Question

The maximum velocity of the photoelectron emitted by the metal surface is $'v'$. Charge and mass of the photoelectron is denoted by $'e'$ and $'m'$ respectively. The stopping potential in volt is

• A. $\frac{v^{2}}{2\left(\frac{m}{e}\right)}$
• B. $\frac{v^{2}}{2\left(\frac{e}{m}\right)}$
• C. $\frac{v^{2}}{\left(\frac{e}{m}\right)}$
• D. $\frac{v^{2}}{\left(\frac{m}{e}\right)}$

Answer 1

Answer: (B)

Given, maximum velocity of a photoelectron $=v$
charge of photoelectron = 9
and mass of photoelectron $=m$.
Let, the stopping potential of the photoelectron $=V$,
Then, the maximum kinetic energy
$K E_{\max }=e V \Rightarrow \frac{1}{2} m v^{2}=e V$
$\left(\because K E=\frac{1}{2} m v^{2}\right)$
$\Rightarrow V=\frac{v^{2} m}{2 e}$
or $V=\frac{v^{2}}{2\left(\frac{e}{m}\right)}$