In Lyman series, series limit of wavelength is \(\lambda_1\). The wavelength of first line of Lyman series is \(\lambda_2\) and in Balmer series, the series limit of wavelength is \(\lambda_3\). Then the relation between \(\lambda_1\), \(\lambda_2\) and \(\lambda_3\) is

According to Rydberg's formula,
\(
\frac{1}{\lambda}=R\left(\frac{1}{\mathrm{n}^2}-\frac{1}{\mathrm{~m}^2}\right)
\)
For series limit of Lyman series, \(\mathrm{n}=1, \mathrm{~m}=\infty, \lambda=\lambda_1\)
\(
\therefore \frac{1}{\lambda_1}=\mathrm{R}
\)
For \(1^{\text {st }}\) line of Lyman series,
\(
\mathrm{n}=1, \mathrm{~m}=2, \lambda=\lambda_2
\)
\(
\therefore \frac{1}{\lambda_2}=\frac{3 R}{4}
\)
For series limit of Balmer series,
\(\mathrm{n}=2, \mathrm{~m}=\infty, \lambda=\lambda_3 \)
\( \therefore \frac{1}{\lambda_3}=\frac{\mathbb{R}}{4} \)
\( \text {Now, } \frac{1}{\lambda_1}-\frac{1}{\lambda_2}=\mathbb{R}-\frac{3 \mathrm{R}}{4}=\frac{\mathbb{R}}{4} \)
\( \therefore \frac{1}{\lambda_1}-\frac{1}{\lambda_2}=\frac{1}{\lambda_3}\)