A hydrogen atom changes its state from n=3 to n=2. Due to recoil, the percentage change in the wave length of emitted light is approximately 1 × 10⁻ⁿ. The value of n is _______.
[Given Rhc =13.6 eV, hc=1242 eV nm, h=6.6 × 10⁻³⁴ J s mass of the hydrogenatom =1.6 × 10⁻²⁷ kg]

Solution

Physics sir · 4 steps

The recoiling atom takes a tiny slice of the energy, so the photon comes out slightly longer.

1. Transition energy. For n=3 → 2, the energy released is Δ E = 13.6(1/4-1/9) = 13.6 × 5/36 ≈ 1.89 eV. Without recoil, all of this would go to the photon.
2. Recoil share. The photon carries momentum p = E/c, and the atom recoils with the same momentum. Its kinetic energy is K = p²/2M = E²/2Mc². The photon loses this energy, so the fractional change in wavelength equals the fractional change in energy.

   Δλ/λ = K/E = E/2Mc²

3. Put in numbers. Mc² = 1.6 × 10⁻²⁷ × (3 × 10⁸)² = 1.44 × 10⁻¹⁰ J = 9 × 10⁸ eV. So Δλ/λ = 1.89/(1.8 × 10⁹) ≈ 1.05 × 10⁻⁹.
4. So. The percentage change is 1.05 × 10⁻⁹ × 100 ≈ 1 × 10⁻⁷ %, so n = 7.

The common slip: 9. Stopping at the fraction 1 × 10⁻⁹ and not multiplying by 100 for the percentage gives 9, not 7.

The idea to keep. For recoil, remember Δλ/λ = E/2Mc². A few eV set against the atom's rest energy of about 10⁹ eV always gives a tiny shift.

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