The shortest wavelength of H – atom in Lyman series is x, longest wavelength In Balmer series of He⁺ is

The shortest Lyman wavelength is determined when n₁ is 1 and n₂ is $\infty .$The shortest wavelength of Lyman series is calculated as x.

$$\begin{gathered} \frac{1}{\mathrm{x}}={\mathrm{R}}_{\mathrm{H}}{1}^2\left[\frac{1}{1^2}-\frac{1}{{\infty }^2}\right] \\ {\mathrm{R}}_{\mathrm{H}}=\frac{1}{\mathrm{x}} \end{gathered}$$

The longest wavelength of the Balmer series is obtained when electron jumps from 3 to 2. 

$$\begin{gathered} \frac{1}{\mathrm{\lambda }}={\mathrm{R}}_{\mathrm{H}}{2}^2\left[\frac{1}{2^2}-\frac{1}{3^2}\right] \\ =\frac{1}{\mathrm{x}}\times 4\times \frac{5}{36} \\ \mathrm{\lambda }=\frac{9\mathrm{x}}{5} \end{gathered}$$