JEE Main · 2026 · 23 Jan 2026 · Shift 1 · Modern Physics

For the Lyman series the longest wavelength is the $n=2\to 1$ line: $\frac{1}{\lambda }=R\left(1-\frac{1}{4}\right)=\frac{3R}{4}$, so ${\lambda }_{\max }=\frac{4}{3R}$, and statement A holds. The Balmer series ($n\to 2$) lies in the visible region, so statement B holds. The shortest wavelength (series limit) of the Paschen series is the $n\to \infty$ transition to $n=3$: $\frac{1}{\lambda }=\frac{R}{9}$, giving ${\lambda }_{\min }=\frac{9}{R}$, so statement C holds. The shortest wavelength of the Lyman series is $\frac{1}{\lambda }=R$, i.e. $\frac{1}{R}$, not $\frac{5}{4R}$, so statement D fails. The correct statements are A, B and C.