First slide

Photo Electric Effect

Question

Two identical metal plates show photoelectric effect by a light of wavelength $λ_{A}$ falling on plate A and $λ_{B}$ on plate $B (λ_{A} = 2 λ_{B}) .$ The maximum kinetic energy of electrons is :

Moderate

A

$2 K_{A} = K_{B}$
B

$K_{A} < K_{B} / 2$
C

$K_{A} = 2 K_{B}$
D

$K_{A} > K_{B} / 2$

Solution

$\frac{hc}{λ} = W_{0} + K_{\max} \Rightarrow \frac{hc}{λ_{A}} = W_{0} + K_{A} . . . . (i)$
$and \frac{hc}{λ_{B}} = W_{0} + K_{B} \dots (ii)$
Subtracting (i) from (ii),
$hc [\frac{1}{λ_{B}} - \frac{1}{λ_{A}}] = K_{B} - K_{A}$
$\Rightarrow hc [\frac{1}{λ_{B}} - \frac{1}{2 λ_{B}}] = K_{B} - K_{A} \Rightarrow \frac{hc}{2 λ_{B}}$
$= K_{B} - K_{A} \dots (iii)$
From (ii) and (iii),
$2 K_{B} - 2 K_{A} = W_{0} + K_{B}$
$\Rightarrow K_{B} - 2 K_{A} = W_{0}$
$\Rightarrow K_{A} = \frac{K_{B}}{2} - \frac{W_{0}}{2} which gives$
$K_{A} < \frac{K_{B}}{2}$

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