Suppose frequency of emitted photon is `f_(0)` when the electron of a stationary hydrogen atom jumps from a higher state `m` to a lower state `n`. If the atom is moving with a velocity `v (lt lt c)` and emits a photon of frequency `f` during the same transition, then which of the following statement are possible?

Options

A. f may be equal to `f_(0)`
B. f may be greater than `f_(0)`
C. f may be less than `f_(0)`
D. f cannot be equal to `f_(0)`

Answer

A, B, C

Explanation

When a stationary hydrogen atom emits a photon, then energy of the emitted photon will be equal to the difference of the energy of the two levels involved in the transition. Hence, energy of emitted photon will be equal to `(E_(m) - E_(n))`
If a hydrogen atom is moving and a photon is emitted by it along the same direction in which it is moving , due to momentum of the emitted photon , the momentum of hydrogen atom will get decreased. Therefore energy of the emitted photon will be equal to `(E_(m) - E_(n)` + "loss of KE of the hydrogen atom").
But if the photon is emitted in a direction normal to the motion, of the hydrogen atom, then the frequency of the emitted photon will be equal to `f_(0)`. Hence, option (a) is correct, Obviously, option (d) is wrong.
If the photon is emitted by the hydrogen atom in the direction opposite to its motion, then frequency of the emitted photon will be less than `f_(0)`. Hence, option (c) is correct.