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 is denoted by 'e' and 'm' respectively. The stopping potential in volt is
  1. $\frac{v^{2}}{\left(\frac{m}{e}\right)}$
  2. $\frac{v^{2}}{\left(\frac{e}{m}\right)}$
  3. $\frac{v^{2}}{2\left(\frac{m}{e}\right)}$
  4. $\frac{v^{2}}{2\left(\frac{e}{m}\right)}$

Solution

$\frac{1}{2} m v^{2}=e v$ $v=\frac{m v^{2}}{2 e}=\frac{v^{2}}{2\left(\frac{e}{m}\right)}$

Asked in: MHT CET 2020 (12 Oct Shift 1)

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