A certain metallic surface is illuminated by monochromatic...

KE = \dfrac{hc}{\lambda} - ϕhc$</p><p>$e\left(3{V}_{0}\right) = \dfrac{hc}{{\lambda}_{0}} - ϕ \ldots \left(i\right)$</p><p>$e{V}_{0} = \dfrac{hc}{2{\lambda}_{0}} - ϕ \ldots \left(ii\right)$</p><p>Using \left(i\right)\text{and}\left(ii\right)$,

$ϕ = \dfrac{hc}{4{\lambda}_{0}} = \dfrac{hc}{{\lambda}_{t}}$

${\lambda}_{t} = 4{\lambda}_{0}$