Question

The \[4eV\] is the energy of an incident photon and the work function is \[2eV\]. The stopping potential will be
A. \[2V\]
B. \[4V\]
C. \[6V\]
D. \[2\sqrt 2 V\]

Answer 1

Hint: Stopping potential is the potential required to bring the electron to rest. Also, we can say that stopping potential is defined as the minimum negative voltage applied to the anode to stop the photocurrent. The maximum kinetic energy of the electrons equals the stopping voltage when measured in electron volt. Here we find the stopping potential by using the values of the energy of the incident photon and the work function. By using Einstein’s equation we can solve this problem.

Formula used
Kinetic energy of photoelectrons is given as:
\[KE = E - \phi \]
Where E is the energy and \[\phi \] is the work function.
Also, \[KE = e{V_0}\]
Where, \[{V_0}\] is the stopping potential.
1eV=\[1.6 \times {10^{ - 19}}J\]

Complete step by step solution:
Given Energy of the incident photon, \[E = 4eV\].
Work function, \[\phi = 2eV\]
By Einstein’s equation,
\[KE = E - \phi \\
\Rightarrow E = \phi + KE\]
Now after putting the value of kinetic energy in terms of stopping potential we get;
\[E = \phi + e{V_0}\\
\Rightarrow {\rm{e}}{{\rm{V}}_0}{\rm{ = E - }}\phi \\
\Rightarrow {V_0} = \dfrac{{E - \phi }}{e}\]
Substituting values
\[{V_0} = \dfrac{{4eV - 2eV}}{e}\\
\therefore {V_0} = 2\,V\]
Therefore, the stopping potential will be 2V.

Hence option A is the correct answer.

Note: On the intensity of incident radiation, stopping potential does not depend. On increasing intensity, the value of saturated current increases, whereas the stopping potential remains unchanged. The minimum amount of energy that is required to eject an electron from the metal surface is called the work function. The stopping voltage can be used to determine the kinetic energy that the electrons have as they are ejected from the metal surface.