The frequency of radiation emitted when the electron falls from $$n = 4$$  to $$n = 1$$  in a hydrogen atom will be (Given ionisation energy of $$H = 2.18 \times {10^{ - 18}}J\,ato{m^{ - 1}}$$      and $$h = 6.625 \times {10^{ - 34}}Js$$    )

Solution :
$$\eqalign{ & {\text{Ionisation energy of}}\,H = 2.18 \times {10^{ - 18}}J\,ato{m^{ - 1}} \cr & \therefore \,\,{E_1}\left( {{\text{Energy of Ist orbit of }}H{\text{ - }}atom} \right) \cr & = - 2.18 \times {10^{ - 18}}J\,ato{m^{ - 1}} \cr & \therefore \,\,{E_n} = \frac{{ - 2.18 \times {{10}^{ - 18}}}}{{{n^2}}}J\,ato{m^{ - 1}} \cr & Z = 1\,{\text{for}}\,H{\text{ - }}atom \cr & \Delta E = {E_4} - {E_1} \cr & = \frac{{ - 2.18 \times {{10}^{ - 18}}}}{{{4^2}}} - \frac{{ - 2.18 \times {{10}^{ - 18}}}}{{{1^2}}} \cr & = - 2.18 \times {10^{ - 18}} \times \left[ {\frac{1}{{{4^2}}} - \frac{1}{{{1^2}}}} \right] \cr & \Delta E = - 2.18 \times {10^{ - 18}} \times - \frac{{15}}{{16}} \cr & \,\,\,\,\,\,\,\,\,\, = + \,2.0437 \times {10^{ - 18\,}}J\,ato{m^{ - 1}} \cr & \therefore \,\,\nu = \frac{{\Delta E}}{h} \cr & \,\,\,\,\,\,\,\,\,\, = \frac{{2.0437 \times {{10}^{ - 18}}J\,ato{m^{ - 1}}}}{{6.625 \times {{10}^{ - 34}}J\,s}} \cr & \,\,\,\,\,\,\,\,\,\, = 3.084 \times {10^{15}}{s^{ - 1}}ato{m^{ - 1}} \cr} $$