Photon Momentum and Atomic Recoil
A photon of wavelength lambda carries momentum h/lambda. Neglecting the small recoil correction to the line energy, a stationary atom of mass M acquires equal opposite momentum and speed h/(M lambda); its recoil energy is p^2/(2M).
Why this shows up in the exam
Finding recoil speed after emission · Comparing momenta of spectral photons · Accounting for energy-momentum conservation in atomic processes
Learn the idea
Emission conserves momentum, so a previously resting atom recoils opposite to its photon. A photon leaving in one direction carries momentum; the atom must move the other way so the original zero momentum stays zero.
🧠 Memory hook: Photon out, atom back with equal momentum.
Get this one clearly and it pays off every single time it shows up in the paper. 🎯
Formulas & facts to keep ready
- p_gamma = h/lambda = E_gamma/c — photon momentum
- v_recoil = h/(M lambda) — atomic recoil speed for initial rest
- K_recoil = p_gamma²/(2M) — non-relativistic recoil energy
How to approach it
- 1Find photon energy or wavelength
- 2Convert it to momentum
- 3Apply equal opposite momentum to the atom and then compute speed or energy
Common slip-ups that cost marks
- •Setting recoil momentum to zero because photon mass is zero
- •Equating photon and atom speeds
- •Using electron mass for recoil of the whole atom
🌟 That's the whole idea — you've got this. Try the practice set below; every question you attempt makes it stick a little harder.
Original chapter practice
Original questions for this chapter, not past-paper questions or an exact mapping to this individual concept.
In hydrogen, an electron transitions from n = 2 to n = 1. Using E_n = -13.6/n^2 eV, find the emitted photon energy.
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