Question

If the maximum kinetic energy of emitted photoelectrons from a metal surface of work function $1.7eV$ is $2.5eV$ . If the wavelength of incident radiation is halved, then stopping potential will be
(A) $2.5V$
(B) $5V$
(C) $6.7V$
(D) $1.1V$

Answer 1

Hint
The given question belongs to the topic photons and photoelectric effect; in the question, we have been provided with the maximum kinetic energy of the emitted photoelectrons from the metal surface and the work function of the metal. We have been asked to find the change in the stopping potential if the wavelength of the radiation is halved. We know that if the wavelength of the radiation decreases, the frequency increases and higher the frequency, greater is the kinetic energy of the emitted photoelectrons. So if the wavelength is to be decreased, then the stopping potential must increase. Let’s head towards a detailed solution to the given question.
Formula Used: $V=\dfrac{E}{e}$ , $E\propto \dfrac{1}{\lambda }$

Complete step by step answer
We know that the equation of photoelectric effect is given as $E-\phi =eV$ where $E$ is the energy of the incident radiation, $\phi$ is the work function of the metal and $eV$ is the product of the elementary charge on the electron and the stopping potential
The stopping potential can be defined as the minimum potential difference it would take to stop the process of photoelectric emission.
Now if you can stop the fastest moving electron, then the other electrons can be very easily stopped. So we can also say that the stopping potential is dependent on the kinetic energy of the fastest moving electron.
The stopping potential for the data given can hence be given as ${{V}_{1}}=\dfrac{E}{e}=\dfrac{2.5eV}{e}=2.5V$
Now we all know that the energy of any electromagnetic radiation can be given as
$\begin{aligned} & E=h\nu =\dfrac{hc}{\lambda }\left[ \because c=\nu \lambda \right] \\\ & \Rightarrow E\propto \dfrac{1}{\lambda } \\\ \end{aligned}$
Where $h$ is the Planck’s constant, $c$ is the speed of light, $\nu$ is the frequency of the radiation and $\lambda$ is the wavelength of the radiation
Since we have established that the energy of the fastest moving electron is inversely proportional to the wavelength of the incident radiation, we can say that if the wavelength is halved, the energy of the fastest moving electron would be ${{E}_{2}}=2\times 2.5eV=5eV$
The new stopping potential can hence be given as ${{V}_{2}}=\dfrac{{{E}_{2}}}{e}=\dfrac{5eV}{e}=5V$
Hence we can say that option (B) is the correct answer.

Note
We have established the general equation for the photoelectric effect but we haven’t used it in our solution. In most of the questions, we use the general equation for the photoelectric effect, hence we have stated it for understanding purpose. This question could have been solved faster with a logical approach hence we did that. You could also think that since the work function of a metal surface is constant, it will not change the stopping potential for incident radiation.