Question

An electron of a stationary hydrogen atom passes from the fifth energy level to the ground level. The velocity that the atom acquired as a result of photon emission will be:

• A. \( \frac{24hR}{25m} \)
• B. \( \frac{25hR}{24m} \)
• C. \( \frac{24m}{25hR} \)
• D. \( \frac{25m}{24hR} \)

Hint:

The velocity of an atom after photon emission can be determined using the conservation of momentum, taking into account the energy and momentum of the emitted photon.

Answer 1

The Correct Option is A

The energy difference between the fifth and ground energy levels in a hydrogen atom is responsible for the emission of a photon.
The velocity of the atom after the emission can be found using conservation of momentum.
The momentum of the photon emitted is equal to the change in momentum of the atom: \[ \Delta p_{\text{photon}} = \frac{h}{\lambda} \] where \( \lambda \) is the wavelength corresponding to the energy difference between the levels. Using energy conservation and the Rydberg formula, we get: \[ v = \frac{24hR}{25m} \] where \( h \) is Planck's constant, \( R \) is the Rydberg constant, and \( m \) is the mass of the electron.
Thus, the correct answer is (a).