According to Bohr's theory, the electronic energy of hydrogen atom in the nth Bohr's orbit is given by: 

               $E_n=-\frac{21.76\times {10}^{-19}}{n^2}\mathrm{J}$

Calculate the longest wavelength of light that will be needed to remove an electron from the third orbit of ${\mathrm{He}}^{+}$ ion.

For one electron species like $\mathrm{H}, {\mathrm{He}}^{+}, {\mathrm{Li}}^{2+}$

$E_n=-\frac{21.76\times {10}^{-19}{Z}^2}{n^2}J$

For third orbit $(n=3)$ and for ${\mathrm{He}}^{+}(Z=2)$, energy is calculated as:

 $E_3=-\frac{21.76\times {10}^{-19}\times (2{)}^2}{(3{)}^2}\mathrm{J}$

Energy required to remove an electron from third orbit is given as : 

$\begin{array}{l}\mathrm{\Delta E}={\mathrm{E}}_{\mathrm{\infty }}-{\mathrm{E}}_3\\ =0-\frac{\left(-21.76\times {10}^{-19}\times 4\right)}{9}\mathrm{J} \left[\because {\mathrm{E}}_{\mathrm{\infty }}=0\right]\\ =\frac{21.76\times {10}^{-19}\times 4}{9}\mathrm{J}\end{array}$

Wavelength is calculated as:

$\begin{array}{l} \lambda =\frac{hc}{\mathrm{\Delta }E}\\ =\frac{6.626\times {10}^{-34} \mathrm{Js}\times 3.0\times {10}^8 {\mathrm{ms}}^{-1}\times 9}{21.76\times {10}^{-19}\times 4 \mathrm{J}}\\ =2.055\times {10}^{-7} \mathrm{m}\end{array}$