The kinetic energy of emitted electron is E when the light incident on the metal has wavelength $λ$ . To double the kinetic energy, the incident light must have wavelength :

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$\frac{hcλ}{Eλ + hc}$

$\frac{hλ}{Eλ + hc}$

$\frac{hc}{Eλ - hc}$

$\frac{hcλ}{Eλ - hc}$

answer is B.

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Detailed Solution

$\begin{aligned} E = \frac{hc}{λ} - ϕ - (i) \\ 2 E = \frac{hc}{λ^{'}} - ϕ (ii) \\ (ii) - (i) \\ E = hc (\frac{1}{λ^{'}} - \frac{1}{λ}) \\ \Rightarrow λ^{'} = \frac{hcλ}{Eλ + hc} \end{aligned}$

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