Transition of neon from ground state to excited state

[popeadam]

Neon has lowest excited energy at 1(s^2)2(s^2)2(p^5)3(s^1) state. And the excitation energy is about 16.9eV.
And next energy is at 1(s^2)2(s^2)2(p^5)3(p^1) state. And the excitation energy is about 19eV.

In Franck Hertz experiment with Neon, current decrease at every 19eV, no 17eV.

Why 19eV? Why not 17eV?
Why neon can't go to the lowest excited state directly. They go to high excited state and transfer to lowest energy. Why?

 

[SpectraCat]

Neon has lowest excited energy at 1(s^2)2(s^2)2(p^5)3(s^1) state. And the excitation energy is about 16.9eV.
And next energy is at 1(s^2)2(s^2)2(p^5)3(p^1) state. And the excitation energy is about 19eV.

In Franck Hertz experiment with Neon, current decrease at every 19eV, no 17eV.

Why 19eV? Why not 17eV?
Why neon can't go to the lowest excited state directly. They go to high excited state and transfer to lowest energy. Why?

It's a good question .. I don't actually know the answer off-hand. My guess is that it has to do with angular momentum conservation during the collision. For the 19 eV transition (actually 18.71 eV), a neon electron gets excited from a 2p to a 3p orbital ... that means that the orbital angular momentum associated with that particular electron doesn't change (\Delta l=0). Also, for that transition, the total angular momentum of the atom does not change (\Delta J=0).

My guess is that those transitions have a higher probability of being excited than ones where there is a net transfer of angular momentum between the incident electron and the neon atom.

Note that there is a transition to a 1s²2s²2p⁵3s¹ state that also corresponds to
\Delta J=0 (the energy is 16.85 eV), however that means that the angular momentum of the excited electron has to change (i.e. \Delta l=-1), which is also probably not very likely.

Anyway, I am not sure this explanation is completely correct .. for example, the angular momentum coupling in the excited states of rare gases does not follow simple L-S coupling rules if I remember correctly. However, I do think it is at least plausible. For reference, I used the table of Ne energy levels found http://physics.nist.gov/PhysRefData/Handbook/Tables/neontable5.htm" .

 

[popeadam]

It's a good question .. I don't actually know the answer off-hand. My guess is that it has to do with angular momentum conservation during the collision. For the 19 eV transition (actually 18.71 eV), a neon electron gets excited from a 2p to a 3p orbital ... that means that the orbital angular momentum associated with that particular electron doesn't change (\Delta l=0). Also, for that transition, the total angular momentum of the atom does not change (\Delta J=0).

My guess is that those transitions have a higher probability of being excited than ones where there is a net transfer of angular momentum between the incident electron and the neon atom.

Note that there is a transition to a 1s²2s²2p⁵3s¹ state that also corresponds to
\Delta J=0 (the energy is 16.85 eV), however that means that the angular momentum of the excited electron has to change (i.e. \Delta l=-1), which is also probably not very likely.

Anyway, I am not sure this explanation is completely correct .. for example, the angular momentum coupling in the excited states of rare gases does not follow simple L-S coupling rules if I remember correctly. However, I do think it is at least plausible. For reference, I used the table of Ne energy levels found http://physics.nist.gov/PhysRefData/Handbook/Tables/neontable5.htm" .

Thank you. But (\Delta l=0) is against the selection rules?

 

[SpectraCat]

Thank you. But (\Delta l=0) is against the selection rules?

What selection rules? These are collisional excitations, not excitations due to absorption of EM radiation ... as far as I am aware, there are no selection rules, you just need to conserve angular momentum (and linear momentum and energy).

Furthermore, the selection rule for EM absorption would be that the TOTAL angular momentum quantum number for the atom would have to change by one (\Delta J=\pm 1) ... that would certainly allow the orbital angular momentum change of an individual electron to be zero .. that is what I was trying to indicate by \Delta l=0 .. sorry if it was confusing.