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MHT CET · Physics · Dual Nature of Matter

When a certain metal surface is illuminated with light of frequency \(v\), the stopping potential for the photoelectric current is \(V_s\). When the same surface is illuminated by a light of frequency \(\frac{v}{2}\), the stopping potential is \(V_s / 4\). The threshold frequency of photoelectric emission is

  1. A \(\frac{4 v}{3}\)
  2. B \(\frac{5 v}{3}\)
  3. C \(\frac{v}{3}\)
  4. D \(\frac{v}{5}\)
Verified Solution

Answer & Solution

Correct Answer

(C) \(\frac{v}{3}\)

Step-by-step Solution

Detailed explanation

Stopping potential is the energy required to stop the moving electrons after photoelectric emission. Hence it must be equal to the maximum kinetic energy of the emitted electron.
Incident energy \(=\) Threshold energy + Kinetic energy
Given:
\(h v=h v_0+e V_0\quad\ldots(1)\)
and
\(\frac{h v}{2}=h v_0+\frac{e V_0}{4}\quad\ldots(2)\)
Multiply equation (2) by 4 and substract equation (1) from it to obtain:
\(
v_0=\frac{v}{3}
\)