The wavelength of the first member of the Balmer series in hydrogen spectrum is x A⁰. Then the wave length (in A⁰) of the first member of Lyman series in the same spectrum is

H- atom Z =1

Balmer series n₁=2

First member of Balmer series n₁=2 ; n₂=3

$$\begin{gathered} \stackrel{ }{\stackrel{-}{\mathrm{v}}=\frac{1}{\mathrm{\lambda }}={\mathrm{R}}_{\mathrm{H}}{\mathrm{Z}}^2(\frac{1}{2^2}-\frac{1}{3^2})} \\ \stackrel{-}{v} =\frac{1}{\mathrm{\lambda }}={\mathrm{R}}_{\mathrm{H}}{\mathrm{Z}}^2(\frac{1}{4}-\frac{1}{9}) \\ \mathrm{\lambda } =\frac{36}{5{\mathrm{R}}_{\mathrm{H}}} =\mathrm{x} {\mathrm{A}}^0 \\ {\mathrm{R}}_{\mathrm{H}} =\frac{36}{5\mathrm{x}}----------\left(1\right) \end{gathered}$$

H- atom Z =1

Lymann series n₁=1

First member of Lymann series n₁=1 ; n₂=2

$$\begin{gathered} \stackrel{ }{\stackrel{-}{\mathrm{v}}=\frac{1}{\mathrm{\lambda }}={\mathrm{R}}_{\mathrm{H}}{\mathrm{Z}}^2(\frac{1}{1^2}-\frac{1}{2^2})} \\ \stackrel{-}{v} =\frac{1}{\mathrm{\lambda }}={\mathrm{R}}_{\mathrm{H}}{\mathrm{Z}}^2(\frac{1}{1}-\frac{1}{4}) \\ \mathrm{\lambda } =\frac{4}{3{\mathrm{R}}_{\mathrm{H}}} \end{gathered}$$

Substituting the value of R_H from (1) equation

$$\begin{gathered} \mathrm{\lambda } =\frac{4}{3\left({\displaystyle \frac{36}{5\mathrm{x}}}\right)} \\ \mathrm{\lambda } =\frac{5\mathrm{x}}{27} \end{gathered}$$