COMEDK · Physics · 27. 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
We know that, \(e V_0=h v-\phi_0\) where, \(\quad V_0=\) stopping potential and \(\quad v=\) frequency of light. Case II \(\frac{e V_0}{4}=\frac{h v}{2}-\phi_0\) From Eqs. (i) and (ii), we get…
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