a
We have to find the frequency of emitted photons. For emission of photons the transition must take place from a higher energy level to a lower energy level which are given only in options \((c)\) and \((d)\).

Frequency is given by

\(h v=-13.6\left(\frac{1}{n_{2}^{2}}-\frac{1}{n_{1}^{2}}\right)\)

For transition from \(n=6\) to \(n=2\)

\({v_1} = \frac{{ - 13.6}}{h}\left( {\frac{1}{{{6^2}}} - \frac{1}{{{2^2}}}} \right)\) \( = \frac{2}{9} \times \left( {\frac{{13.6}}{h}} \right)\)

For transition from \(n=2\) to \(n=1\)

\(v_{2}=\frac{-13.6}{h}\left(\frac{1}{2^{2}}-\frac{1}{1^{2}}\right)=\frac{3}{4} \times\left(\frac{13.6}{h}\right)\)

\(\therefore v_{1}>v_{2}\)