In a photoelectric experiment, with light of wavelength $λ$ , the fastest electron has speed v. If the exciting wavelength is changed to $3 λ / 4$ , the speed of the fastest emitted electron will become-

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$v \sqrt{\frac{3}{4}}$

$less than v \sqrt{\frac{4}{3}}$

$\sqrt[v]{\frac{4}{3}}$

$greater than v \sqrt{\frac{4}{3}}$

answer is D.

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Detailed Solution

$\begin{array}{l} \frac{1}{2} {mv}^{2} = \frac{hc}{λ} - ϕ \\ \frac{1}{2} {mv}^{' 2} = \frac{hc}{(3 λ / 4)} - ϕ = \frac{4 hc}{3 λ} - ϕ \\ {Clearly, v}^{'} > \sqrt{\frac{4}{3}} v \end{array}$

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