When light of frequency $v_{1}$ is incident on a metal with work function $W$ (where $h v_{1} > W)$ , the 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

\frac{W (v_{2} + v_{1})}{v_{1} V_{2} + v_{2} V_{1}}

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\frac{W (v_{2} + v_{1})}{v_{1} V_{1} + v_{2} V_{2}}

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\frac{W (v_{2} - v_{1})}{v_{1} V_{2} - v_{2} V_{1}}

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\frac{W (v_{2} - v_{1})}{v_{2} V_{2} - v_{1} V_{1}}

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Solution

The correct option is B $\frac{W (v_{2} - v_{1})}{v_{1} V_{2} - v_{2} V_{1}}$
$h v_{2} = w + e V_{2} . . . (i i)$
$\frac{v_{1}}{v_{2}} = \frac{w + e V_{1}}{w + e V_{2}} \Rightarrow W v_{1} + e V_{2} v_{1} = W v_{2} + e V_{1} v_{2}$
$\Rightarrow e = \frac{W (v_{2} - v_{1})}{v_{1} V_{2} - v_{2} V_{1}}$

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When light of frequency v1 is incident on a metal with work function W (where hv1>W), the photocurrent falls to zero at a stopping potential of V1. If the frequency of light is increased to v2, the stopping potential changes to V2. Therefore, the charge of an electron is given by

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