Let $K_1$ be the maximum kinetic energy of photoelectrons emitted by light of wavelength ${\lambda }_1$ and $K_2$ corresponding to wavelength ${\lambda }_2$. If ${\lambda }_1=2{\lambda }_2$ then

Using Einstein's photoelectricity equation,

$$\begin{gathered} K_1=\frac{hc}{{\lambda }_1}-{\varphi }_0 \cdots \left(i\right) \\ {K}_2=\frac{hc}{{\lambda }_2}-{\varphi }_0 \cdots \left(ii\right) \end{gathered}$$

Since, ${\lambda }_1=2{\lambda }_2$, equation (i) becomes

$K_1=\frac{hc}{2{\lambda }_2}-W=\frac{1}{2}\left(\frac{hc}{{\lambda }_2}\right)-{\varphi }_0$

$\Rightarrow K_1=\frac{1}{2}\left(K_2+{\varphi }_0\right)-{\varphi }_0$

$\Rightarrow K_1=\frac{K_2}{2}-\frac{{\varphi }_0}{2}$

$\therefore K_1<\frac{K_2}{2}$