Photoelectric emission is observed from a surface for frequencies $v_{1}$ and $v_{2}$ of the incident radiation $(v_{1} > v_{2})$ . If maximum kinetic energies of the phot electrons in the two eases are in the ratio 1 : K, then the threshold frequency is given by

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$\frac{K v_{1} - v_{2}}{K - 1}$

$\frac{K v_{2} - v_{1}}{K - 1}$

$\frac{K v_{2} - v_{1}}{K}$

$\frac{v_{2} - v_{1}}{K - 1}$

answer is D.

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Detailed Solution

It is known that;

$\begin{aligned} K E_{1} = h v_{1} - h v_{0}; K E_{2} = h v_{2} - h v_{0} \\ \frac{K E_{1}}{K E_{2}} = \frac{1}{K} = \frac{v_{1} - v_{0}}{v_{2} - v_{0}} \Rightarrow v_{0} = \frac{K v_{1} - v_{2}}{K - 1} \end{aligned}$

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