If $p$ is the momentum of the fastest electron ejected from a metal surface after the irradiation of light having wavelength $λ$ , then for $1.5 p$ momentum of the photoelectron, the wavelength of the light should be(Assume kinetic energy of ejected photoelectron to be very high in comparison to work function)

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$\frac{4}{9} λ$

$\frac{3}{4} λ$

$\frac{1}{2} λ$

$\frac{2}{3} λ$

answer is D.

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Detailed Solution

$h v - ϕ = K E$

$\begin{array}{l} \Rightarrow {(\frac{h c}{λ})}_{incident} = K E + ϕ \\ {(\frac{h c}{λ})}_{incident} ≃ K E \end{array}$

$K E = \frac{p^{2}}{2 m} = \frac{h c}{λ_{incident}} = \frac{h c}{λ}$ …..(i)

$\Rightarrow \frac{p^{2} \times (1.5)^{2}}{2 m} = \frac{h c}{λ^{'}}$ .... (ii)

Divide (i) and (ii)

$\begin{aligned} (1.5)^{2} = \frac{λ}{λ^{'}} \\ \Rightarrow λ^{'} = \frac{4 λ}{9} . \end{aligned}$

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