Hydrogen Spectrum

NCERT describes the hydrogen emission spectrum through the Rydberg formula $\frac{1}{\lambda }=R\left(\frac{1}{n_1^2}-\frac{1}{n_2^2}\right)$, where $n_1$ fixes the series. The series limit is the shortest wavelength, obtained when $n_2\to \infty$, giving $\frac{1}{\lambda }=\frac{R}{n_1^2}$. For the Lyman series $n_1=1$, so $\lambda =\frac{1}{R}=\frac{1}{1.097\times 10^7}\approx 91$ nm, which lies in the ultraviolet. The Balmer series has $n_1=2$ with a limit near 365 nm, in the near ultraviolet-visible boundary, so it is wrong. The Paschen series with $n_1=3$ has its limit near 820 nm in the infrared, which is wrong. The Brackett series with $n_1=4$ sits even deeper in the infrared, so it is also wrong. As stated in NCERT Class 12, Chapter 12 (Atoms), the Lyman series arises from transitions ending at the ground state and is entirely ultraviolet, while the Balmer series ending at the second level produces the visible lines. Because the series limit corresponds to an electron falling from \infty, it carries the maximum energy for that series, and the ground-state-terminating Lyman series therefore has the most energetic limit of all. The value $\frac{1}{R}$ evaluates to just under 91.2 nm, firmly in the far ultraviolet, matching the given figure. A magnitude check confirms that the smallest $n_1$ produces the highest energy and hence the shortest wavelength limit, consistent with 91 nm being the most energetic series edge.