Answer
122 nm, ultraviolet region.
Work Step by Step
$\Delta E_{atom}=-2.18\times10^{-18}\,J(\frac{1}{n_{f}^{2}}-\frac{1}{n_{i}^{2}})$
$=-2.18\times10^{-18}\,J(\frac{1}{1^{2}}-\frac{1}{2^{2}})$
$=-1.635\times10^{-18}\,J$
$E_{photon}=-\Delta E_{atom}=1.635\times10^{-18}\,J$
$E=\frac{hc}{\lambda}\implies \lambda=\frac{hc}{E}$
$=\frac{(6.626\times10^{-34}\,J\cdot s)(3.00\times10^{8}\,m/s)}{1.635\times10^{-18}\,J}$
$=1.22\times10^{-7}\,m=122\,nm$
This light is found in the ultraviolet region (wavelength ranges from 10 nm to 400 nm).
