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

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

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

$g r e a t e r t h a n v \sqrt{\frac{4}{3}}$

$l e s s t h a n v \sqrt{\frac{3}{4}}$

answer is D.

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Detailed Solution

$\frac{1}{2} m v^{2} = \frac{h c}{λ} - φ$
$\frac{1}{2} m v^{2} = \frac{h c}{(3 λ / 4)} - ϕ = \frac{4 h c}{3 λ} - ϕ$
$C l e a r l y V_{\max} > \sqrt{\frac{4}{3}} V$

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