JEE Main · 2020 · 04 Sep 2020 · Shift 1 · Modern Physics

For hydrogen $\frac{1}{\lambda }=R\left(\frac{1}{n_1^2}-\frac{1}{n_2^2}\right)$. In the Lyman series ($n_1=1$) the shortest wavelength comes from $n_2=\infty$, $\frac{1}{{\lambda }_{\min }}=R$, and the longest from $n_2=2$, $\frac{1}{{\lambda }_{\max }}=\frac{3R}{4}$; their difference is ${\lambda }_{\max }-{\lambda }_{\min }=\frac{4}{3R}-\frac{1}{R}=\frac{1}{3R}=304$ \AA. In the Paschen series ($n_1=3$) the shortest wavelength ($n_2=\infty$) is $\frac{9}{R}$ and the longest ($n_2=4$) is $\frac{144}{7R}$, so their difference is $\frac{144}{7R}-\frac{9}{R}=\frac{81}{7R}$. Dividing the two differences removes $R$: $\frac{\Delta {\lambda }_{\text{Paschen}}}{\Delta {\lambda }_{\text{Lyman}}}=\frac{81/7}{1/3}=\frac{243}{7}$, so $\Delta {\lambda }_{\text{Paschen}}=304\times \frac{243}{7}\approx 10553$ \AA.