In a photo–emissive cell, with exciting wavelength $λ,$ the maximum kinetic energy of the electron is K. If the exciting wavelength is changed to $3 λ / 4,$ the kinetic energy of the fastest emitted electron will be

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$\frac{3 K}{4}$

$\frac{4 K}{3}$

More than $\frac{4 K}{3}$

Less than $\frac{4 K}{3}$

answer is D.

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Detailed Solution

$K = \frac{h c}{λ} - ϕ_{0} (i)$

and $K' = \frac{4 h c}{3 λ} - ϕ_{0} (i i)$

From Eq/s $(i) & (i i)$ , we get

$\Rightarrow K' - K = \frac{4 h c}{3 λ} - \frac{h c}{λ}$

$K' - K = \frac{h c}{3 λ}$

But from Eq.(i) $\frac{h c}{λ} = K + ϕ_{0}$

$\Rightarrow K' - K = \frac{4 K}{3} + \frac{ϕ_{0}}{3}$ $∴ K' - K = \frac{K}{3} + \frac{ϕ_{0}}{3}$

Or $K' > \frac{4 K}{3}$

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