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

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

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

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

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

answer is B.

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Detailed Solution

$E_{1} = h (v_{1} - v_{0}) and E_{2} = h (v_{2} - v_{0})$

Dividing them, we get

$\frac{E_{2}}{E_{1}} = \frac{v_{2} - v_{0}}{v_{1} - v_{0}} = \frac{n}{1}$

Hence, we have:

$n = \frac{v_{2} - v_{0}}{v_{1} - v_{0}}$

which gives $v_{0} = \frac{n v_{1} - v_{2}}{(n - 1)}$

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