JEE Main 2024 — 6 April, Shift 2 — Question 31

For longest wavelength in Paschen's series:

$\frac{1}{\lambda}=\mathrm{R}\left[\frac{1}{\mathrm{n}_{1}{ }^{2}}-\frac{1}{\mathrm{n}_{2}{ }^{2}}\right]$

For longest $\mathrm{n}_{1}=3$

$\mathrm{n}_{2}=4$

$\frac{1}{\lambda}=\mathrm{R}\left[\frac{1}{(3)^{2}}-\frac{1}{(4)^{2}}\right]$

$\frac{1}{\lambda}=\mathrm{R}\left[\frac{1}{9}-\frac{1}{16}\right]$

$\frac{1}{\lambda}=\mathrm{R}\left[\frac{16-9}{16 \times 9}\right]$

$\Rightarrow \lambda=\frac{16 \times 9}{7 \mathrm{R}}=\frac{16 \times 9}{7 \times 1.097 \times 10^{7}}$ $\lambda=1.876 \times 10^{-6} \mathrm{~m}$