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Photo electric emission is observed from a metallic surface for frequencies $ν_{1}$ and $ν_{2}$ of the incident light rays $(ν_{1} > ν_{2})$ . If the maximum values of kinetic energy of the photo elections emitted in the two cases are in the ratio of $1 : K$ , then the threshold frequency of the metallic surface is

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$\frac{ν_{2} - ν_{1}}{K - 1}$

$\frac{K ν_{1} - ν_{2}}{K - 1}$

$\frac{K ν_{2} - ν_{1}}{K - 1}$

$\frac{ν_{2} - ν_{1}}{K}$

answer is B.

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Detailed Solution

$K E_{max} = h (ν - ν_{o})$

$\begin{aligned} ∴ \frac{K_{1}}{K_{2}} = \frac{ν_{1} - ν_{o}}{ν_{2} - ν_{o}} \\ ∴ \frac{1}{K} = \frac{ν_{1} - ν_{o}}{ν_{2} - ν_{o}} \\ ∴ ν_{o} = \frac{K ν_{1} - ν_{2}}{K - 1} \end{aligned}$

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