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 1 : k , then the threshold frequency of metallic surface is

TestHub · Accepted Answer

According to Einstein's photoelectric equation, the maximum kinetic energy of emitted photoelectrons is given by:

$K_{m a x} = h v - h v_{0}$

where $v$ is the frequency of incident light and $v_{0}$ is the threshold frequency.

For the first case with frequency $v_{1}$ :

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

For the second case with frequency $v_{2}$ :

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

Given that the ratio of maximum kinetic energies in the two cases is $1 : k$ :

$\frac{K _{1}}{K _{2}} = \frac{1}{k}$

$\Rightarrow k K_{1} = K_{2}$

Substituting the expressions for $K_{1}$ and $K_{2}$ :

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

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

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

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

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

Multiplying the numerator and the denominator by $- 1$ , we get:

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

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

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