Dual Nature of Radiation and Matter — Page 12 | JEE Physics

Q111

An electron having de-Broglie wavelength

\lambda

is incident on a target in a X-ray tube. Cut-off wavelength of emitted X-ray is :

A 0

B

{{2{m^2}{c^2}{\lambda ^2}} \over {{h^2}}}

C

{{2mc{\lambda ^2}} \over h}

D

{{hc} \over {mc}}

Correct Answer

Option C

Solution

\lambda = {h \over {mv}}

kinetic energy,

{{{P^2}} \over {2m}} = {{{h^2}} \over {2m{\lambda ^2}}} = {{hc} \over {{\lambda _c}}}

{\lambda _c} = {{2mc{\lambda ^2}} \over h}

Q112

The de-Broglie wavelength of a particle having kinetic energy E is

\lambda

. How much extra energy must be given to this particle so that the de-Broglie wavelength reduces to 75% of the initial value?

A

{1 \over 9}

E

B

{7 \over 9}

E

C E

D

{16 \over 9}

E

Correct Answer

Option B

Solution

\lambda = {h \over {mv}} = {h \over {\sqrt {2mE} }}

,

mv = \sqrt {2mE}

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

{{{\lambda _2}} \over {{\lambda _1}}} = \sqrt {{{{E_1}} \over {{E_2}}}} = {3 \over 4}

,

{\lambda _2} = 0.75{\lambda _1}

{{{E_1}} \over {{E_2}}} = {\left( {{3 \over 4}} \right)^2}

{E_2} = {{16} \over 9}{E_1} = {{16} \over 9}E

(E1 = E) Extra energy given =

{{16} \over 9}E - E = {7 \over 9}E

Q113

In a photoelectric experiment ultraviolet light of wavelength 280 nm is used with lithium cathode having work function

\phi

= 2.5 eV. If the wavelength of incident light is switched to 400 nm, find out the change in the stopping potential. (h = 6.63

\times

10

-

34 Js, c = 3

\times

108 ms

-

1)

A 1.3 V

B 1.1 V

C 1.9 V

D 0.6 V

Correct Answer

Option A

Solution

K{E_{\max }} = e{V_s} = {{hc} \over \lambda } - \phi

\Rightarrow e{V_s} = {{1240} \over {280}} - 2.5 =

1.93 eV

\Rightarrow {V_{{s_1}}} =

1.93 V .... (i)

\Rightarrow e{V_{{s_2}}} = {{1240} \over {400}} - 2.5 =

0.6 eV

\Rightarrow {V_{{s_2}}} =

0.6 V .... (ii)

\Delta

V =

{V_{{s_1}}} - {V_{{s_2}}}

= 1.93 $-$ 0.6 = 1.33 V

Q114

Consider two separate ideal gases of electrons and protons having same number of particles. The temperature of both the gases are same. The ratio of the uncertainty in determining the position of an electron to that of a proton is proportional to :-

A

{\left( {{{{m_p}} \over {{m_e}}}} \right)^{3/2}}

B

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

C

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

D

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

Correct Answer

Option C

Solution

\Delta x\,.\,\Delta p \ge {h \over {4\pi }}

\Delta x = {h \over {4\pi m\Delta v}}

v = \sqrt {{{3KT} \over m}}

{{\Delta {x_e}} \over {\Delta {x_p}}} = \sqrt {{{{m_p}} \over {{m_e}}}}

Q115

Let K1 and K2 be the maximum kinetic energies of photo-electrons emitted when two monochromatic beams of wavelength

\lambda

1 and

\lambda

2, respectively are incident on a metallic surface. If

\lambda

1 = 3

\lambda

2 then :

A

{K_1} > {{{K_2}} \over 3}

B

{K_1} < {{{K_2}} \over 3}

C

{K_1} = {{{K_2}} \over 3}

D

{K_2} = {{{K_1}} \over 3}

Correct Answer

Option B

Solution

{K_1} = {{hc} \over {{\lambda _1}}} - \phi = {{hc} \over {3{\lambda _2}}} - \phi

..... (i) and

{K_2} = {{hc} \over {{\lambda _2}}} - \phi

..... (ii) from (i) and (ii) we can say

3{K_1} = {K_2} - 2\phi

$${K_1}

Q116

The work function of a substance is

3.0 \mathrm{~eV}

. The longest wavelength of light that can cause the emission of photoelectrons from this substance is approximately;

A 215 nm

B 400 nm

C 414 nm

D 200 nm

Correct Answer

Option C

Solution

\begin{aligned} & \text{ For P.E.E. : } \lambda \leq \frac{h c}{W_e} \\ & \lambda \leq \frac{1240 \mathrm{~nm}-\mathrm{eV}}{3 \mathrm{eV}} \\ & \lambda \leq 413.33 \mathrm{~nm} \\ & \lambda_{\max } \approx 414 \mathrm{~nm} \text{ for P.E.E. } \end{aligned}

Q117

An electron with speed v and a photon with speed c have the same de-Broglie wavelength. If the kinetic energy and momentum of electron are Ee and pe and that of photon are Eph and pph respectively. Which of the following is correct?

A

{{{E_e}} \over {{E_{ph}}}} = {{2c} \over v}

B

{{{E_e}} \over {{E_{ph}}}} = {v \over {2c}}

C

{{{p_e}} \over {{p_{ph}}}} = {{2c} \over v}

D

{{{p_e}} \over {{p_{ph}}}} = {v \over {2c}}

Correct Answer

Option B

Solution

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

Now, A/Q,

{h \over {{P_e}}} = {h \over {{P_{photon}}}}

\Rightarrow {P_e} = {P_{photon}}

....... (i) Now,

{K_e} = {1 \over 2}M{v^2} = {{Pv} \over 2}

{K_{ph}} = m{c^2} = Pc

..... (ii)

{{{K_e}} \over {{K_{eq}}}} = {v \over {2c}}

Q118

A metal surface is illuminated by a radiation of wavelength 4500

\mathop A\limits^o

. The ejected photo-electron enters a constant magnetic field of 2 mT making an angle of 90

^\circ

with the magnetic field. If it starts revolving in a circular path of radius 2 mm, the work function of the metal is approximately :

A 1.36 eV

B 1.69 eV

C 2.78 eV

D 2.23 eV

Correct Answer

Option A

Solution

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

...... (i)

R = {{mv} \over {Bq}} = {{\sqrt {2m(KE)} } \over {Bq}}

...... (ii) Putting the values,

\phi \simeq 1.36

eV

Q119

A metal exposed to light of wavelength

800 \mathrm{~nm}

and and emits photoelectrons with a certain kinetic energy. The maximum kinetic energy of photo-electron doubles when light of wavelength

500 \mathrm{~nm}

is used. The workfunction of the metal is : (Take hc

=1230 \,\mathrm{eV}-\mathrm{nm}

).

A 1.537 eV

B 2.46 eV

C 0.615 eV

D 1.23 eV

Correct Answer

Option C

Solution

$\because$

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

\Rightarrow K = {{1230} \over {800}} - \phi

and,

2K = {{1230} \over {500}} - \phi

\Rightarrow 2 \times {{1230} \over {800}} - 2\phi = {{1230} \over {500}} - \phi

\Rightarrow \phi = 0.615\,eV

Q120

Two streams of photons, possessing energies equal to five and ten times the work function of metal are incident on the metal surface successively. The ratio of maximum velocities of the photoelectron emitted, in the two cases respectively, will be

A 1 : 2

B 1 : 3

C 2 : 3

D 3 : 2

Correct Answer

Option C

Solution

{1 \over 2}mv_1^2 = 5\phi - \phi

And,

{1 \over 2}mv_2^2 = 10\phi - \phi

\Rightarrow {\left( {{{{v_1}} \over {{v_2}}}} \right)^2} = {4 \over 9}

\Rightarrow {{{v_1}} \over {{v_2}}} = {2 \over 3}