(b) Here, $\lambda_{0} = 200nm;\ \lambda = 100nm;$
$hc/e\ = \ 1240eV\ nm$
maximum KE $= \frac{hc}{\lambda e} - \frac{hc}{\lambda_{0}e}(in\ eV)$
$= \frac{hc}{e}\left( \frac{1}{\lambda} - \frac{1}{\lambda_{0}} \right)$
$= 1240\left( \frac{1}{100} - \frac{1}{200} \right)$
$= 6.2\ eV$