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MHT CET · Physics · Dual Nature of Matter

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

  1. A \(\frac{v^2 e}{m}\)
  2. B \(\frac{v^2 m}{2 e}\)
  3. C \(\frac{v^2 m}{e}\)
  4. D \(\frac{v^2 e}{2 m}\)
Verified Solution

Answer & Solution

Correct Answer

(B) \(\frac{v^2 m}{2 e}\)

Step-by-step Solution

Detailed explanation

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
\(\therefore V=\frac{v^2 m}{2 e}\)
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