Answer: 10553.14

Solution

From Bohr’s formula, $\frac{1}{\lambda} = R\left(\frac{1}{n_1^2} - \frac{1}{n_2^2}\right)$. For Lyman series: $\frac{1}{\lambda_{min}} = R(1) \Rightarrow n_2 = \infty, n_1 = 1$. $\frac{1}{\lambda_{max}} = R\left(1 - \frac{1}{4}\right) = \frac{3R}{4}$ ($n_1 = 1, n_2 = 2$). $\lambda_{max} - \lambda_{min} = \frac{4}{3R} - \frac{1}{R} = \frac{1}{3R} = 304$ A. For Paschen series: $\lambda’{min} = R\left(\frac{1}{9}\right)$ and $\lambda’{max} = R\left(\frac{1}{9} - \frac{1}{16}\right) = \frac{7R}{144}$. $\lambda’{max} - \lambda’{min} = \frac{9 \times 16}{7R} - \frac{9}{R} = \frac{81}{7R} = \frac{81 \times 3}{7 \times 3R} = \frac{81}{7} \times 304 \times 3 = 10553.14$ A.