Question

Photons with energy $5\,eV$ are incident on a cathode $C$ in a photoelectric cell. The maximum energy of emitted photoelectrons is $2\,eV$. When photons of energy $6\,eV$ are incident on$C$, no photoelectrons will reach the anode $A$, if the stopping potential of $A$ relative to $C$ is:
(A) $ - 3\,V$
(B) $ + 3\,V$
(C) $ + 4\,V$
(D) $ - 1\,V$

Answer 1

Hint
The stopping potential energy of the $A$ relative to $C$ can be determined by using the kinetic energy of the photo electron, by using the kinetic energy formula the work function is determined by using the first energy of photon, then by using the work function and the second given energy of photon, the stopping potential can be determined.
The kinetic energy of the photoelectron is given by,
$\Rightarrow K = E - \phi $
Where, $K$ is the kinetic energy of the photoelectron, $E$ is the energy of the photon and $\phi $ is the work function of the photon.

Complete step by step answer
Given that, The energy of the photon is, $E = 5\,eV$,
The kinetic energy of the photon is, $\phi = 2\,eV$,
The energy of the photon in the second case is, $E = 6\,eV$.
Now, The kinetic energy of the photoelectron is given by,
$\Rightarrow K = E - \phi \,...................\left( 1 \right)$
By substituting the kinetic energy of the photoelectron and the energy of the photon in the above equation (1), then the above equation (1) is written as,
$\Rightarrow 2 = 5 - \phi $
By rearranging the terms in the above equation, then the above equation is written as,
$\Rightarrow \phi = 5 - 2$
By subtracting the terms in the above equation, then the above equation is written as,
$\Rightarrow \phi = 3\,eV$
Now the energy of the electron $6\,eV$ is incident on the $C$ , then
By substituting the energy of the photon and the work function in the equation (1), then the equation (1) is written as,
$\Rightarrow K = 6 - 3$
By subtracting the terms in the above equation, then the above equation is written as,
$\Rightarrow K = 3\,eV$
The kinetic energy is written as,
$\Rightarrow e{V_0} = 3\,eV$
By cancelling the same terms, then the stopping potential of the $A$ relative to $C$ is $ - 3\,V$.
Hence, the option (C) is the correct answer.

Note
The stopping potential is negative because it is the potential which acts in the opposite direction to the normal potential so that it is called a stopping potential. If the stopping potential is maximum in the system, the system is not efficient.