\(\frac{1}{\lambda}=R\left[\frac{1}{n_{f}^{2}}-\frac{1}{n_{i}^{2}}\right]\)

In first case, \(n_{f}=3, n_{i}=4\)

\(\therefore \quad \frac{1}{\lambda_{1}}=R\left[\frac{1}{3^{2}}-\frac{1}{4^{2}}\right]=R\left[\frac{1}{9}-\frac{1}{16}\right]=\frac{7}{144} R\)    .... \((i)\)

In second case, \(n_{f}=2, n_{i}=3\)

\(\therefore \quad \frac{1}{\lambda_{2}}=R\left[\frac{1}{2^{2}}-\frac{1}{3^{2}}\right]=R\left[\frac{1}{4}-\frac{1}{9}\right]=\frac{5}{36} R\)    .... \((ii)\)

Divide \((ii)\) by \((i)\), we get

\(\frac{\lambda_{1}}{\lambda_{2}}=\frac{5}{36} \times \frac{144}{7}=\frac{20}{7}\)