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When a hydrogen atom drops from n = 2 to n = 1 it sends out a photon, and that photon carries momentum. So the atom must kick back the other way. Questions ask for the recoil speed, the recoil energy, or how far the photon's energy falls short of the gap between the levels. This post sets up the two conservation laws, works the Lyman-alpha case with CODATA constants, and shows why the correction is usually ignored.
Where does this sit in the syllabus?
The NCERT Class 12 chapter Dual Nature of Radiation and Matter (Reprint 2026-27) gives each photon "energy E (=hν) and momentum p (= hν/c)" and says that in a photon-particle collision "the total energy and total momentum are conserved". The Atoms chapter gives the hydrogen ground-state energy as −13.6 eV, with levels E = −13.6/n² eV.
The JEE (Main) 2026 syllabus lists the "particle nature of light" and "Bohr model, energy levels, hydrogen spectrum", and the NEET (UG) 2026 syllabus lists the same topics. The JEE (Advanced) 2026 brochure lists "Bohr's theory of hydrogen-like atoms". None of them names recoil; it is momentum conservation applied to these topics. Check the syllabus for your exam year.
How do you set up the problem?
Take the atom, of mass M, at rest before emission. Afterwards the photon has energy E and momentum E/c, and the atom moves the opposite way with speed v.
Momentum: Mv = E/c, so v = E/(Mc).
Energy: ΔE = E + ½Mv², where ΔE is the gap between the two levels.
Substituting v gives the recoil energy K = ½Mv² = E²/(2Mc²). Because K is tiny compared with E, you can put E ≈ ΔE on the right-hand side without noticeable error.
Worked example: the Lyman-alpha line
From CODATA values published by NIST, the proton's rest energy is 938.272 MeV and the electron's is 0.511 MeV, so Mc² ≈ 938.783 MeV for the atom. Take c = 2.998 × 10⁸ m/s.
Gap: ΔE = 13.6 × (1 − 1/4) = 10.2 eV. Wavelength λ = 1239.84/10.2 ≈ 121.6 nm, in the ultraviolet.
Recoil speed: v = (E/Mc²) × c = (10.2/938.783 × 10⁶) × 2.998 × 10⁸ ≈ 3.26 m/s.
Recoil energy: K = E²/(2Mc²) = 10.2²/(2 × 938.783 × 10⁶) ≈ 5.54 × 10⁻⁸ eV.
Fraction lost to recoil: K/E ≈ 5.4 × 10⁻⁹, about five parts in a billion.
Check in SI units. With the Planck constant h = 6.626 × 10⁻³⁴ J s, the photon momentum is p = h/λ = 6.626 × 10⁻³⁴/1.2156 × 10⁻⁷ = 5.45 × 10⁻²⁷ kg m/s. Dividing by the atom's mass, 1.6735 × 10⁻²⁷ kg, gives 3.26 m/s again. The atom moves off at about 12 km/h, a jogging pace.
How do other lines compare?
Hα (n = 3 → 2): E = 1.89 eV, λ ≈ 656 nm, v ≈ 0.60 m/s.
Lyman-gamma (n = 4 → 1): E = 12.75 eV, λ ≈ 97.2 nm, v ≈ 4.07 m/s.
A neat result hides here. The recoil shifts the wavelength by Δλ = λ × (K/E) = (hc/E)(E/2Mc²) = h/(2Mc). The photon energy cancels, so every line shifts by the same amount, about 6.6 × 10⁻¹⁶ m.
Why is recoil usually ignored?
Compare 3.26 m/s with the thermal motion of hydrogen atoms in a gas at 300 K. Using the Boltzmann constant k = 1.381 × 10⁻²³ J/K, the root-mean-square speed is √(3kT/M) ≈ 2720 m/s, more than 800 times the recoil speed. Doppler shifts from that motion swamp the recoil shift in an ordinary discharge tube.
Absorption works the same way in reverse. To lift an atom at rest from n = 1 to n = 2, the photon must supply 10.2 eV plus the recoil energy, so the emission and absorption lines are separated by 2K, about 1.1 × 10⁻⁷ eV. Because K grows as the square of the photon energy, recoil matters far more for high-energy photons such as gamma rays from nuclei.
What mistakes cost marks?
Giving the photon a kinetic energy ½mv². A photon has no rest mass. Use E = pc.
Answering ΔE when the question asks for the photon energy "taking recoil into account". That is ΔE − K.
Using the electron's mass. The whole atom recoils, and its mass is about 1837 times the electron's.
Mixing units. Keep E and Mc² both in eV, or both in joules.
Forgetting direction. The atom moves opposite to the photon. In a cascade of two photons, add the momenta as vectors.
Our posts on photon momentum and radiation pressure and hydrogen spectrum problems with the Bohr model cover the two ideas this problem joins.
A practice routine
Derive v = E/(Mc) and K = E²/(2Mc²) from the two conservation laws without notes.
Find v for the first three Balmer lines and check that it rises in step with E.
Show that Δλ = h/(2Mc) for every line, and evaluate it.
An atom at rest absorbs a photon and goes from n = 1 to n = 2. Explain why the photon needs slightly more than 10.2 eV, and find how much more. (Answer: about 5.5 × 10⁻⁸ eV.)
If you set up both equations and still get stuck partway, the Fermi Problem Library has worked problems with a step-by-step solver that explains the reasoning rather than printing an answer.
Today, rework the Lyman-alpha example for deuterium, whose atom is about twice as heavy, and check that the recoil speed roughly halves.
