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
- A \(\frac{4 v}{3}\)
- B \(\frac{5 v}{3}\)
- C \(\frac{v}{3}\)
- D \(\frac{v}{5}\)
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:
\(
v_0=\frac{v}{3}
\)
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:
\(
v_0=\frac{v}{3}
\)
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