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I was solving this question:

Here is part of the energy level diagram of hydrogen:

n=4 --> -0.85eV

n=3 --> -1.50eV

n=2 --> -3.40eV

n=1 --> -13.6eV

When an electron of energy 12.1eV collides with this atom, photons of three different energies are emitted. Show on the diagram (with arrows) the transitions responsible for these three photons.

I worked out that $E_3-E_1=12.1$ and $(E_3-E_2)+(E_2-E_1)=12.1$.

What I really want to know is what happens when the electron collides with the atom. I would really appreciate if someone can give a detailed description of this in simple language. My attempt would be:

The electron collides with the atom and goes to one of the energy levels. Since only $E_3-E_1$ and $(E_3-E_2)+(E_2-E_1)$ have a value of $12.1$ it follows that the electron can either have ended up in $n=3$ and then dropped to $n=1$, or it could have ended up in $n=3$ then dropped to $n=2$ then to $n=1$.

Is this correct? If so, how does the electron escape from the atom? It can't stay in the ground state $n=1$ after emitting the photon(s) because that is already "full".

Please correct me on anything incorrect that I said.

Thanks

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  • $\begingroup$ Well, the question is kind of stupid. It should be worded, "a colliding electron looses 12.1eV to an atom", if both electrons (the one colliding and the one that already was bound in the hydrogen atom) stuck with the atom, the hydrogen energy levels would not describe the system! $\endgroup$
    – Sebastian Riese
    Commented May 10, 2015 at 22:27
  • $\begingroup$ @SebastianRiese: Thanks. Sorry though, I'm not sure what "hydrogen energy levels would not describe the system" means. (To clarify, I am in AP physics and have very limited knowledge about this stuff, which is probably why the wording of the question is stupid, they dumb down stuff a lot at this level) $\endgroup$
    – user45220
    Commented May 10, 2015 at 22:31
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    $\begingroup$ A bound quantum mechanical system has sharp energy levels (as given in the energy level diagram). So only transitions between these levels are allowed (which gives the sharp spectral lines of atoms). A bound state of a proton and an electron is a hydrogen atom, if you bind another electron to this you get the $H^-$ ion, whose energy spectrum (set of allowed energies) is rather like that of helium, than that of hydrogen. $\endgroup$
    – Sebastian Riese
    Commented May 10, 2015 at 22:34
  • $\begingroup$ Correction: while it has an excitation spectrum similar to helium, only the ground state is a bound state (meaning, on cannot excite the $H^-$ ion without separating one electron). $\endgroup$
    – Sebastian Riese
    Commented May 10, 2015 at 22:45
  • $\begingroup$ @user45220: Your assessment of the transitions which can occur, and hence the photons which can be emitted, is correct. However, the colliding electron does not go to one of the energy levels in the atom (as Sebastian already correctly pointed out). What happens is that the colliding electron can deposit its energy in the bound electron, 'promoting' it from the ground state to the $n=3$ level. It is the subsequent decay of this electron, which remains bound throughout the whole process, which leads to photon emission. The incoming electron remains free, albeit with zero kinetic energy. $\endgroup$
    – tok3rat0r
    Commented May 10, 2015 at 23:02

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Your assessment of the transitions which can occur, and hence the photons which can be emitted, is correct. However, the colliding electron does not go to one of the energy levels in the atom (as Sebastian already correctly pointed out). What happens is that the colliding electron can deposit its energy in the bound electron, 'promoting' it from the ground state to the $n=3$ level. It is the subsequent decay of this electron, which remains bound throughout the whole process, which leads to photon emission. The incoming electron remains free, albeit with zero kinetic energy.

But to second what Sebastian Riese said, the question is extremely poor. Such a situation would not arise in reality, since the free electron of zero energy would combine with the atom to form a $\mathrm{H}^−$ ion. This ion would then have different energy levels from the basic hydrogen atom. Also, such a (single electron) impact could only emit at most two photons, since the excited electron could decay either via $n=3-2-1$ or via $n=3-1$, not both! To emit all three possible photons, two separate 12.1 eV electron impacts would be required.

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