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The kinetic energy of the most energetic photoelectrons emitted from a metal surface is doubled when the wavelength of the incident radiation is reduced from $λ_{1}$ to $λ_{2} .$ The work function of the metal is

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$\frac{2 h c}{λ_{1} λ_{2}} (λ_{1} + λ_{2})$

$\frac{h c}{λ_{1} λ_{2}} (2 λ_{2} - λ_{1})$

$\frac{2 h c}{λ_{1} λ_{2}} (2 λ_{1} - λ_{2})$

$\frac{2 h c}{λ_{1} λ_{2}} (λ_{1} - λ_{2})$

answer is A.

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Detailed Solution

Given $K_{max} = \frac{h c}{λ_{1}} - W_{0}$ (i)

and $2 K_{max} = \frac{h c}{λ_{2}} - W_{0}$ (ii)

Dividing (ii) by (i), we get

$2 = \frac{\frac{h c}{λ_{2}} - W_{0}}{\frac{h c}{λ_{1}} - W_{0}}$

which gives $W_{0} = \frac{h c}{λ_{1} λ_{2}} (2 λ_{2} - λ_{1})$

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