0
$\begingroup$

Suppose a free electron meets a hydrogen ion. The electron drops to the ground state and emits a photon. The energy of the photon will be equal to the ionisation energy of the hydrogen atom, 13.6 eV.

Next, suppose a free electron having a kinetic energy of 0.1 eV meets a hydrogen ion. Is it possible the energy of the emitted photon is 13.7 eV, due to the initial kinetic energy of the electron?

$\endgroup$
1
$\begingroup$

Yes, it is entirely possible. This is why emission bands in real life have a spectral width to them; it is because of the doppler shift of the light that is emitted from moving ions. Light emitted from an ion that is moving towards you will have a slightly higher frequency than light emitted from an ion moving away. This effect is called Doppler Broadening.

Edit: Ok, I misunderstood the original question. If the electron is moving before being captured by the ion, then this will also change the energy of the emitted photon as the OP suggested. The emitted photon is the difference in energy between the electron before capture and after capture, so if the electron has some initial kinetic energy then there is a bigger energy difference between that energy state and the lowest energy state of the atom as compared to the difference between an electron at rest and the lowest energy state of the atom. Of course, it's possible that the electron will graze by the ion on the first couple passes and lose energy gradually through emitting lower frequency photons, so that by the time it's captured its kinetic energy is close to zero. Also, if the electron's kinetic energy is greater than the potential well of the ion, then the electron will just be scattered rather than captured.

$\endgroup$
  • $\begingroup$ Ok, my question was unclear, I should not have used the word blueshift because it suggests a hydrogen atom is moving with respect to the observer. I meant the hydrogen atom is at rest, only the electron is moving. $\endgroup$ – jkien Mar 28 '17 at 17:41
  • $\begingroup$ Ok, I made an edit $\endgroup$ – Travis Mar 28 '17 at 18:05

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.