# States of an atom or electron

I can state my question with an example. We say that the hydrogen atom is in the ground or excited state. But should it be the electron which is in the ground or the excited state. Like a question states that an electron collides & excites a ground state hydrogen atom to the 3rd excited state. What does that mean? - That the electron collides with the hydrogen atom and knocks its 1st shell electron( which is in the ground state ,-13.6 eV) to the 3rd the shell.

Isn't that right ?If it is the why do we talk of the state of atom when it should be the state of the electron Or have I got it wrong ? If yes then what is right.

• Well, the electron is part of the atom. This seems to be a language nuance at best. Commented Feb 2, 2017 at 15:51

No it is the whole atom that is in a certain state i.e. has a certain energy.

The proton is about 2000 times heavier than the electron, so at least when first teaching students about the hydrogen atom we tend to assume the proton is fixed. That is the electron moves in a fixed Coulomb potential. If we make this assumption then the only variable in the system is the electron and it makes sense to talk about the states of the electron.

However the proton isn't fixed and it responds to the electron just as the electron responds to the proton, and this shifts the energy levels by a small but measurable amount. For example the ground state wavefunction is:

$$\psi_{1s} = \frac{e^{-r/a_0}}{\sqrt{\pi}a_0^{3/2}}$$

where the Bohr radius $a_0$ is:

$$a_0 = \frac{\hbar^2}{me^2}$$

And that gives the ground state energy as:

$$E_{1s} = \frac{me^4}{8h^2\varepsilon_0^2}$$

So far so good, but $m$ in these equations is not the electron mass. If the proton really were fixed it would be the electron mass, but it's actually the reduced mass of the electron-proton system:

$$m = \frac{m_e m_p}{m_e + m_p}$$

And this works out to be about $0.9995m_e$. So if we treat the system considering only the electron we're going to get energies that are too high by about $0.05\%$ and that's easily measurable.

The same argument tells us that the energy states of the deuterium and tritium atoms are going to be different again because the reduced mass will be different. The wavefunctions and energies really are properties of the whole atom not just the electron.

• I would like to ask , that in the example @E2n gives , it is the electron (of hydrogen) which which gets excited to the 3rd shell from the ground state when the electron collides with the hydrogen atom. If we neglect the COM calculation , then we are correct in saying that it's the electron which has excited , but in the process account for a small error . If we account for the COM , then we get the precise result. But it's the electron which jumps to the 3rd shell & the total gain of energy in this process is excitation ! Is what I have written correct ? Commented Feb 2, 2017 at 18:08
• The same pattern of approximations holds for multi-electron atoms. It's a convenient fiction to talk about, say, carbon as having two $1s$ electrons, two $2s$ electrons, and two $2p$ electrons. But the wavefunctions which actually give good predictions of the behavior of carbon have other combinations as well --- as long as it has the same overall quantum numbers, it contributes. You can't really point to an electron in a carbon atom and say "that's the $2p$ electron."
– rob
Commented Feb 2, 2017 at 19:07
• So its the electron which collides and excites the electron of hydrogen atom to go to an excited state but since the nucleus also moves around the center of mass its the excitation of the entire atom
– E2n
Commented Feb 6, 2017 at 15:36
• @E2n: yes, that's a pretty good way of putting it. Commented Feb 6, 2017 at 16:21

I'm not sure but I think that in the nucleus itself there are energy levels so maybe the fact that the electron moves up an energy level, that affects the nucleus and raises the protons for example by an energy level, because remember if the electrons need more em force to keep them from escaping the atom because they are now farther away from the attractive nucleus.

Don't quote me, this is just speculation, but it would be intresting to see if i'm right (probably not).