From Einstein photoelectric equation.

\(h . v=\phi+ K _{\max }\)

\(\frac{h c}{\lambda}=\phi+e v_0\) \(\quad \left(v_6=\right.\) stopping potential)

\(\frac{h c}{\lambda}=\phi+e x\)

\(\frac{h c}{2 \lambda}=\phi+e \frac{x}{3}\)

On solving

(work function) \(\phi=\frac{h c}{4 \lambda}\)

\(\frac{h c}{\lambda_0}=\frac{h c}{4 \lambda}\)

\(\lambda_0=4 \lambda\)