Photoelectric emission is observed from a metallic surface for frequencies v 1 and v 2 of the incident light rays ( v 1 v 2 ). If the maximum values of kinetic energy of the photoelectrons e

Question

Photoelectric emission is observed from a metallic surface for frequencies v 1 ​ and 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 k : 1 , then what is the threshold frequency of the metallic surface?

TestHub · Accepted Answer

Using Einstein's photoelectric equation, the maximum kinetic energy of photoelectrons is given by $K = h (v - v_{0})$ , where $v_{0}$ is the threshold frequency.

For frequency $v_{1}$ , $K_{1} = h (v_{1} - v_{0})$

For frequency $v_{2}$ , $K_{2} = h (v_{2} - v_{0})$

Given that the ratio of maximum kinetic energies is $k : 1$ , we have:

$\frac{K _{1}}{K _{2}} = \frac{h ( v _{1} - v _{0} )}{h ( v _{2} - v _{0} )} = k$

$\frac{v _{1} - v _{0}}{v _{2} - v _{0}} = k$

$v_{1} - v_{0} = k (v_{2} - v_{0})$

$v_{1} - v_{0} = k v_{2} - k v_{0}$

$k v_{0} - v_{0} = k v_{2} - v_{1}$

$v_{0} (k - 1) = k v_{2} - v_{1}$

$v_{0} = \frac{k v _{2} - v _{1}}{k - 1}$

Answer: $\frac{k v _{2} - v _{1}}{k - 1}$

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