Light of wavelength $\lambda$, strikes a photoelectric surface and electrons are ejected with an energy E. If E is to be increased to exactly twice its original value, the wavelength changes to $\lambda \text{'}$, where

Energy of photoelectron,

$E=\frac{hc}{\lambda }-{\varphi }_0$

$\Rightarrow \frac{hc}{\lambda }=E+{\varphi }_0$       …(i)

where  ${\varphi }_0$ is the work function for the metal surface (constant).

When $E\text{'}$ = 2E, then: 

$2E=\frac{hc}{{\lambda }^{\text{'}}}-{\varphi }_0$

 $\Rightarrow$  $\frac{hc}{{\lambda }^{\text{'}}}=2E+{\varphi }_0$          …(ii)

Dividing Eq. (i) by Eq. (ii), we get:

$\frac{\lambda \text{'}}{\lambda }=\frac{E+{\varphi }_0}{2E+{\varphi }_0}$

$\frac{\lambda \text{'}}{\lambda }=\frac{\left(E+{\varphi }_0\right)}{2\left(E+\frac{{\varphi }_0}{2}\right)}$

$\therefore$ $\frac{\lambda \text{'}}{\lambda }>\frac{1}{2}\text{ or }\lambda \text{'} >\frac{\lambda }{2}$

or   $\lambda >\lambda \text{'}>\frac{\lambda }{2}$