Answer
(e) 122 nm
Work Step by Step
$n_{i}=2$ and $n_{f}=1$
Recall that
$\frac{1}{\lambda}=\frac{2.18\times10^{-18}\,J}{hc}(\frac{1}{n_{f}^{2}}-\frac{1}{n_{i}^{2}})$ where
$h=6.626\times10^{-34}\,J\cdot s$ and $c=3.00\times10^{8}\,m/s$
Then,
$\frac{1}{\lambda}=\frac{2.18\times10^{-18}\,J}{(6.626\times10^{-34}\,J\cdot s)(3.00\times10^{8}\,m/s)}(\frac{1}{1^{2}}-\frac{1}{2^{2}})$
$=8.225\times10^{6}\,m^{-1}$
$\lambda=\frac{1}{8.225\times10^{6}\,m^{-1}}=122\times10^{-9}\,m$
$=122\,nm$
