Question

Two identical metal plates show photoelectric effect by a light of wavelength ${\lambda }_A$ falls on plate A and ${\lambda }_B$ on plate B $({\lambda }_A=2{\lambda }_B)$. The maximum kinetic energy is

• A. $2K_A=K_B$
• B. $K_A<\frac{K_B}{2}$
• C. $K_A=2K_B$
• D. $K_A=\frac{K_B}{2}$

Answer 1

The correct option is B

$K_A<\frac{K_B}{2}$

$\frac{hc}{\lambda }=W_0+K_{max}\Rightarrow \frac{hc}{{\lambda }_A}=W_0+K_A$ . . .. . . .(i)
and $\frac{hc}{{\lambda }_B}=W_0+K_B$ . . . .. . . . .(ii)
Subtracting (i) from (ii),
$hc[\frac{1}{{\lambda }_B}-\frac{1}{{\lambda }_A}]=K_B-K_A$
$\Rightarrow hc[\frac{1}{{\lambda }_B}-\frac{1}{2{\lambda }_B}]=K_B-K_A\Rightarrow \frac{hc}{2{\lambda }_B}=K_B-K_A$ . . . . .(iii)
From (ii) and (iii), $2K_B-2K_A=W_0+K_B$
$\Rightarrow K_B-2K_A=W_0$
$\Rightarrow K_A=\frac{K_B}{2}-\frac{W_0}{2}whichgives{K}_A<\frac{K_B}{2}$