Explanation

(b) Stopping potential
${\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ v}_{0} = \frac{hc}{e}\left( \frac{1}{\lambda} - \frac{1}{\lambda_{0}} \right)$
Where $\lambda_{0}$= stopping potential
Ist case, $4.8\ = \frac{hc}{e}\left( \frac{1}{\lambda} - \frac{1}{\lambda_{0}} \right)\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ldots(i)$
IInd case, $1.6\ = \ \frac{hc}{e}(\frac{1}{2\lambda} - \frac{1}{\lambda_{0}})\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ldots(ii)$
Dividing Eq. (i) by Eq. (ii)
3 =$\ \frac{\left( \frac{1}{\lambda} - \frac{1}{\lambda_{0}} \right)}{\left( \frac{1}{2\lambda} - \frac{1}{\lambda_{0}\ } \right)}$
$\ \ \frac{3}{2\lambda} - \frac{3}{\lambda_{0}}\ = \frac{1}{\lambda} - \frac{1}{\lambda_{0}}$
$\frac{1}{\lambda_{0}} - \frac{3}{\lambda_{0}} = \frac{1}{\lambda} - \frac{3}{2\lambda}$
$\frac{- 2}{\lambda_{0}} = \frac{2 - 3}{2\lambda}$
$\frac{2}{\lambda_{0}} = \frac{1}{2\lambda}$
$\lambda_{0} = \ 4\lambda$