Question

When photons of wavelength $\lambda_1$ are incident on an isolated sphere, the corresponding stopping potential is found to be V. When photons of wavelength $\lambda_2$are used, the corresponding stopping potential is thrice that of the above value. If light of wavelength $\lambda_3$ is used then the stopping potential for this case:
$\text{A.}\quad \dfrac{hc}{e} \left[ \dfrac{1}{\lambda_3}+\dfrac{1}{\lambda_2} -\dfrac{1}{\lambda_1} \right]$
$\text{B.}\quad \dfrac{hc}{e} \left[ \dfrac{1}{\lambda_3}+\dfrac{1}{2\lambda_2} -\dfrac{3}{2\lambda_1} \right]$
$\text{C.}\quad \dfrac{hc}{e} \left[ \dfrac{1}{\lambda_3}-\dfrac{1}{\lambda_2} -\dfrac{1}{\lambda_1} \right]$
$\text{D.}\quad \dfrac{hc}{e} \left[ \dfrac{1}{\lambda_3}+\dfrac{1}{2\lambda_2} -\dfrac{1}{\lambda_1} \right]$

Answer 1

Hint: Different material has different values of work function and stopping potential. It is the property of the material which defines how much energy is needed to eject an electron from the surface of a metal. For example, ejecting electrons out is easiest in metals such as tungsten.
Formula used:
$\dfrac{hc}{\lambda} = W_\circ + eV$

Complete answer:
Work function – It is defined as the minimum energy required to eject an electron from the surface of the material.
Threshold wavelength – It is defined as the maximum wavelength of the incident photon/ energy so that the electron can be ejected out of the surface.
Stopping potential – The minimum electric potential needed to stop the moving electron after getting ejected from the surface of the metal.

Now, for the first case, $\lambda = \lambda_1$
Thus using $\dfrac{hc}{\lambda} = W_\circ + eV$, we get
$\dfrac{hc}{\lambda_1} = W_\circ + eV$ . .. . 1
Now, for second case, $\lambda = \lambda_2 \ and \ V_o = 3V$
$\implies \dfrac{hc}{\lambda_2} = W_\circ + 3eV$ . .. . 2
Now, multiplying equation (1) with 3, we get;
$\dfrac{3hc}{\lambda_1} = 3W_\circ + 3eV$ . .. . 3
Now, subtracting (2) from (3), we get;
$hc\left[\dfrac{3}{\lambda_1} - \dfrac{1}{\lambda_2} \right]= 2W_\circ$
$\implies W_\circ = \dfrac{hc}{2}\left[ \dfrac{3}{\lambda_1} - \dfrac{1}{\lambda_2} \right]$
Now, in case (3), using $\dfrac{hc}{\lambda} = W_\circ + eV$, we get;
 $\dfrac{hc}{\lambda_3} = W_\circ + eV_x$
Putting $\implies W_\circ = \dfrac{hc}{2}\left[ \dfrac{3}{\lambda_1} - \dfrac{1}{\lambda_2} \right]$, we get;
$\dfrac{hc}{\lambda_3} = \dfrac{hc}{2}\left[ \dfrac{3}{\lambda_1} - \dfrac{1}{\lambda_2} \right] + eV_x$
$\implies eV_x = \dfrac{hc}{1} \left[ \dfrac{1}{\lambda_3}+\dfrac{1}{2\lambda_2} -\dfrac{3}{2\lambda_1} \right]$
$\implies V_x = \dfrac{hc}{e} \left[ \dfrac{1}{\lambda_3}+\dfrac{1}{2\lambda_2} -\dfrac{3}{2\lambda_1} \right]$

So, the correct answer is “Option B”.

Note:
Now, according to the law of conservation of energy, the energy of the incident photon will go in first ejecting the electron and then in the kinetic energy of the electron. Now, stopping potential is the minimum potential required to stop the most energetic photo electron. Hence in order to stop the moving electron, all the kinetic energy of the photoelectron must get converted to the potential energy of the electron so that it can come to stop. So the magnitude of kinetic energy and potential energy will be the same.