Question

Sol. [Key Concept: For Balmer lines $p = \frac{h}{\lambda}$ with $\frac{1}{\lambda} = R(\frac{1}{4} - \frac{1}{n^{2}})$; the ratio for $n = 3$ and $n = 4$ gives $\alpha:\beta = 20:27$.]

$${p = \frac{h}{\lambda} }{\frac{1}{\lambda_{1}} = R\left( \frac{1}{2^{2}} - \frac{1}{3^{2}} \right) = R\left( \frac{1}{4} - \frac{1}{9} \right) }{\frac{1}{\lambda_{1}} = R\frac{5}{36} }{\frac{1}{\lambda_{2}} = R\left( \frac{1}{2^{2}} - \frac{1}{4^{2}} \right) = R\left( \frac{1}{4} - \frac{1}{16} \right) }{\frac{1}{\lambda_{2}} = \frac{3R}{16} }{\frac{p_{1}}{p_{2}} = \frac{\lambda_{2}}{\lambda_{1}} = \frac{\frac{16}{3R}}{\frac{36}{5R}} = \frac{20}{27} }{\alpha = 20\&\beta = 27 }$$