When a photon is emitted by a hydrogen atom, it carries momentum. A...

The momentum of the emitted photon is given by $p = \dfrac{h}{\lambda}$

By conservation of linear momentum, the recoil momentum of the hydrogen atom is equal in magnitude to the momentum of the photon.

The kinetic energy of recoil of the atom is $K = \dfrac{p^2}{2m} = \dfrac{h^2}{2m\lambda^2}$

Substituting the given values:

$K = \dfrac{(6.63 \times 10^{-34})^2}{2 \times 1.67 \times 10^{-27} \times (656.3 \times 10^{-9})^2} \text{ J}$

$K = \dfrac{43.96 \times 10^{-68}}{3.34 \times 10^{-27} \times 4.3 \times 10^{-13}} \text{ J} \approx 3.05 \times 10^{-28} \text{ J}$

Converting the energy into electron-volts (eV):

$K = \dfrac{3.05 \times 10^{-28}}{1.6 \times 10^{-19}} \text{ eV} \approx 1.9 \times 10^{-9} \text{ eV}$