Dual Nature of Radiation and Matter — Page 4 | NEET Physics

Q31

If the kinetic energy of the particle is increased to 16 times its previous value, the percentage change in the de Broglie wavelength of the particle is

A 25

B 75

C 60

D 50

Correct Answer

Option B

Solution

de-Broglie wavelength of particle, $\lambda$ =

{h \over p}

=

{h \over {\sqrt {2mE} }}

$\therefore$

{{\lambda '} \over \lambda } = \sqrt {{E \over {E'}}}

$\Rightarrow$ $\lambda$ ' =

{\lambda \over 4}

So, % change in the de Broglie wavelength =

{{\lambda - {\lambda \over 4}} \over \lambda } \times 100

= 75%

Q32

When the energy of the incident radiation is increased nby 20%, the kinetic energy of the photoelectrons emitted from a metal surface increased from 0.5 eV to 0.8 eV. The work function of the metal is

A 0.65 eV

B 1.0 eV

C 1.3 eV

D 1.5 eV

Correct Answer

Option B

Solution

0.5 = E – $\phi$ $\Rightarrow$ 0.8 = 1.2 E – $\phi$ From above expressions, work function $\phi$ = 1 eV

Q33

The de-broglie wavelength of neutrons in thermal equilibrium at temperature T is

A

{{3.08} \over {\sqrt T }}\mathop A\limits^ \circ

B

{{0.308} \over {\sqrt T }}\mathop A\limits^ \circ

C

{{0.0308} \over {\sqrt T }}\mathop A\limits^ \circ

D

{{30.8} \over {\sqrt T }}\mathop A\limits^ \circ

Correct Answer

Option D

Solution

de Broglie wavelength of neutrons in thermal equilibrium at temperature T is

\lambda = {h \over {\sqrt {2m{k_B}T} }}

=

{{6.63 \times {{10}^{ - 34}}} \over {\sqrt {2 \times 1.67 \times {{10}^{ - 27}} \times 1.38 \times {{10}^{ - 23}} \times T} }}

=

{{30.8} \over {\sqrt T }}\mathop A\limits^ \circ

Q34

A source of light is placed at a distance of 50 cm from a photo cell and the stopping potential is found to be V 0 . If the distance between the light source and photo cell is made 25 cm, the new stopping potential will be :

A V 0 /2

B V 0

C 4V 0

D 2V 0

Correct Answer

Option B

Solution

Since, stopping potential is independent of distance hence new stopping potential will remain unchanged i.e., new stopping potential = V 0 .

Q35

The wavelength

\lambda

e of an electron and

\lambda

p of a photon of same energy E are related by

A

{\lambda _p} \propto \sqrt {{\lambda _e}}

B

{\lambda _p} \propto {1 \over {\sqrt {{\lambda _e}} }}

C

{\lambda _p} \propto {\lambda _e}^2

D

{\lambda _p} \propto {\lambda _e}

Correct Answer

Option C

Solution

Wavelength of an electron of energy E is

{\lambda _e} = {h \over {\sqrt {2{m_e}E} }}

$\Rightarrow$

\lambda _e^2 = {{{h^2}} \over {2{m_e}E}}

.....(1) Wavelength of a photon of same energy E is

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

$\Rightarrow$

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

.....(2) Equating (1) and (2), we get

{{hc} \over {{\lambda _p}}} = {{{h^2}} \over {2{m_e}\lambda _e^2}}

$\Rightarrow$

{\lambda _p} = {{2{m_e}c} \over h}\lambda _e^2

$\Rightarrow$

{\lambda _p} \propto {\lambda _e}^2

Q36

For photoelectric emission from certain metal the cutoff frequency is

\upsilon

. If radiation of frequency 2

\upsilon

impinges on the metal plate, the maximum possible velocity of the emitted electron will be (m is the electron mass)

A

\sqrt {{{2h\upsilon } \over m}}

B

2\sqrt {{{h\upsilon } \over m}}

C

\sqrt {{{h\upsilon } \over {\left( {2m} \right)}}}

D

\sqrt {{{h\upsilon } \over m}}

Correct Answer

Option A

Solution

Work function, $\phi$ = h $\nu$ According to Einstein’s photoelectric equation

{1 \over 2}mv_{\max }^2

= h(2 $\nu$ ) - h $\nu$ $\Rightarrow$

{1 \over 2}mv_{\max }^2 = h\nu

$\Rightarrow$

v_{\max }^2 = {{2h\nu } \over m}

$\Rightarrow$ v max =

\sqrt {{{2h\upsilon } \over m}}

Q37

A parallel beam of fast moving electrons is incident normally on a narrow slit. A fluorescent screen is placed at a large distance from the slit. If the speed of the electrons is increased, which of the following statements is correct ?

A The angular width of the central maximum will decrease.

B The angular wifth of the central maximum will be unaffected.

C Diffraction pattern is not observed on the screen in the case of electrons.

D The angular width of the central maximum of the diffraction pattern will increase.

Correct Answer

Option A

Solution

As speed of an electron increases. Its de-Broglie wavelength decreases

\lambda = {h \over {mv}}

and angular width for central maxima is $\omega$ =

{{2\lambda } \over d}

\omega \propto \lambda \propto {1 \over v}

Q38

Two radiations of photons energies 1 eV and 2.5 eV, successively illuminate a photosensitive metallic surface of work function 0.5 eV. The ratio of the maximum speeds of the emitted electrons is

A 1 : 4

B 1 : 2

C 1 : 1

D 1 : 5

Correct Answer

Option B

Solution

According to Einsten’s photoelectric effect, the K.E. of the radiated electrons K.E max = E – W

{1 \over 2}mv_1^2

= (1 - 0.5) eV = 0.5 eV

{1 \over 2}mv_2^2

= (2.5 - 0.5) eV = 2 eV

{{{v_1}} \over {{v_2}}} = \sqrt {{{0.5} \over 2}}

= 1 : 2

Q39

If the momentum of an electron is changed by P, then the de Broglie wavelength associated with changes by 0.5%. The initial momentum of electron will be

A 200 P

B 400 P

C

{P \over {200}}

D 100 P

Correct Answer

Option A

Solution

de Broglie wavelength associated with an electron is

\lambda = {h \over P}

$\Rightarrow$ P =

{h \over \lambda }

{{\Delta P} \over P} = - {{\Delta \lambda } \over \lambda }

{P \over {{P_{initial}}}} = {{0.5} \over {100}}

P initial = 200P

Q40

An

\alpha

-particle moves in a circular path of radius 0.83 cm in the presence of a magnetic field of 0.25 Wb/m 2 . The de Broglie wavelength associated with the particle will be

A 1

\mathop A\limits^ \circ

B 0.1

\mathop A\limits^ \circ

C 10

\mathop A\limits^ \circ

D 0.01

\mathop A\limits^ \circ

Correct Answer

Option D

Solution

Wavelength $\lambda$ =

{h \over p}

$\Rightarrow$

{h \over p}

=

{h \over {mv}}

For circular motion = F c = qvB $\Rightarrow$

{{m{v^2}} \over r}

= qvB $\Rightarrow$ mv = rBq $\Rightarrow$ mv = (0.83 × 10 –2 )(0.25)(2 × 1.6 × 10 –19 ) de Broglie wavelength, $\lambda$ =

{h \over {mv}}

=

{{6.6 \times {{10}^{ - 34}}} \over {0.83 \times {{10}^{ - 2}} \times 0.25 \times 2 \times 1.6 \times {{10}^{ - 19}}}}

= 0.01

\mathop A\limits^ \circ