a
(a) \({E_{{\rm{ionisation}}}} = {E_\infty } - {E_n} = \frac{{13.6Z_{eff}^2}}{{{n^2}}}\,eV\) = \(\left[ {\frac{{13.6{Z^2}}}{{n_2^2}} - \frac{{13.6{Z^2}}}{{n_1^2}}} \right]\) \(E = h\nu = \frac{{13.6 \times {1^2}}}{{{{(1)}^2}}} - \frac{{13.6 \times {1^2}}}{{{{(4)}^2}}}\); \(h\nu = 13.6 - 0.85\) \(h = 6.625 \times {10^{ - 34}}\) \(\nu = \frac{{13.6 - 0.85}}{{6.625 \times {{10}^{ - 34}}}} \times 1.6 \times {10^{ - 19}}\) = \(3.08 \times {10^{15}}{s^{ - 1}}\).