When light of frequency v_{1} is incident on | vedclass

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(C) From Einstein's photoelectric equation, $h v = W + e V$, where $W$ is the work function, $v$ is the frequency of incident light, and $V$ is the st

Vedclass · Accepted Answer

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

Vedclass · Answer

(C) From Einstein's photoelectric equation, $h v = W + e V$, where $W$ is the work function, $v$ is the frequency of incident light, and $V$ is the stopping potential.For frequency $v_{1}$, we have: $h v_{1} = W + e V_{1}$  --- $(i)$For frequency $v_{2}$, we have: $h v_{2} = W + e V_{2}$  --- (ii)Subtracting equation $(i)$ from equation (ii):$h(v_{2} - v_{1}) = e(V_{2} - V_{1})$$e = \frac{h(v_{2} - v_{1})}{V_{2} - V_{1}}$However, we need to express $e$ in terms of $W$. From $(i)$, $h = \frac{W + e V_{1}}{v_{1}}$. Substituting this into (ii):$(\frac{W + e V_{1}}{v_{1}}) v_{2} = W + e V_{2}$$W v_{2} + e V_{1} v_{2} = W v_{1} + e V_{2} v_{1}$$e(V_{1} v_{2} - V_{2} v_{1}) = W v_{1} - W v_{2}$$e(V_{1} v_{2} - V_{2} v_{1}) = -W(v_{2} - v_{1})$$e = \frac{W(v_{2} - v_{1})}{V_{2} v_{1} - V_{1} v_{2}}$This matches option $C$.

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