The recoil speed of a hydrogen atom after it emits a photon in going from n = 5 state to n = 1 state will be :

Solution

Physics sir · 4 steps

The photon carries momentum E/c, so the atom must kick back with exactly that much momentum.

1. Photon energy. For hydrogen, Eₙ = -13.6/n² eV. Going from n = 5 to n = 1 releases E = 13.6(1 - 1/25) = 13.6 × 24/25 = 13.056 eV, which is 13.056 × 1.6 × 10⁻¹⁹ = 2.09 × 10⁻¹⁸ J.
2. Conserve momentum. The atom starts at rest, so its momentum after emission equals the photon's momentum E/c in the opposite direction. The recoil energy is tiny, so the photon takes almost all of the 13.056 eV.

   Mv = E/c ⇒ v = E/Mc

3. Put in numbers. With M = 1.67 × 10⁻²⁷ kg and c = 3 × 10⁸ m s⁻¹, Mc = 5.01 × 10⁻¹⁹ kg m s⁻¹. So v = (2.09 × 10⁻¹⁸)/(5.01 × 10⁻¹⁹) ≈ 4.17 m s⁻¹.
4. So. The hydrogen atom recoils at about 4.17 m/s, option (d).

The idea to keep. A photon of energy E has momentum E/c. Whenever an atom emits one, use momentum conservation: recoil speed is E/(Mc), with E taken from the actual energy-level difference.

Answer from the NTA final answer key. Solution by PrepWiser.