A

Step 1: When an electron in a hydrogen atom transitions from a higher energy level to a lower one (in this case from n = 4 to n = 1), it emits a photon.
Step 2: The energy of the emitted photon can be calculated using the Rydberg formula:
$$ E = -13.6 \times \left(\frac{1}{n_f^2} - \frac{1}{n_i^2}\right) $$
where n_f = 1 and n_i = 4.
Step 3: Substituting the values:
$$ E = -13.6 \times \left(\frac{1}{1^2} - \frac{1}{4^2}\right) = -13.6 \times (1 - \frac{1}{16}) = -13.6 \times (\frac{15}{16}) = -12.75 ext{ eV} $$
Step 4: The momentum of a photon is given by the relation:
$$ p = \frac{E}{c} $$
where E is the energy of the photon and c is the speed of light (approximately $3 \times 10^8$ m/s).
Step 5: Therefore:
$$ p = \frac{12.75 \times 1.6 \times 10^{-19} \text{ J}}{3 \times 10^8} \approx 6.8 \times 10^{-27} \text{ kg m/s} $$
Step 6: By conservation of momentum, the recoiling hydrogen atom will have momentum equal in magnitude but opposite in direction to the photon emitted.
Therefore, the momentum of the recoiled hydrogen atom will be approximately $6.8 \times 10^{-27} \text{ kg m/s}$.