b
\({k} {E}_{\max }=\frac{{hc}}{\lambda_{{i}}}+\phi\)

or \({eV}_{{o}}=\frac{{hc}}{\lambda_{{i}}}+\phi\)

\(\text { when } \lambda_{{i}}=670.5\, {nm} ; \quad {V}_{0}=0.48\)

\(\text { when } \lambda_{{i}}=474.6\, {nm} ; \quad {V}_{0}=?\)

So, \({e}(0.48)=\frac{1240}{670.5}+\phi...(1)\)

\({e}\left({V}_{{o}}\right)=\frac{1240}{474.6}+\phi...(2)\)

\((2)-(1)\)

\({e}\left({V}_{0}-0.48\right)=1240\left(\frac{1}{474.6}-\frac{1}{670.5}\right) \,{eV}\)

\({V}_{0}=0.48+1240\left(\frac{670.5-474.6}{474.6 \times 670.5}\right) {Volts}\)

\({V}_{0}=0.48+0.76\)

\({V}_{0}=1.24 {V} \simeq 1.25\, {V}\)