Question

When light of frequency \(v_{1}\) is incident on a metal with work function \(W(w h e r e, h>W)\), the photocurrent falls to zerd at a stopping potential of \(\mathrm{V}_{1} .\) If the frequency of light is increased to \(\mathrm{v}_{2}\), the stopping potential changes to \(\mathrm{V}_{2}\). Therefore, the charge of an electron is given by
(A) \(\frac{\mathrm{W}\left(\mathrm{v}_{2}+\mathrm{v}_{1}\right)}{\mathrm{v}_{1} \mathrm{~V}_{2}+\mathrm{v}_{2} \mathrm{~V}_{1}}\)
(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 1

Ans : (C)
$$
\begin{aligned}
&h v_{1}=w+e v_{1} \ldots . .(i) \\
&h v_{2}=w+e V_{2} \ldots(\text { ii })
\end{aligned}
$$
$$
\begin{aligned}
&\frac{v_{1}}{v_{2}}=\frac{W \mathrm{eV}_{1}}{W \in \mathrm{V}_{2}} \\
&\Rightarrow W_{+} 1 \mathrm{eV}_{2} \mathrm{v}_{1}=W v_{2}+e V_{1} v \\
&\Rightarrow \mathrm{e} \frac{W\left(v_{2} v_{1}\right)}{V_{2} v_{1}-V_{1} v}
\end{aligned}
$$

Answer 2

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Answer 3

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