WBJEE · Physics · Dual Nature of Matter
When a certain metal surface is illuminated with light of frequency \(v\), the stopping potential for photoelectric current is \(V_{0}\). When the same surface is illuminated by light of frequency \(\frac{v}{2}\) the stopping potential is \(\frac{V_{0}}{4} .\) The threshold
frequency for photoelectric, emission is
- A \(\frac{v}{6}\)
- B \(\frac{v}{3}\)
- C \(\frac{2 v}{3}\)
- D \(\frac{4 v}{3}\)
Answer & Solution
Correct Answer
(B) \(\frac{v}{3}\)
Step-by-step Solution
Detailed explanation
Ist case \[ \begin{array}{l} h v=h v_{0}+e V_{0} \\ \frac{h v}{2}=h v_{0}+\frac{e V_{0}}{4} \end{array} \] Solving Eqs. (i) and (ii), we get \[ v_{0}=\frac{v}{3} \]
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