Question

The maximum velocity of the photoelectron emitted by the metal surface is ${}^{\prime }{v}^{\prime }$. Charge and mass of the photoelectron is denoted by ${}^{\prime }{e}^{\prime }$ and ${}^{\prime }{m}^{\prime }$ 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
$K{E}_{\max }=eV\Rightarrow \frac{1}{2}m{v}^2=eV$
$\left(\because KE=\frac{1}{2}m{v}^2\right)$
$\Rightarrow V=\frac{v^{2}m}{2e}$
or $V=\frac{v^2}{2\left(\frac{e}{m}\right)}$