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

Q11

The minimum wavelength of

X

-rays produced by an electron accelerated through a potential difference of

V

volts is proportional to :

A

\dfrac{1}{V}

B

\dfrac{1}{\sqrt{V}}

C

V^{2}

D

\sqrt{V}

Correct Answer

Option A

Solution

The minimum wavelength,

\lambda_{\min}

, of X-rays produced by an electron can be determined using the equation derived from the energy of a photon and the energy given to an electron by an accelerating voltage,

V

. The energy of a photon is given by

E = hc/\lambda

, where

h

is Planck's constant,

c

is the speed of light, and $\lambda$ is the wavelength of the photon.

When an electron is accelerated through a potential difference of

V

volts, it gains kinetic energy equal to

eV

, where

e

is the electron charge.

This energy is then converted into a photon's energy when the electron collides with a target in an X-ray tube, resulting in X-rays of wavelength $\lambda$ .

Setting the kinetic energy equal to the photon energy gives

eV = hc/\lambda

. Solving for $\lambda$ gives

\lambda = hc/(eV)

.

Therefore, the minimum wavelength (corresponding to the maximum energy photon produced when all the kinetic energy is converted into photon energy) is inversely proportional to the voltage,

V

. Thus,

\lambda_{\min} \propto 1/V

.

Q12

The light rays having photons of energy 4.2 eV are falling on a metal surface having a work function of 2.2 eV. The stopping potential of the surface is

A 6.4 V

B 2 eV

C 2 V

D 1.1 V

Correct Answer

Option C

Solution

We know,

K{E_{\max }} = hv - h{v_0}

e{V_0} = hv - h{v_0}

( $\because$

K{E_{\max }} = e{V_0}

)

e{V_0} = 4.2\,eV - 2.2\,eV

$\therefore$

{V_0} = 2\,V

Q13

The threshold frequency of a photoelectric metal is v 0 . If light of frequency 4v 0 is incident on this metal, then the maximum kinetic energy of emitted electrons will be :

A 4 hv 0

B hv 0

C 2 hv 0

D 3 hv 0

Correct Answer

Option D

Solution

According to Einstein's photoelectric equation

{(K.E)_{\max }} = hv - h{v_0}

{(K.E)_{\max }} = h(4{v_0}) - h{v_0}

{(K.E)_{\max }} = 3h{v_0}

Q14

When two monochromatic lights of frequency, v and

{v \over 2}

are incident on a photoelectric metal, their stopping potential becomes

{{{V_s}} \over 2}

and V s respectively. The threshold frequency for this metal is

A 2v

B 3v

C

{2 \over 3}

v

D

{3 \over 2}

v

Correct Answer

Option D

Solution

Since

{k_{\max }} = e{V_s} = hv - \phi

{{e{V_s}} \over 2} = hv - h{v_0}

....... (i)

e{V_s} = {{hv} \over 2} - h{v_0}

...... (ii)

{1 \over 2}\left[ {{{hv} \over 2} - h{v_0}} \right] = hv - h{v_0}

\Rightarrow h{v_0} - {{h{v_0}} \over 2} = hv - {{hv} \over 4}

\Rightarrow {{h{v_0}} \over 2} = {{3hv} \over 4}

{v_0} = {{3v} \over 2}

* Language of question is wrongly framed. The values of stopping potentials should be interchanged.

Q15

The number of photons per second on an average emitted by the source of monochromatic light of wavelength 600nm, when it delivers the power of 3.3

\times

10

-

3 watt will be : (h = 6.6

\times

10

-

34 Js)

A 10 15

B 10 18

C 10 17

D 10 16

Correct Answer

Option D

Solution

p = {{nhc} \over \lambda } \Rightarrow n = {{p\lambda } \over {hc}}

n = {{3.3 \times {{10}^{ - 3}} \times 600 \times {{10}^{ - 9}}} \over {6.6 \times {{10}^{ - 34}} \times 3 \times {{10}^8}}} = {10^{16}}

Q16

An electromagnetic wave of wavelength '

\lambda

' is incident on a photosensitive surface of negligible work function. If 'm' mass is of photoelectron emitted from the surface has de-Broglie wavelength

\lambda

d , then :

A

\lambda = \left( {{{2h} \over {mc}}} \right){\lambda _d}^2

B

\lambda = \left( {{{2m} \over {hc}}} \right){\lambda _d}^2

C

{\lambda _d} = \left( {{{2mc} \over h}} \right){\lambda ^2}

D

\lambda = \left( {{{2mc} \over h}} \right){\lambda _d}^2

Correct Answer

Option D

Solution

{{hc} \over \lambda } = {k_{\max }} + \phi

[given $\phi$ is negligible] So,

{{hc} \over \lambda } = {K_{\max }}

{\lambda _d} = {h \over {\sqrt {2m{K_{\max }}} }} \Rightarrow {K_{\max }} = {{{h^2}} \over {2m\lambda _d^2}}

\left( {{{hc} \over \lambda }} \right) = {{{h^2}} \over {2m\lambda _d^2}} \Rightarrow \lambda = \left( {{{2mc} \over h}} \right)\lambda _d^2

Q17

Light of frequency 1.5 times the threshold frequency is incident on a photosensitive material. What will be the photoelectric current if the frequency is halved and intensity is doubled?

A four times

B one-fourth

C zero

D doubled

Correct Answer

Option C

Solution

v = {3 \over 2}{v_0}

and

{v'} = {v \over 2} = {3 \over 4}{v_0}

Here we see

v' < v' < {v_0}

Below threshold frequency, no photo-electric emission takes place.

Q18

Light with an average flux of 20 W/cm 2 falls on a non-reflecting surface at normal incidence having surface area 20 cm 2 . The energy received by the surface during time span of 1 minute is :

A 12

\times

10 3 J

B 24

\times

10 3 J

C 48

\times

10 3 J

D 10

\times

10 3 J

Correct Answer

Option B

Solution

Energy = intensity $\times$ time $\times$ area = 20 $\times$ 60 $\times$ 20 E = 24 $\times$ 10 3 J

Q19

An electron is accelerated from rest through a potential difference of V volt. If the de-Broglie wavelength of the electron is 1.227

\times

10 -2 nm, the potential difference is :

A 10 2 V

B 10 3 V

C 10 4 V

D 10 V

Correct Answer

Option C

Solution

The de-Broglie wavelength and the accelerated potential is related as follows :

\lambda = {{12.27} \over {\sqrt V }}

A^\circ

\sqrt V = {{12.27 \times {{10}^{ - 10}}} \over {1.227 \times {{10}^{ - 11}}}} = {10^2}

V = {10^4}

volts

Q20

An electron is accelerated through a potential difference of 10,000 V. Its de Broglie wavelength is (nearly) : (m e = 9

\times

10 –31 kg)

A 12.2 × 10 –12 m

B 12.2 × 10 –13 m

C 12.2 × 10 –14 m

D 12.2 nm

Correct Answer

Option A

Solution

Expression for an electron accelerated through a potential V is given as $\lambda$ =

{{12.27} \over {\sqrt V }}\mathop A\limits^o

=

{{12.27 \times {{10}^{ - 10}}} \over {\sqrt {10000} }}

= 12.2 × 10 –12 m