The maximum velocity of the photoelectron emitted by the metal surface is $v$, charge and mass of the…
The maximum velocity of the photoelectron emitted by the metal surface is $v$, charge and mass of the photoelectron are denoted by $e$ and $m$ respectively. The stopping potential in volt is
$\frac{v^2 e}{m}$
$\frac{v^2 m}{2 e}$
$\frac{v^2 m}{e}$
$\frac{v^2 e}{2 m}$
Solution
We have, potential energy to stop the fastest moving electrons is equal to the kinetic energy:
$\frac{1}{2} m v^2=e V$
$V$ is the stopping potential
$\therefore V=\frac{v^2 m}{2 e}$
.