Figure shows the graph of stopping potential versus the frequency of radiation incident on a photosensitive metal. Threshold frequency of the metal is

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$\frac{(V_{2} v_{2} + V_{1} v_{1})}{V_{2} + V_{1}}$

$\frac{(V_{2} v_{1} + V_{1} v_{2})}{V_{2} + V_{1}}$

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

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

answer is D.

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Detailed Solution

$V e = h υ + h υ_{0}$

$V_{1} e = h υ_{1} - h υ_{0}______(1) V_{2} e = h υ_{2} - h υ_{0}______(2) \frac{(1)}{(2)} \Rightarrow \frac{V_{1}}{V_{2}} = \frac{υ_{1} - υ_{0}}{υ_{2} - υ_{0}} \Rightarrow (υ_{2} - υ_{0}) V_{1} = (υ_{1} - υ_{0}) V_{2} \Rightarrow υ_{0} = \frac{V_{2} v_{1} - V_{1} v_{2}}{V_{2} - V_{1}}$