Dual Nature of Radiation and Matter — Page 11 | JEE Physics

Q101

Radiation, with wavelength 6561

\mathop A\limits^o

falls on a metal surface to produce photoelectrons. The electrons are made to enter a uniform magnetic field of 3 × 10–4 T. If the radius of the largest circular path followed by the electrons is 10 mm, the work function of the metal is close to :

A 1.8eV

B 0.8eV

C 1.1eV

D 1.6eV

Correct Answer

Option C

Solution

Let the work function be $\phi$ . $\therefore$ KEmax =

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

We know r =

{{mv} \over {qB}}

and p = mv = rqB $\therefore$ KEmax =

{{{p^2}} \over {2m}}

=

{{{q^2}{r^2}{B^2}} \over {2m}}

=

{{{{\left( {1.6 \times {{10}^{ - 19}}} \right)}^2}{{\left( {10 \times {{10}^{ - 3}}} \right)}^2}{{\left( {3 \times {{10}^{ - 4}}} \right)}^2}} \over {2 \times 9 \times {{10}^{ - 31}} \times 1.6 \times {{10}^{ - 19}}}}

= 0.8 eV

{{hc} \over \lambda }

=

{{12420} \over {6561}}

= 1.9 eV $\therefore$ $\phi$ = 1.9 - 0.8 = 1.1 eV

Q102

An electron of mass m and magnitude of charge |e| initially at rest gets accelerated by a constant electric field E. The rate of change of de-Broglie wavelength of this electron at time t ignoring relativistic effects is :

A

{{ - h} \over {\left| e \right|Et}}

B

{{ - h} \over {\left| e \right|E\sqrt t }}

C

{{ - h} \over {\left| e \right|E{t^2}}}

D

{{\left| e \right|Et} \over h}

Correct Answer

Option C

Solution

F = |e| E

a = {F \over m}

=

{{\left| e \right|E} \over m}

V =

at =

{{\left| e \right|E} \over m}t

$\lambda$ =

{h \over {mV}}

=

{h \over {\left| e \right|Et}}

{{d\lambda } \over {dt}}

=

{{ - h} \over {\left| e \right|E{t^2}}}

Q103

Given below are two statements : Statement I : Two photons having equal linear momenta have equal wavelengths. Statement II : If the wavelength of photon is decreased, then the momentum and energy of a photon will also decrease. In the light of the above statements, choose the correct answer from the options given below.

A Statement I is false but Statement II is true

B Both Statement I and Statement II are false

C Both Statement I and Statement II are true

D Statement I is true but Statement II is false

Correct Answer

Option D

Solution

As we know,

\lambda = {h \over p} = {h \over {\sqrt {2mK} }}

If linear momenta of two photons are equal, then their wavelengths is also equal.

Also, if the wavelength is decreased, then the momentum and energy of photon will increase.

Hence, option (d) is correct.

Q104

The de-Broglie wavelength of a proton and

\alpha

-particle are equal. The ratio of their velocities is :

A 4 : 2

B 4 : 3

C 4 : 1

D 1 : 4

Correct Answer

Option C

Solution

Let $\lambda$ p, $\lambda$ $\alpha$ , mp, m $\alpha$ , vp, v $\alpha$ , pp and p $\alpha$ be the wavelength, mass, velocity and momentum of proton and $\alpha$ -particle, respectively.

Given, $\lambda$ p = $\lambda$ $\alpha$ As we know that, $\lambda$ = h/p $\therefore$

{h \over {{p_p}}} = {h \over {{p_\alpha }}}

$\Rightarrow$ pp = p $\alpha$ $\Rightarrow$ mpvp = m $\alpha$ v $\alpha$ $\Rightarrow$ mpvp = 4mpv $\alpha$ ( $\because$ m $\alpha$ = 4mp) $\Rightarrow$

{{{v_p}} \over {{v_\alpha }}} = {4 \over 1}

or 4 : 1

Q105

An electron of mass me and a proton of mass mp = 1836 me are moving with the same speed. The ratio of their de Broglie wavelength

{{{}^\lambda electron} \over {{}^\lambda proton}}

will be :

A 1

B 1836

C

{1 \over {1836}}

D 918

Correct Answer

Option B

Solution

Given mass of electron = me Mass of proton = mp $\therefore$ given mp = 1836 me From de-Broglie wavelength

\lambda = {h \over p} = {h \over {mv}}

{{{\lambda _e}} \over {{\lambda _p}}} = {{{m_p}} \over {{m_e}}}

= {{1836{m_e}} \over {{m_e}}}

{{{\lambda _e}} \over {{\lambda _p}}} = 1836

Q106

The stopping potential for electrons emitted from a photosensitive surface illuminated by light of wavelength 491 nm is 0.710 V. When the incident wavelength is changed to a new value, the stopping potential is 1.43 V. The new wavelength is :

A 400 nm

B 329 nm

C 309 nm

D 382 nm

Correct Answer

Option D

Solution

From the photoelectric effect equation

{{hc} \over \lambda } = \phi + e{v_s}

so,

e{v_{{s_1}}} = {{hc} \over {{\lambda _1}}} - \phi

.....(i)

e{v_{{s_2}}} = {{hc} \over {{\lambda _2}}} - \phi

......(ii) Subtract equation (i) from equation (ii)

e{v_{{s_1}}} - e{v_{{s_2}}} = {{hc} \over {{\lambda _1}}} - {{hc} \over {{\lambda _2}}}

{v_{{s_1}}} - {v_{{s_2}}} = {{hc} \over e}\left( {{1 \over {{\lambda _1}}} - {1 \over {{\lambda _2}}}} \right)

(0.710 - 1.43) = 1240\left( {{1 \over {491}} - {1 \over {{\lambda _2}}}} \right)

{{ - 0.72} \over {1240}} = {1 \over {491}} - {1 \over {{\lambda _2}}}

{1 \over {{\lambda _2}}} = {1 \over {491}} + {{0.72} \over {1240}}

{1 \over {{\lambda _2}}} = 0.00203 + 0.00058

{1 \over {{\lambda _2}}} = 0.00261

{\lambda _2} = 383.14

{\lambda _2} \simeq 382

nm

Q107

Given below are two statements : one is labeled as Assertion A and the other is labelled as Reason R. Assertion A : An electron microscope can achieve better resolving power than an optical microscope. Reason R : The de Broglie's wavelength of the electrons emitted from an electron gun is much less than wavelength of visible light. In the light of the above statements, choose the correct answer from the options given below :

A A is false but R is true.

B Both A and R are true but R is NOT the correct explanation of A.

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

D A is true but R is false.

Correct Answer

Option C

Solution

Resolution limit (

\Delta

$\theta$ ) =

{{1.22\lambda } \over d}

Resolution power =

{1 \over {{\mathop{\rm Re}\nolimits} solution\,\lim it}}

$\therefore$ Resolution power $\propto$

{1 \over \lambda }

Since, wavelength of electron is much less than visible light, its resolving power will be much more.

Q108

An electron of mass me and a proton of mass mp are accelerated through the same potential difference. The ratio of the de-Broglie wavelength associated with the electron to that with the proton is

A

{{{m_e}} \over {{m_p}}}

B 1

C

{{{m_p}} \over {{m_e}}}

D

\sqrt {{{{m_p}} \over {{m_e}}}}

Correct Answer

Option D

Solution

\lambda = {h \over p} = {h \over {\sqrt {2km} }}

{{{\lambda _e}} \over {{\lambda _p}}} = \sqrt {{{{k_p}{m_p}} \over {{k_e}{m_e}}}} = \sqrt {{{{m_p}} \over {{m_e}}}}

Q109

A particle is travelling 4 time as fast as an electron. Assuming the ratio of de-Broglie wavelength of a particle to that of electron is 2 : 1, the mass of the particle is :

A

{1 \over {16}}

times the mass of e

-

B 8 times the mass of e

-

C 16 times the mass of e

-

D

{1 \over {8}}

times the mass of e

-

Correct Answer

Option D

Solution

\lambda = {h \over p}

{{{\lambda _p}} \over {{\lambda _e}}} = {{{p_e}} \over {{p_p}}} = {{{m_e}{v_e}} \over {{m_p}{v_p}}}

2 = {{{m_e}} \over {{m_p}}}\left( {{{{v_e}} \over {4{v_e}}}} \right)

$\therefore$

{m_p} = {{{m_e}} \over 8}

Q110

A particle of mass 4M at rest disintegrates into two particles of mass M and 3M respectively having non zero velocities. The ratio of de-Broglie wavelength of particle of mass M to that of mass 3M will be :

A 1 : 3

B 3 : 1

C 1 :

\sqrt 3

D 1 : 1

Correct Answer

Option D

Solution

\lambda = {h \over p}

both the particles will move with momentum same in magnitude & opposite in direction.

So De-Broglie wavelength of both will be same i.e. ratio 1 : 1