Photoelectric emission is observed from a metallic surface for frequencies $v_{1} a n d v_{2}$ of the incident light rays $(v_{1} > v_{2})$ . If the maximum values of kinetic energy of the photoelectrons 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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$\frac{υ_{1} - ν_{2}}{k - 1}$

$\frac{k υ_{1} - υ_{2}}{k - 1}$

$\frac{k υ_{2} - υ_{1}}{k - 1}$

$\frac{υ_{2} - υ_{1}}{k}$

answer is B.

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Detailed Solution

From given details
$1 = h v_{1} - h v_{0} \Rightarrow h v_{0} = h v_{1} - 1$
$v_{0} = v_{1} - \frac{1}{h}$ ...........(1)
and $\begin{array}{l} k = h v_{2} - h v_{0} \\ k = h v_{2} - h v_{1} + 1 \\ h (v_{2} - v_{1}) = k - 1 \end{array}$
$h = \frac{k - 1}{v_{2} - v_{1}}$ ...........(2)
put value of h in equation (1)
$\begin{array}{l} v_{0} = v_{1} - \frac{υ_{2} - υ_{1}}{k - 1} \\ v_{0} = \frac{k v_{1} - v_{1} - v_{2} + v_{1}}{k - 1} \end{array}$