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.