\(h v=h v_{0}+K E_{\max }\)

\(\frac{h c}{\lambda_{\text {incident }}}=\frac{h c}{\lambda_{0}}+ eV _{ s }\)

When \(\lambda_{\text {incident }}=\lambda,\) then \(V_{s}=3 V\)

\(\frac{h c}{\lambda}=\frac{h c}{\lambda_{0}}+3 eV \ldots .\) \((i)\)

And for \(\lambda_{\text {incident }}=2 \lambda,\) then \(V_{z}=1 V\)

\(\frac{h c}{\lambda}=\frac{h c}{\lambda_{0}}+1 eV \ldots\) \((ii)\) 

On solving equations \((i)\) and \((ii)\) we get,

\(\frac{2 h c}{\lambda_{0}}=\frac{1}{2} \frac{h c}{\lambda}\)

\(\lambda_{0}=4 \lambda\)