JEE Main · 2026 · 06 Apr 2026 · Shift 1 · Atomic Structure

The shortest wavelength of a series arises from the transition that ends on the series level coming from $n=\infty$. For the Lyman series of hydrogen ($Z=1$, final level $n=1$),

The longest wavelength of a series arises from the smallest-energy jump, from the level just above the series limit. For the Balmer series of $\mathrm{H}{\mathrm{e}}^{+}$ ($Z=2$, final level $n=2$) this is $n=3\to n=2$:

Substituting $R=1/x$ gives $\lambda =\frac{9}{5R}=\frac{9x}{5}$.
The longest Balmer wavelength of $\mathrm{H}{\mathrm{e}}^{+}$ is $\frac{9x}{5}$.