Question

If the frequency of light in a photoelectric experiment is doubled, the stopping potential will –
A) Be doubled.
B) Be halved
C) Becomes more than double.
D) Becomes less than double.

Answer 1

Hint: We need to understand the relation between the frequency of a light or the photon incident on the metal, its work function and the stopping potential to be applied to the electron which escapes from the metal as a result of the photoelectric effect.

Complete answer:
We know that a photon with a threshold frequency when incident on a metal surface can cause an electron to jump off the metal atom. The threshold frequency of the electron is defined as the minimum frequency required to pull the electron from the atom. This energy which is equivalent to the threshold frequency is known as the work function which is given as –
\[{{\phi }_{0}}=h{{\nu }_{0}}\]
Where \[{{\phi }_{0}}\] is the work function, \[{{\nu }_{0}}\] is the threshold frequency and ‘h’ is the Plank’s constant.
Now, consider a photon of frequency \[\nu \] which is greater than the threshold frequency hits the metal. An electron will be jumped from the metal surface with a kinetic energy which is due to the extra energy from the frequency, i.e.,
\[KE=h\nu -h{{\nu }_{0}}\]
Now, a stopping potential is the potential to be provided to the electron to stop its motion. This potential should be therefore, equal and opposite to the kinetic energy, which can be given as –
\[\begin{align}
  & KE=h\nu -h{{\nu }_{0}} \\
 & \therefore e{{V}_{0}}=h(\nu -{{\nu }_{0}}) \\
\end{align}\]
Where, \[{{V}_{0}}\] is the stopping potential and e is the electronic charge.
Now, we are given that another photon with double the value of the initial frequency hits the same metal surface. We can see the stopping potential will become –
\[\begin{align}
  & KE=h(2\nu )-h{{\nu }_{0}} \\
 & \Rightarrow e{{V}_{0}}'=h(2\nu -{{\nu }_{0}}) \\
 & \therefore e{{V}_{0}}'=h\nu +e{{V}_{0}} \\
\end{align}\]
We know that the energy possessed by the electron will have energy more than double the initial energy.
\[\therefore {{V}_{0}}'>2{{V}_{0}}\]

The correct answer is option C.

Note:
The work function of a metal is its characteristic property, i.e., it differs for each metal depending on the energy required to ionise the metal by removing an electron from the valence shell. Also, one photon can remove only one electron, i.e., the photon is quantised.