Hydrogen spectrum · Bohr's quantized atom

Why does hot hydrogen emit a few sharp colours — a barcode — instead of a smooth rainbow, and what does that say about the atom?

▶ Run the simulationSee the measured result

Units: m⁻¹ (Rydberg constant R_∞, CODATA 2018)

How the lab tests it

From Bohr's postulate that angular momentum is quantized (L = nℏ), derive the energy ladder from first principles — balance the Coulomb pull against the centripetal demand, giving E_n = −μe⁴/(8ε₀²h²n²) = −13.6 eV/n² (μ = the electron–proton reduced mass). A jump n₂→n₁ emits a photon E_{n₂}−E_{n₁}; collect twelve lines from three series (Lyman, Balmer, Paschen) and least-squares fit 1/λ against (1/n₁²−1/n₂²).

What it checks

the Rydberg constant R_H = 1.0968×10⁷ m⁻¹ — recovered as the slope of a SINGLE straight line through the origin (intercept ≈ 0), so all three series obey one law; the Bohr radius a₀ = 5.292×10⁻¹¹ m and ground-state energy E₁ = −13.6 eV, from the constants alone; and the four visible Balmer lines (Hα 656, Hβ 486, Hγ 434, Hδ 410 nm), matching the textbook. Bohr's 1913 model — the first time a quantization rule explained a spectrum, pairing with blackbody and the photoelectric effect as the third pillar of early quantum theory

Rydberg equation, Balmer-line, energy-level and series-limit calculator

Why hot hydrogen emits a barcode instead of a rainbow, priced from first principles rather than from a table. Not one number on this page is stored — not even the Rydberg constant. R_∞ is ASSEMBLED as mₑe⁴/(8ε₀²h³c) out of the five constants in the fields below, and comes out 10973731.568073 m⁻¹ against the CODATA value they were taken from, a check of the combination rather than a quotation of it; the Bohr radius is assembled from a different combination of the same numbers and reads no speed of light at all; a line is the reciprocal of a difference of two inverse squares; the series limit is the n₂ → ∞ end of that same difference; and the classical collapse time comes from a third combination entirely. h, e, ε₀, mₑ, c and the two nuclear masses are all FIELDS, which is exactly why no physical constant could be typed into the arithmetic. The defaults are this world's own bench, the Hα line: n₂ = 3 → n₁ = 2 on a proton, λ = 656.469606 nm at 1.888651009 eV, with R_M = R_∞/(1 + mₑ/M) = 10967758.340193 m⁻¹ — the same R_H the simulation above fits out of its twelve synthesized lines, and 1.759498e-11 from the spectroscopic value. Five things here the finding never computes. The ionization energy assembled from this ladder is 13.598287264 eV against NIST's 13.598434599702, and that 1.08e-5 deficit is not an error but the fine structure and QED deliberately left out, disclosed here at the size α² predicts. The whole infinite Balmer series is squeezed into the 291.764269 nm between its head and its limit at 364.705337 nm, and seven of its lines fall in a 380–750 nm window even though a discharge tube shows four — the difference is brightness, not arithmetic. Run the formula BACKWARDS and any wavelength you type is searched against every (n₁, n₂) pair, which is what a table used to be for. Swap the proton for a deuteron and Hα moves 0.178576 nm to the blue, which is how Urey found deuterium in 1932. And Balmer's 1885 formula λ = B n²/(n²−4) is not a rival at all: it is the n₁ = 2 slice of this one, with B = 4/R, agreeing to 1.7e-16 — while the rival that does die, the classical radiating atom, collapses in 1.556177e-11 s.