_P}}$ .