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 ______.

Options

• ${\displaystyle \frac{{\text{v}}^2}{2\left(\frac{\text{m}}{\text{e}}\right)}}$

• ${\displaystyle \frac{{\text{v}}^2}{2\left(\frac{\text{e}}{\text{m}}\right)}}$

• ${\displaystyle \frac{{\text{v}}^2}{\left(\frac{\text{e}}{\text{m}}\right)}}$

• ${\displaystyle \frac{{\text{v}}^2}{\left(\frac{\text{m}}{\text{e}}\right)}}$

Answer 1

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 ${\displaystyle \underline{\frac{{\text{v}}^2}{2\left(\frac{\text{e}}{\text{m}}\right)}}}$.

Explanation:

Given, maximum velocity of a photoelectron = v

charge of photoelectron = e

and mass of photoelectron = m.

Let, the stopping potential of the photoelectron = V,

Then, the maximum kinetic energy

${\displaystyle {\text{KE}}_{\text{max}}=\text{eV}\Rightarrow \frac{1}{2}{\text{mv}}^2=\text{eV} (\because \text{KE}=\frac{1}{2}{\text{mv}}^2)}$

${\displaystyle \Rightarrow \text{V}=\frac{{\text{v}}^2\text{m}}{2\text{e}}}$ or

V = ${\displaystyle \frac{{\text{v}}^2}{2\left(\frac{\text{e}}{\text{m}}\right)}}$