Energy of an electron is given by $\mathrm{E}=-2.178\times {10}^{-18} \mathrm{J}\left(\frac{{\mathrm{Z}}^2}{{\mathrm{n}}^2}\right)$

Wavelength of light required to excite an electron in an hydrogen atom from level n = 1 to n = 2 will be

$\left(\mathrm{h}=6.62\times {10}^{-34}\mathrm{Js}\text{ and }\mathrm{c}=3.0\times {10}^8{ \mathrm{ms}}^{-1}\right)$

 

$\text{ Given, in the question }\mathrm{E}=-2.178\times {10}^{-8} \mathrm{J}\left[\frac{{\mathrm{Z}}^2}{{\mathrm{n}}^2}\right]$

For hydrogen Z = 1,

$\text{ So, } {\mathrm{E}}_1=-2.178\times {10}^{-18} \mathrm{J}\left[\frac{1}{1^2}\right]$

${\mathrm{E}}_2=-2.178\times {10}^{-18} \mathrm{J}\left[\frac{1}{2^2}\right]$

$\text{ Now, }{\mathrm{E}}_1-{\mathrm{E}}_2$

$\text{ i.e. } \mathrm{\Delta E}=2.178\times {10}^{-18}\left(\frac{1}{1^2}-\frac{1}{2^2}\right)=\frac{\mathrm{hc}}{\mathrm{\lambda }}$

$2.178\times {10}^{-18}\left(\frac{1}{1^2}-\frac{1}{2^2}\right)=\frac{6.62\times {10}^{-34}\times 3.0\times {10}^8}{\mathrm{\lambda }}$

$\therefore \lambda \approx 1.21\times {10}^{-7} \mathrm{m}$