An electron of an atom transits from $n_{1}$ to $n_{2}$. In which of the following maximum frequency of photon will be emitted?

For emission of energy, electron of H-atom must fall from higher energy state to lower energy state. Hence, options (a) and (c) are not possible. In option (b), $n_{1}=2, n_{2}=1$ $\therefore$ Energy of emitted photon is given as $$ \begin{aligned} E^{\prime} &=-13.6 Z^{2}\left(\frac{1}{n_{1}^{2}}-\frac{1}{n_{2}^{2}}\right) \\ \Rightarrow \quad h v^{\prime} &=-13.6 Z^{2}\left(\frac{1}{2^{2}}-\frac{1}{1^{2}}\right) \\ v^{\prime} &=\frac{13.6 Z^{2}}{h} \times \frac{3}{4} \end{aligned} $$ In option (c), $n_{1}=2, n_{2}=6$, hence energy of emitted photon is given as $$ \begin{aligned} & E^{\prime \prime}=-13.6 Z^{2}\left(\frac{1}{6^{2}}-\frac{1}{2^{2}}\right) \\ \Rightarrow & h v^{\prime \prime}=13.6 Z^{2} \times \frac{8}{36} \\ \Rightarrow & \quad v^{\prime \prime}=\frac{13.6 Z^{2}}{h} \times \frac{8}{36} \end{aligned} $$ Hence, from Eqs. (i) and (ii), we get $$ v^{\prime}>v^{\prime \prime} $$