A metallic surface is illuminated with radiation of wavelength λ ,...

Let the threshold frequency be ${\lambda}_{0}$.

By the equation of photoelectric effect, for wavelength $\lambda$,

$e{V}_{0} = hc\left(\dfrac{1}{\lambda} - \dfrac{1}{{\lambda}_{0}}\right) . . . ( i )$

For the wavelength $2\lambda$,

$\dfrac{e{V}_{0}}{4} = hc\left(\dfrac{1}{2\lambda} - \dfrac{1}{{\lambda}_{0}}\right) . . . ( ii )$

Dividing (i) by (ii)

$4 = \dfrac{hc ( \dfrac{{\lambda}_{0} - \lambda}{\lambda{\lambda}_{0}} )}{hc\left(\dfrac{{\lambda}_{0} - 2\lambda}{2\lambda{\lambda}_{0}}\right)} ⇒ 4 = \dfrac{2 ( {\lambda}_{0} - \lambda )}{{\lambda}_{0} - 2\lambda} ⇒ 2{\lambda}_{0} = 6\lambda ⇒ {\lambda}_{0} = 3\lambda$