Question

Find the frequency of light that ejects an electron from a metal surface fully stopped by a retarding potential of 3V. The photoelectric effect begins in this metal at a frequency of $6 \times {10^{14}}Hz$.
A. $1.324 \times {10^{15}}Hz$
B. $2.295 \times {10^{16}}Hz$
C. $3.678 \times {10^{18}}Hz$
D. $2.7 \times {10^{14}}Hz$

Answer 1

Hint: In order to solve this problem, Einstein's photoelectric equation must be used. It says that the incident light energy is spent on uplifting an electron from its ground state and the kinetic energy of the moving electron after it is ejected from the metal plate.

Formula used:
Kinetic energy, $KE = h\upsilon - \phi $
where $h\upsilon $ is the incident light energy and $\phi $ is called the work function of the material.

Complete Step by Step Answer:
The photoelectric effect is the phenomenon in which the electrons are emitted from a metal surface when the light with a sufficient frequency is incident upon. The concept of photoelectric effect was first documented in the year 1887 by Heinrich Hertz and later by Hallwach and Lenard in their famous experiment in 1902.
Einstein was finally able to explain this phenomenon of photoelectric effect by using Planck’s idea that light consists of discrete energy packets known as photons. These photons carry an energy equivalent to the frequency of light.
Energy carried by photon, $E = h\upsilon $ where h = Planck’s constant = $6.624 \times {10^{ - 34}}J - s$
The work function of a material is defined as the minimum energy that is required to eject the electrons from inside of the material to be ready to accelerate.
By Einstein's equation, we have –
Kinetic energy, $KE = h\upsilon - \phi $
where $h\upsilon $ is the incident light energy and $\phi $ is called the work function of the material.
The stopping potential is defined as the negative potential to be applied on the plates to completely stop the motion of the negative electrons.
In order to stop the electrons with some kinetic energy, a potential energy equal to the kinetic energy must be applied.
If ${V_0}$ is the stopping potential,
$K{E_{\max }} = e{V_0}$
Substituting in the above equation of photoelectric effect, we have –
$ KE = h\upsilon - \phi $
$ e{V_0} = h\upsilon - \phi $
Given,
Stopping potential, ${V_0} = 3V$
Frequency at which photoelectric begins, ${\upsilon _\phi } = 6 \times {10^{14}}Hz$
Work function , $\phi = h{\upsilon _\phi } = 6.626 \times {10^{ - 34}} \times 6 \times {10^{15}} = 39.756 \times {10^{ - 19}}J$
Elementary charge, $e = 1.602 \times {10^{ - 19}}C$
$ e{V_0} = h\upsilon - \phi $
$ e{V_0} = h\upsilon - h{\upsilon _\phi }\because \phi = h{\upsilon _\phi } $
Substituting and rearranging, we get –
$
  h\upsilon - h{\upsilon _\phi } = e{V_0} \\
 \Rightarrow \upsilon - {\upsilon _\phi } = \dfrac{{e{V_0}}}{h} \\
 \Rightarrow \upsilon = \dfrac{{e{V_0}}}{h} + {\upsilon _\phi } $
On substituting the corresponding values,
$ \upsilon = \dfrac{{1.602 \times {{10}^{ - 19}} \times 3}}{{6.626 \times {{10}^{ - 34}}}} + 6 \times {10^{14}} \\
\Rightarrow \upsilon = \dfrac{{4.806 \times {{10}^{ - 19}}}}{{6.626 \times {{10}^{ - 34}}}} + 6 \times {10^{14}} \\
\Rightarrow \upsilon = 0.725 \times {10^{34 - 19}} + 6 \times {10^{14}} \\
\Rightarrow \upsilon = 0.725 \times {10^{15}} + 6 \times {10^{14}} \\
\Rightarrow\upsilon = 7.25 \times {10^{14}} + 6 \times {10^{14}} \\
\Rightarrow \upsilon = 13.24 \times {10^{14}}Hz \\
\Rightarrow \upsilon = 1.324 \times {10^{15}}Hz \\
 $
$\therefore$ The frequency of light is the frequency of light is $\upsilon = 1.324 \times {10^{15}}$Hz . Hence, option (A) is the correct answer.

Note:
The photoelectric effect of Einstein gave a solid basis for the scientific theory of Max Planck that the light is not a continuous wave but a discrete wave with energy packets, each called a photon.