\(\phi=\frac{\mathrm{hc}}{\lambda_0}\)
\(\mathrm{~K}_{\max }=\mathrm{eV}_0\)
\(8 \mathrm{e}=\frac{\mathrm{hc}}{\lambda}-\frac{\mathrm{hc}}{\lambda_0} \ldots . . \text { (i) }\)
\(2 \mathrm{e}=\frac{\mathrm{hc}}{3 \lambda}-\frac{\mathrm{hc}}{\lambda_0} \ldots . . . \text { (ii) }\)
\(\text { on solving (i) & (ii) }\)
\(\lambda_0=9 \lambda\)