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

When a hydrogen atom transitions from the $$n = 5$$ state to the $$n = 1$$ state, it emits a photon. By conservation of momentum, the atom recoils with momentum equal in magnitude to the photon's momentum.

The energy of the emitted photon is $$E = 13.6 \left(\frac{1}{1^2} - \frac{1}{5^2}\right) = 13.6 \times \left(1 - \frac{1}{25}\right) = 13.6 \times \frac{24}{25} = 13.056$$ eV.

Converting to joules: $$E = 13.056 \times 1.6 \times 10^{-19} = 2.089 \times 10^{-18}$$ J.

The momentum of the photon is $$p = \frac{E}{c} = \frac{2.089 \times 10^{-18}}{3 \times 10^8} = 6.963 \times 10^{-27}$$ kg m/s.

The recoil speed of the hydrogen atom (mass $$m = 1.67 \times 10^{-27}$$ kg) is $$v = \frac{p}{m} = \frac{6.963 \times 10^{-27}}{1.67 \times 10^{-27}} = 4.17$$ m/s.