In Balmer series, wavelength of the $2^{\text {nd }}$ line is ' $\lambda_1$ ' and for Paschen series,…
In Balmer series, wavelength of the $2^{\text {nd }}$ line is ' $\lambda_1$ ' and for Paschen series, wavelength of the $1^{\text {st }}$ line is ' $\lambda_2$ ', then the ratio ' $\lambda_1$ ' to ' $\lambda_2$ ' is
$5: 128$
$5: 81$
$7: 27$
$9: 132$
Solution
For spectral series, $\frac{1}{\lambda}=R Z^2\left(\frac{1}{\mathrm{n}_1^2}-\frac{1}{\mathrm{n}_2^2}\right)$
For the Balmer series, $\mathrm{n}_1=2$
The wavelength for $2^{\text {nd }}$ line of the Balmer series is
$\begin{aligned}
& \frac{1}{\lambda_1}=\mathrm{RZ}^2\left(\frac{1}{2^2}-\frac{1}{4^2}\right) \\
& \frac{1}{\lambda_1}=\mathrm{RZ}^2\left(\frac{1}{4}-\frac{1}{16}\right) \\
& \frac{1}{\lambda_1}=\mathrm{RZ}^2\left(\frac{3}{16}\right) \Rightarrow \lambda_1=\left[\frac{16}{3}\right]
\end{aligned}$
For the Paschen series, $\mathrm{n}_1=3$
The wavelength for $1^{\text {st }}$ line of the Paschen series is
$\begin{aligned}
\frac{1}{\lambda_2} & =\mathrm{RZ}^2\left(\frac{1}{3^2}-\frac{1}{4^2}\right) \\
\frac{1}{\lambda_2} & =\mathrm{RZ}^2\left(\frac{1}{9}-\frac{1}{16}\right) \\
\lambda_2 & =\frac{144}{7} \\
\frac{1}{\lambda_2} & =\mathrm{RZ}^2\left(\frac{7}{144}\right) \\
\therefore \quad \frac{\lambda_1}{\lambda_2} & =\frac{16}{3} \times \frac{7}{144}=\frac{7}{27}
\end{aligned}$