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
- A \(\frac{v^2 e}{m}\)
- B \(\frac{v^2 m}{2 e}\)
- C \(\frac{v^2 m}{e}\)
- D \(\frac{v^2 e}{2 m}\)
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
\(\therefore V=\frac{v^2 m}{2 e}\)
\(\frac{1}{2} m v^2=e V\)
\(V\) is the stopping potential
\(\therefore V=\frac{v^2 m}{2 e}\)
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