In Lyman series, shortest wavelength of H-atom appears at x meter , then longest wavelength in Balmer series of He⁺ appear at

For the spectral line in H-atom and H-like species (one electron) 

$\frac{1}{\mathrm{\lambda }}=\overline{\mathrm{v}}={\overline{\mathrm{R}}}_{\mathrm{H}}{\mathrm{Z}}^2\left[\frac{1}{{\mathrm{n}}_1^2}-\frac{1}{{\mathrm{n}}_2^2}\right]$

For Lyman series

For shortest wavelength (maximum wave number) ${\mathrm{n}}_2\to \infty$

$\therefore \frac{1}{{\mathrm{\lambda }}_{\min }}={\overline{\mathrm{v}}}_{\max }={\overline{\mathrm{R}}}_{\mathrm{H}}{\left(1\right)}^2\left[\frac{1}{1}\right]$

For longest wavelength (minimum wave number), ${\mathrm{n}}_2=\left({\mathrm{n}}_1+1\right)$

For Balmer series, ${\mathrm{n}}_1=2$

$\therefore \frac{1}{{\mathrm{\lambda }}_{\max }}={\overline{\mathrm{v}}}_{\min }={\overline{\mathrm{R}}}_{\mathrm{H}}{\left(\mathrm{Z}\right)}^2\left[\frac{1}{2}\right]$

For He⁺ , Z = 2

$$\begin{gathered} \frac{1}{{\mathrm{\lambda }}_{\max }}=\frac{1}{\mathrm{x}}{\left(2\right)}^2\left[\frac{5}{36}\right]\Rightarrow \frac{1}{{\mathrm{\lambda }}_{\max }}=\frac{5}{9\mathrm{x}} \\ \therefore {\mathrm{\lambda }}_{\max }=\frac{9\mathrm{x}}{5} \end{gathered}$$