An electron with some Kinetic energy say $K$ hits a hydrogen nuclei and goes to a stable orbit $(n=1)$ releasing a photon while doing so. What will be the energy of photon emitted?

I am limited to Bohr's model of atom. Here is the trouble, if the electron had zero Kinetic energy, the photon would simply have energy $13.6eV$, the ionisation energy of hydrogen atom. By conserving energy here, the photon energy comes out to be $13.6eV + K$, this was done in class, but the momentum is not conserved in doing so as final Kinetic energy is zero everything would be at rest, the photon may not carry the whole momentum.

So, what is the right thing to do in this question?


1 Answer 1


You conserve momentum by making the atom recoil as needed, when the photon is emitted. The recoil will involve negligible kinetic energy, but you can always solve for it self-consistently if you want. Given a stationary ion and a moving electron, you start with a total momentum and energy, and you end with a photon and a neutral atom, so you still have two variables in the final state as well, and can conserve both energy and momentum. But you're right, it does mean the photon energy is not exactly 13.6+K, but it is very close to that. In fact, the possibility for recoil means that the photon does not need to be emitted along the same direction that the electron was originally moving.


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