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

$\frac{\mathrm{hc}}{\mathrm{\lambda }}={\mathrm{W}}_0+{\mathrm{K}}_{\max }\Rightarrow \frac{\mathrm{hc}}{{\mathrm{\lambda }}_{\mathrm{A}}}={\mathrm{W}}_0+{\mathrm{K}}_{\mathrm{A}}.... \left(\mathrm{i}\right)$

$\text{ and }\frac{\mathrm{hc}}{{\mathrm{\lambda }}_{\mathrm{B}}}={\mathrm{W}}_0+{\mathrm{K}}_{\mathrm{B}}\dots \text{ (ii) }$

Subtracting (i) from (ii),

$\mathrm{hc}\left[\frac{1}{{\mathrm{\lambda }}_{\mathrm{B}}}-\frac{1}{{\mathrm{\lambda }}_{\mathrm{A}}}\right]={\mathrm{K}}_{\mathrm{B}}-{\mathrm{K}}_{\mathrm{A}}$

$\Rightarrow \mathrm{hc}\left[\frac{1}{{\mathrm{\lambda }}_{\mathrm{B}}}-\frac{1}{2{\mathrm{\lambda }}_{\mathrm{B}}}\right]={\mathrm{K}}_{\mathrm{B}}-{\mathrm{K}}_{\mathrm{A}}\Rightarrow \frac{\mathrm{hc}}{2{\mathrm{\lambda }}_{\mathrm{B}}}$

$={\mathrm{K}}_{\mathrm{B}}-{\mathrm{K}}_{\mathrm{A}}\dots \left(\mathrm{iii}\right)$

From (ii) and (iii),

$2{\mathrm{K}}_{\mathrm{B}}-2{\mathrm{K}}_{\mathrm{A}}={\mathrm{W}}_0+{\mathrm{K}}_{\mathrm{B}}$

$\Rightarrow {\mathrm{K}}_{\mathrm{B}}-2{\mathrm{K}}_{\mathrm{A}}={\mathrm{W}}_0$

$\Rightarrow {\mathrm{K}}_{\mathrm{A}}=\frac{{\mathrm{K}}_{\mathrm{B}}}{2}-\frac{{\mathrm{W}}_0}{2}\text{ which gives }$

${\mathrm{K}}_{\mathrm{A}}<\frac{{\mathrm{K}}_{\mathrm{B}}}{2}$