If light of wavelength $λ_{1}$ is allowed to fall on a metal, then kinetic energy of photoelectrons emitted is $E_{1}$ . If wavelength of light changes to $λ_{2}$ then kinetic energy of electrons changes to $E_{2}$ . Then work function of the metal is

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$\frac{λ_{1} λ_{2} E_{1} E_{2}}{(λ_{2} - λ_{1})}$

$\frac{E_{1} E_{2} (λ_{1} - λ_{2})}{λ_{1} λ_{2}}$

$\frac{E_{1} λ_{1} - E_{2} λ_{2}}{(λ_{2} - λ_{1})}$

$\frac{E_{1} λ_{1} - E_{2} λ_{2}}{(λ_{1} - λ_{2})}$

answer is C.

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Detailed Solution

$\begin{array}{l} E = W_{o} + K_{\max} \Rightarrow \frac{hc}{λ_{1}} = W_{o} + E_{1} and \frac{hc}{λ_{2}} = W_{o} + E_{2} \\ \Rightarrow hc = W_{o} λ_{1} + E_{1} λ_{1} and hc = W_{o} λ_{2} + E_{2} λ_{2} \\ \Rightarrow W_{o} λ_{1} + E_{1} λ_{1} = W_{o} λ_{2} + E_{2} λ_{2} \Rightarrow W_{o} = \frac{E_{1} λ_{1} - E_{2} λ_{2}}{(λ_{2} - λ_{1})} . \end{array}$