COMEDK · Physics · 27. Dual Nature of Matter
When a certain metal surface is illuminated with light of frequency \(\nu\), the stopping potential for photoelectric current is \(V_0\). When the same surface is illuminated by light of frequency \(\dfrac{\nu}{2}\), the stopping potential is \(\dfrac{V_0}{4}\). The threshold frequency for photoelectric emission is
- A \(\dfrac{4\nu}{3}\)
- B \(\dfrac{\nu}{3}\)
- C \(\dfrac{2\nu}{3}\)
- D \(\dfrac{\nu}{6}\)
Answer & Solution
Correct Answer
(B) \(\dfrac{\nu}{3}\)
Step-by-step Solution
Detailed explanation
According to Einstein's photoelectric equation, the maximum kinetic energy \(K_{max}\) is given by \(K_{max} = h\nu - \phi\), where \(\phi = h\nu_0\) is the work function and \(\nu_0\) is the threshold frequency. The stopping potential \(V_0\) is related to the maximum kinetic…
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