WBJEE · Physics · Dual Nature of Matter
The stopping potential for photoelectrons from a metal surface is \(V_{1}\) when monochromatic light of frequency \(v_1\) is incident on it. The stopping potential becomes \(V_{2}\) when monochromatic light of another frequency is incident on the same metal surface. If \(h\) be the Planck's constant and \(e\) be the charge of an electron, then the frequency of light in the second case is
- A \(v_{1}-\frac{e}{h}\left(V_{2}+V_{1}\right)\)
- B \(v_{1}+v \frac{e}{h}\left(V_{2}+V_{1}\right)\)
- C \(v_{1}-\frac{e}{h}\left(V_{2}-V_{1}\right)\)
- D \(v_{1}+\frac{e}{h}\left(V_{2}-V_{1}\right)\)
Answer & Solution
Correct Answer
(D) \(v_{1}+\frac{e}{h}\left(V_{2}-V_{1}\right)\)
Step-by-step Solution
Detailed explanation
Here, \(h v_{1}=\phi_{0}=e V_{1}\) ...(i) and \(h v_{2}=\phi_{0}+e V_{2}\)...(ii) From Eqs. (i) and (ii), we have \(\begin{aligned} & h\left(v_{2}-v_{1}\right)=e\left(V_{2}-v_{1}\right) \\ \Rightarrow \quad & v_{2}=\frac{e}{h}\left(V_{2}-v_{1}\right)+v_{1} \end{aligned}\)
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