A non-monochromatic light is used in an experiment on the photoelectric effect. The stopping potential
A. Is related to the mean wavelength
B. Is related to the longer wavelength
C. Is related to the shortest wavelength
D. Is not related to the wavelength.

Hint: In photoelectric effect the stopping potential is the potential that is required for stopping the electrons having maximum kinetic energy. The wave with maximum frequency will emit an electron with maximum kinetic energy. We know that the frequency and wavelength are inversely related. Using this relation, we can arrive at the correct answer.

Complete step by step answer:
The photoelectric effect is the ejection of electrons from a metal surface when light radiation falls on it.
The energy of the incoming photon is used for pulling out an electron from the surface and extra energy is converted into the kinetic energy of the electron. So, if a photon has higher energy then the kinetic energy of the electron emitted will be higher. Since non-monochromatic light is used, the electron with maximum kinetic energy will be produced by the component of light with maximum frequency.
This is because the energy of a photon is given as
$E = h\upsilon $
where $h$ is the planck's constant and $\upsilon $ is the frequency of the wave.
The higher the frequency higher will be the energy of the photon.
The photoelectric equation is given as
$\Rightarrow h\upsilon = \phi + \dfrac{1}{2}m{v^2}$
Where $\phi $ is the work function and $\dfrac{1}{2}m{v^2}$ is the kinetic energy.
From this, we can write,
$\Rightarrow \dfrac{1}{2}m{v^2} = h\upsilon - \phi $
In the photoelectric effect, the stopping potential is the potential required for stopping the electrons having maximum kinetic energy. Thus, we can write
$\Rightarrow e{V_0} = h\upsilon - \phi $ ……………..(1)
Where ${V_o}$ is the stopping potential, $e$ is the charge of an electron.
We know that frequency is related to wavelength by the following relation
$c = \upsilon \lambda $
On substituting for $\upsilon $ in equation (1) we get
$\Rightarrow e{V_0} = \dfrac{{hc}}{\lambda } - \phi $
So, we can see that the stopping potential depends on the wavelength.
In the given question non-monochromatic light is used. Which means that light consists of several wavelengths. We know that frequency is inversely related to wavelength. So, the shortest wavelength will have the maximum frequency. Hence it will emit photoelectrons with maximum kinetic energy. Stopping potential would be the potential required to stop these electrons. Therefore, the wavelength $\lambda $ corresponds to the shortest wavelength. Since kinetic energy is maximum for the shortest wavelength.

Hence the correct answer is option C.

Note:
Remember that frequency is inversely related to wavelength. A wave with higher frequency will have higher energy. So, the kinetic energy of emitted electrons will be maximum for maximum frequency. When the frequency is the maximum wavelength will be minimum.