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The kinetic energy of electron is E when the incident light has wavelength $λ$ . To increase the KE from E to 2E, the incident light must have wavelength

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$\frac{h c}{E λ - h c}$

$\frac{h c λ}{E λ + h c}$

$\frac{h λ}{E λ + h c}$

$\frac{h c λ}{E λ - h c}$

answer is B.

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Detailed Solution

$E = \frac{h c}{λ} - ϕ$

& $2 E = \frac{h c}{λ^{⊤}} - ϕ$

$\Rightarrow 2 E - E = h c (\frac{1}{λ^{'}} - \frac{1}{λ}) \Rightarrow E + \frac{h c}{λ} = \frac{h c}{λ^{'}}$

$\Rightarrow \frac{E λ + h c}{λ} = \frac{h c}{λ^{'}} \Rightarrow λ^{'} = \frac{h c λ}{E λ + h c}$

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