JEE Main 2026 Jan 22 Shift 2, Physics Q7: The Line Spectra of the Hydrogen Atom

Smallest Lyman wavelength corresponds to $n_1=1,n_2=\infty$: $\frac{1}{\lambda }=R(1)\Rightarrow R=\frac{1}{91}n{m}^{-1}$.

Largest Balmer wavelength is the $n=3\to 2$ transition: $\frac{1}{{\lambda }_B}=R\left(\frac{1}{4}-\frac{1}{9}\right)=\frac{R\cdot 5}{36}$, so ${\lambda }_B=\frac{36\cdot 91}{5}\approx 655.2$ nm.

Largest Paschen wavelength is the $n=4\to 3$ transition: $\frac{1}{{\lambda }_P}=R\left(\frac{1}{9}-\frac{1}{16}\right)=\frac{7R}{144}$, so ${\lambda }_P=\frac{144\cdot 91}{7}\approx 1872$ nm.

Difference $\approx 1872-655.2\approx 1217$ nm.