# Hydrogen Emission Spectra > Hydrogen emission spectra for JEE Physics: Rydberg formula, Lyman-Balmer-Paschen series limits, line counting and He+ Z-squared scaling. - Canonical URL: https://prepelephant.com/topics/jee/physics/emission-spectra-hydrogen - Exam / course: JEE · Subject: Physics - Publisher: PrepElephant (https://prepelephant.com) — Prepared and reviewed by the PrepElephant Academic Review Team - First published: 2026-10-02 - Last updated: 2026-10-02 - How to cite: "Hydrogen Emission Spectra", PrepElephant, https://prepelephant.com/topics/jee/physics/emission-spectra-hydrogen ## Direct answer One electron around one proton emits dozens of sharp lines, and every one of them is a solution of the Rydberg formula 1/λ = RZ²(1/n1² − 1/n2²), with R = 1.097 × 10⁷ m⁻¹ and Z = 1 for hydrogen. The series are named by the lower level: Lyman (n1 = 1) in the ultraviolet with limit 91.2 nm, Balmer (n1 = 2) — the only visible series, Hα at 656.3 nm through Hδ at 410.2 nm with limit 364.6 nm — then Paschen, Brackett and Pfund marching deeper into the infrared from n1 = 3, 4 and 5. An atom excited to level n can emit n(n − 1)/2 distinct lines, and any hydrogen-like ion (He⁺, Li²⁺) reproduces the entire pattern with wavelengths shrunk by Z²: He⁺'s analogue of the red Hα line sits at 656.3/4 = 164.1 nm, deep in the ultraviolet. ## What you must remember - **Rydberg formula:** 1/λ = RZ²(1/n1² − 1/n2²), R = 1.097 × 10⁷ m⁻¹; n2 > n1, and the photon carries the energy difference 13.6Z²(1/n1² − 1/n2²) eV. - **Series by lower level:** Lyman n1 = 1 (UV, first line 121.6 nm, limit 91.2 nm); Balmer n1 = 2 (visible, Hα 656.3 nm, limit 364.6 nm); Paschen n1 = 3, Brackett n1 = 4, Pfund n1 = 5 — all infrared. - **Only Balmer is visible:** the four named lines Hα 656.3 (red), Hβ 486.1 (blue-green), Hγ 434.0 (blue-violet), Hδ 410.2 (violet) converge to the 364.6 nm limit. - **Line counting:** maximum distinct lines from level n = n(n − 1)/2; from n = 5 that is 10, of which exactly 3 fall in the Balmer series (from 3, 4, 5 down to 2). - **Hydrogen-like scaling:** wavelength ∝ 1/Z² and photon energy ∝ Z²; He⁺ (Z = 2) shows the hydrogen pattern at one-quarter the wavelengths. - **Limit energies:** Lyman limit = 13.6 eV (ionisation energy of hydrogen); Balmer limit corresponds to 3.4 eV. - **Convergence logic:** lines crowd together as n2 grows because energy levels themselves crowd — the spacing shrinks as 1/n². ## Counting lines: the transition arithmetic Excite hydrogen to n = 4 and predict the spectrum before touching a calculator. Possible drops: 4→3, 4→2, 4→1, 3→2, 3→1, 2→1 — six lines, matching 4 × 3/2. One is Lyman-family? three are (4→1, 3→1, 2→1), two are Balmer (4→2, 3→2), one is Paschen (4→3). Now compute the 4→2 wavelength: 1/λ = 1.097 × 10⁷ (1/4 − 1/16) = 2.057 × 10⁶ m⁻¹, so λ = 486.1 nm — Hβ, the blue-green line. The examiner's second move is to count emitted photons per atom versus distinct spectral lines. A single atom excited to n = 4 emits at most three photons (cascade 4→2→1 or 4→3→1), while a discharge tube containing billions of atoms shows all six lines at once because different atoms cascade differently. That distinction — per-atom photon count versus population-level line count — is a standard reason-based question. For hydrogen-like ions, the same arithmetic with Z² = 4 (He⁺) moves Hβ to 121.5 nm in the ultraviolet, despite identical structure. ## Spectra questions in JEE The recurring one-mark trap: identifying which series lies in the visible region — answer, always, Balmer alone; Lyman is ultraviolet and everything above n1 = 2 is infrared, and options mix them freely. Second, wavelength versus energy ordering: students rank series by wavelength but answer an energy question; Lyman photons are the most energetic (up to 13.6 eV) yet have the shortest wavelengths, and the two rankings invert. Third, the counting formula applied to the wrong quantity: n(n − 1)/2 counts distinct lines in emission; if the question asks for lines in one named series from level n, the answer is n − n1. Fourth, Z-scaling: He⁺ and Li²⁺ lines shrink by Z² in wavelength, not by Z, a factor-of-two-versus-four option split. Main tests plug-and-identify; Advanced couples the Rydberg formula with the Bohr energy levels, asking which transition gives a photon able to photoionise hydrogen from n = 2, or matching a given wavelength to its transition without warning. ## Frequently asked questions ### Which series of hydrogen lies in the visible region? Only the Balmer series (n1 = 2), from Hα at 656.3 nm to the 364.6 nm limit; Lyman is ultraviolet and Paschen onward are infrared. ### How many spectral lines can hydrogen emit from the n = 5 level? Ten distinct lines in total, given by n(n − 1)/2 = 5 × 4/2, of which three are Balmer lines (from n = 3, 4, 5 down to n = 2). ### Why do the lines of a series crowd together near the limit? Because energy levels converge as 1/n², transitions from very high n2 differ minutely in energy, so their wavelengths pack toward the series limit. ### How do hydrogen-like ion spectra differ from hydrogen's? Wavelengths shrink by a factor Z² and photon energies grow by Z²; He⁺ repeats the hydrogen pattern at one-quarter the wavelengths. ### What is the highest-energy photon hydrogen can emit? The Lyman-limit photon of 13.6 eV at 91.2 nm, from a free electron falling to the ground state — equal to the ionisation energy of the ground-state atom.