WBJEE · Physics · Dual Nature of Matter
When light of frequency \(v_{1}\) is incident on a metal with work function \(W\) (where \(h v_{1}>w\) ), then photocurrent falls to zero at a stopping potential of \(V_{1}\). If the frequency of light is increased to \(v_{2},\) the stopping potential changes to \(V_{2}\). Therefore, the charge of an electron is given by
- A \(\frac{W\left(v_{2}+v_{1}\right)}{v_{1} V_{2}+v_{2} V_{1}}\)
- B \(\frac{W\left(v_{2}+v_{1}\right)}{v_{1} V_{1}+v_{2} V_{2}}\)
- C \(\frac{W\left(v_{2}-v_{1}\right)}{v_{1} V_{2}-v_{2} V_{1}}\)
- D \(\frac{W\left(v_{2}-v_{1}\right)}{v_{2} V_{2}-v_{1} V_{1}}\)
Answer & Solution
Correct Answer
(C) \(\frac{W\left(v_{2}-v_{1}\right)}{v_{1} V_{2}-v_{2} V_{1}}\)
Step-by-step Solution
Detailed explanation
From Einstein's photoelectric equation, \[ \begin{array}{l} h v=W+e V \\ W=\text { Work function } \end{array} \] \(v=\) Frequency of incident photon \(V=\) Stopping potential For a photon of frequency \(v_{1},\) \[ h v_{1}=W+e V_{1} \] For photon of frequency \(v_{2}\)…
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