Question

The maximum velocity of the photoelectrons emitted from the surface is $v$ when light of frequency $n$ falls on a metal surface if the incidence frequency is increased in $3n$, the maximum velocity of the ejected photoelectron will be:
(A) Equal to $\sqrt 3 V$
(B) Equal to $2V$
(C) More than $\sqrt 3 V$
(D) Less than $\sqrt 3 V$

Answer 1

Hint: The maximum velocity and the frequency of a photo electron is given, we have to find the maximum velocity of the photoelectron if the frequency is increased. Use Einstein's photoelectric equation to find the answer. It will be better if you know the whole derivation of the photoelectric equation.

Complete step by step answer
Einstein conducted experiments on the photons and formed the photoelectric equation that gives the energy of the photon.
Photons are allowed to hit on the metal surface, when the photon hits the surface of the metal electron from the metal surface it gets ejected, it is known as the photoelectron using this mechanism photoelectric current is produced . Based on this experiment, the energy of the incident photon is calculated.
The incident energy of the photon is equal to the sum of work function and the kinetic energy of the photon.
$ \Rightarrow {\text{I}}{\text{.E = }}\Phi {\text{ + K}}{\text{.E}}$
$ \Rightarrow K.{\text{E = I}}{\text{.E - }}\Phi $
Where,
$\Phi = h{\nu _0}{\text{ }} \to {\text{1}}$
$I.E = h\nu {\text{ }} \to {\text{2}}$
Then the kinetic energy becomes
$ \Rightarrow K.{\text{E = }}h\nu {\text{ - h}}{\nu _0}$
$ \Rightarrow K.{\text{E = }}h(\nu {\text{ - }}{\nu _0})$
$K.E = \dfrac{1}{2}m{v^2}$
$ \Rightarrow \dfrac{1}{2}m{v^2}{\text{ = }}h(\nu {\text{ - }}{\nu _0})$
Where,
I.E is the energy of incident electron
$\Phi $ is the work function of the metal surface
K.E is the kinetic energy of the incident photon
h is the Planck constant
${\nu _0}$ is the threshold energy
$\nu $ is the frequency of incident photon
m is the mass of the photon
V is the velocity of the incident photon
Given,
The maximum velocity of the photoelectrons emitted from the surface is $v$
Frequency of the incident light is $n$
When,
Frequency of the incident photon is increased to $3n$
The maximum velocity of the photoelectrons emitted from the surface =?
The maximum kinetic energy of the photon is given by
$ \Rightarrow \dfrac{1}{2}m{v^2}_{\max }{\text{ = }}h(\nu {\text{ - }}{\nu _0})$
$ \Rightarrow \dfrac{1}{2}m{v^2}_{\max }{\text{ = }}(h\nu {\text{ - h}}{\nu _0})$
From equation 1
$ \Rightarrow \dfrac{1}{2}m{v^2}_{\max }{\text{ = }}(h\nu {\text{ - }}\phi )$
When,
The maximum velocity of the photoelectrons emitted from the surface is $v$
Frequency of the incident light is $n$
Then,
$ \Rightarrow \dfrac{1}{2}m{v^2}_{\max }{\text{ = }}(hn{\text{ - }}\phi ){\text{ }} \to {\text{3}}$
When,
Frequency of the incident photon is increased to $3n$
Let the maximum velocity of the photoelectrons emitted from the surface be ${v_x}$
$ \Rightarrow \dfrac{1}{2}m{v_x}{^2_{\max }}{\text{ = }}(3hn{\text{ - }}\phi )$
$ \Rightarrow \dfrac{1}{2}m{v_x}{^2_{\max }}{\text{ = 3}}(hn{\text{ - }}\phi )$
From equation 3, we get
\[ \Rightarrow \dfrac{1}{2}m{v_x}{^2_{\max }}{\text{ = 3}}\dfrac{1}{2}m{v^2}_{\max }\]
\[ \Rightarrow {v_x}{^2_{\max }}{\text{ = 3}}{v^2}_{\max }\]
Hence the maximum velocity of photoelectrons emitted from the surface when the frequency of the incident photon is increased to $3n$ is \[{\text{3}}{v^2}_{\max }\]
It is clearly seen that the velocity of photoelectron of 3n frequency is more than $\sqrt 3 V$

Hence the correct answer is option (C) more than $\sqrt 3 V$

Note: In between the derivation a word called work function is used. Work function is defined as the amount of energy required to eject an electron from the metal surface. If the energy of the incident electron is less than the work function then the photoelectron will not be ejected from the metal surface.