Let the series limit for Balmer series be \({ }^{\prime} \lambda_{1}^{\prime}\) and the longest wavelength for Brackett series be \({ }^{\prime} \lambda_{2}{ }^{\prime} .\) Then \(\lambda_{1}\) and \(\lambda_{2}\) are related as

We know,
\(\frac{1}{\lambda}=\mathrm{R}\left(\frac{1}{2^{2}}-\frac{1}{\mathrm{n}^{2}}\right)\) for Balmer series \(\frac{1}{\lambda}=\mathrm{R}\left(\frac{1}{4^{2}}-\frac{1}{\mathrm{n}^{2}}\right)\) for Brackett series
Now,\(\frac{1}{\lambda_{1}}=\frac{R}{4}\)
\(\frac{1}{\lambda_{2}}=\mathrm{R}\left[\frac{1}{16}-\frac{1}{25}\right]=\mathrm{R}\left[\frac{25-16}{16 \times 25}\right]=\frac{9 \mathrm{R}}{16 \times 25}\)
\(\therefore \frac{\lambda_{1}}{\lambda_{2}}=\frac{\mathrm{R}}{4} \times \frac{16 \times 25}{9 \mathrm{R}}=\frac{100}{9}\)
\(\therefore \frac{9}{100} \times \lambda_{1}=\lambda_{2}\)
\(\therefore \lambda_{2}=0.09 \times \lambda_{1}\)