Dual Nature of Radiation and Matter — Page 2 | JEE Physics

Q11

The de Broglie wavelengths for an electron and a photon are

\lambda

e and

\lambda

p respectively. For the same kinetic energy of electron and photon, which of the following presents the correct relation between the de Broglie wavelengths of two ?

A

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

B

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

C

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

D

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

Correct Answer

Option A

Solution

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

...... (i)

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

...... (ii) From (i) and (ii)

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

Q12

A particle A of mass 'm' and charge 'q' is accelerated by a potential difference of 50 V. Another particle B of mass ' 4 m' and charge 'q' is accelerated by a potential difference of 2500 V. The ratio of de-Broglie wavelengths

{{{\lambda _A}} \over {{\lambda _B}}}

is close to :

A 4.47

B 10.00

C 14.14

D 0.07

Correct Answer

Option C

Solution

K.E. acquired by charge = K = qV $\lambda$ =

{h \over P}

=

{h \over {\sqrt {2mK} }}

=

{h \over {\sqrt {2mqV} }}

$\therefore$

{{{\lambda _A}} \over {{\lambda _B}}} = {{\sqrt {2m{}_B{q_B}{V_B}} } \over {\sqrt {2m{}_A{q_A}{V_A}} }} = \sqrt {{{4m.q.2500} \over {m.q.50}}} = 2\sqrt {50}

= 2 \times 7.07 = 14.14

Q13

In an electron microscope, the resolution that can be achieved is of the order of the wavelength of electrons used. To resolve a width of 7.5 × 10–12 m, the minimum electron energy required is close to -

A 25 keV

B 500 keV

C 100 keV

D 1 keV

Correct Answer

Option A

Solution

$\lambda$ =

{h \over p}

{ $\lambda$ = 7.5 $\times$ 10 $-$ 12} P =

{h \over \lambda }

KE =

{{{P^2}} \over {2m}} = {{{{\left( {h/\lambda } \right)}^2}} \over {2m}}

= {{\left\{ {{{6.6 \times {{10}^{ - 34}}} \over {7.5 \times {{10}^{ - 12}}}}} \right\}} \over {2 \times 9.1 \times {{10}^{ - 31}}}}

J KE = 25 Kev

Q14

The ratio of wavelengths of proton and deuteron accelerated by potential Vp and Vd is 1 :

\sqrt2

. Then the ratio of Vp to Vd will be :

A 1 : 1

B

\sqrt2

: 1

C 2 : 1

D 4 : 1

Correct Answer

Option D

Solution

\lambda = {h \over {mv}} = {h \over {\sqrt {2m\,eV} }}

so

{{{\lambda _p}} \over {{\lambda _d}}} = {{\sqrt {{m_d}{V_d}} } \over {\sqrt {{m_p}{V_p}} }} = {1 \over {\sqrt 2 }}

{{2{V_d}} \over {{V_p}}} = {1 \over 2}

{{{V_p}} \over {{V_d}}} = {4 \over 1}

Q15

Proton

(\mathrm{P})

and electron (e) will have same de-Broglie wavelength when the ratio of their momentum is (assume,

\mathrm{m}_{\mathrm{p}}=1849 \mathrm{~m}_{\mathrm{e}}

):

A 1 : 1

B 1 : 43

C 1 : 1849

D 43 : 1

Correct Answer

Option A

Solution

The de Broglie wavelength of a particle is given by the formula:

\lambda = \frac{h}{p}

where $h$ is Planck's constant and $p$ is the momentum of the particle.

If the de Broglie wavelengths of the proton and electron are the same, then:

\frac{h}{p_p} = \frac{h}{p_e}

where $p_p$ and $p_e$ are the momenta of the proton and electron, respectively.

Solving this equation for the ratio of their momenta gives:

\frac{p_p}{p_e} = 1

So, the ratio of their momenta is 1:1

Q16

An

\alpha

particle and a carbon 12 atom has same kinetic energy K. The ratio of their de-Broglie wavelengths

({\lambda _\alpha }:{\lambda _{C12}})

is :

A

1:\sqrt 3

B

\sqrt 3 :1

C

3:1

D

2:\sqrt 3

Correct Answer

Option B

Solution

{K_\alpha } = {K_C}

{{p_\alpha ^2} \over {2{m_\alpha }}} = {{p_C^2} \over {2{m_C}}}

{{{p_\alpha }} \over {{p_C}}} = \sqrt {{{{m_\alpha }} \over {{m_C}}}}

So

{{{\lambda _\alpha }} \over {{\lambda _C}}} = {{h/{p_\alpha }} \over {h/{p_C}}} = \sqrt {{{{m_C}} \over {{m_\alpha }}}}

So

{{{\lambda _\alpha }} \over {{\lambda _C}}} = \sqrt 3

Q17

The threshold wavelength for photoelectric emission from a material is 5500

\mathop A\limits^o

. Photoelectrons will be emitted, when this material is illuminated with monochromatic radiation from a A. 75 W infra-red lamp B. 10 W infra-red lamp C. 75 W ultra-violet lamp D. 10 W ultra-violet lamp Choose the correct answer from the options given below :

A C only

B A and D only

C C and D only

D B and C only

Correct Answer

Option C

Solution

Wavelength of infra-red $=700 \mathrm{~nm}$ (minimum) Wavelength of UV $=100-400 \mathrm{~nm}$ Since we need $\lambda<5000$ Å $\Rightarrow$ Only UV would be able to emit photoelectrons.

Q18

If a source of electromagnetic radiation having power

15 \mathrm{~kW}

produces

10^{16}

photons per second, the radiation belongs to a part of spectrum is. (Take Planck constant

h=6 \times 10^{-34} \mathrm{Js}

)

A Gamma rays

B Radio waves

C Micro waves

D Ultraviolet rays

Correct Answer

Option A

Solution

\begin{aligned} & \text{ Energy of one photon }=\frac{\text{ Power }}{\text{ Photon frequency }} \\\\ & \mathrm{E}=\mathrm{h} v=\frac{15 \times 10^3}{10^{16}} \\\\ & \Rightarrow v=\frac{15 \times 10^{-13}}{6 \times 10^{-34}}=2.5 \times 10^{21} \end{aligned}

So gamma Rays.

Q19

Photon of frequency

v

has a momentum associated with it. If

c

is the velocity of light, the momentum is

A

hv/c

B

v/c

C

h

v

c

D

hv/{c^2}

Correct Answer

Option A

Solution

Energy of a photon of frequency

v

is given by

E = hv.

Also,

E = m{c^2},\,\,m{c^2} = hv

\Rightarrow mc = {{hv} \over C} \Rightarrow p = {{hv} \over c}

Q20

A Laser light of wavelength 660 nm is used to weld Retina detachment. If a Laser pulse of width 60 ms and power 0.5 kW is used the approximate number of photons in the pulse are : [Take Planck's constant h

=

6.62

\times

10

-

34 Js]

A 1020

B 1018

C 1022

D 1019

Correct Answer

Option A

Solution

The power of the given laser light is expressed as

P = {{nhc} \over {\lambda t}}

from which the number of photons per second is given by

n = P\left( {{{\lambda t} \over {hc}}} \right) = (5 \times {10^2}) \times \left[ {{{(660 \times {{10}^{ - 9}})(60 \times {{10}^{ - 3}})} \over {(6.6 \times {{10}^{ - 34}})(3 \times {{10}^8})}}} \right]

= 100 \times {10^{18}} = {10^{20}}