Photoelectric emission is observed from a metallic surface for frequencies ν 1 and ν 2 of the incident light rays ( ν 1 ν 2 ). If the ratio of maximum value of kinetic energy of the photoele

Question

Photoelectric emission is observed from a metallic surface for frequencies ν 1 ​ and ν 2 ​ of the incident light rays ( ν 1 ​ > ν 2 ​ ). If the ratio of maximum value of kinetic energy of the photoelectron emitted in first case to that in second case 3 : K, then the threshold frequency of the metallic surface is

TestHub · Accepted Answer

Using Einstein's photoelectric equation, the maximum kinetic energy of emitted photoelectrons is given by $K_{ma x} = h ν - h ν_{0}$ , where $ν_{0}$ is the threshold frequency.

For the first case:
$K_{1} = h ν_{1} - h ν_{0}$

For the second case:
$K_{2} = h ν_{2} - h ν_{0}$

Given the ratio of maximum kinetic energies is $3 : K$ :
$\frac{K _{1}}{K _{2}} = \frac{3}{K}$

Substituting the expressions for $K_{1}$ and $K_{2}$ :
$\frac{h ν _{1} - h ν _{0}}{h ν _{2} - h ν _{0}} = \frac{3}{K}$

Canceling $h$ and cross-multiplying:
$K (ν_{1} - ν_{0}) = 3 (ν_{2} - ν_{0})$

$K ν_{1} - K ν_{0} = 3 ν_{2} - 3 ν_{0}$

Rearranging to solve for $ν_{0}$ :
$3 ν_{0} - K ν_{0} = 3 ν_{2} - K ν_{1}$

$ν_{0} (3 - K) = 3 ν_{2} - K ν_{1}$

$ν_{0} = \frac{K ν _{1} - 3 ν _{2}}{K - 3}$

Answer: $\frac{K ν _{1} - 3 ν _{2}}{K - 3}$

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