When a metal surface is illuminated by light of wavelength λ , the...

From photoelectric effect equation, we can write E = ϕ + {K}_\max

and work function can be written as $ϕ = \dfrac{hc}{{\lambda}_{0}}$.

Also, maximum kinetic energy {K}_\max = e{V}_{0}

For first case:

$8e = \dfrac{hc}{\lambda} - \dfrac{hc}{{\lambda}_{0}}$   ... (i)

For second case:

$2e = \dfrac{hc}{3\lambda} - \dfrac{hc}{{\lambda}_{0}}$    ... (ii)

After using equation (ii) in (i), we get

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

$⇒ {\lambda}_{0} = 9\lambda$