Transition of neon from ground state to excited state

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Discussion Overview

The discussion revolves around the transition of neon from its ground state to excited states, particularly focusing on the excitation energies observed in the Franck-Hertz experiment. Participants explore the reasons behind the specific energy levels at which current decreases, questioning why transitions occur at 19 eV rather than 17 eV, and the implications of angular momentum conservation during these transitions.

Discussion Character

  • Exploratory
  • Technical explanation
  • Debate/contested

Main Points Raised

  • Some participants note that the lowest excited energy state of neon is at 1(s^2)2(s^2)2(p^5)3(s^1) with an excitation energy of about 16.9 eV, and the next state at 1(s^2)2(s^2)2(p^5)3(p^1) with about 19 eV.
  • There is a question regarding why the Franck-Hertz experiment shows current decreases at 19 eV and not at 17 eV, leading to speculation about the nature of the transitions involved.
  • One participant suggests that the angular momentum conservation during collisions might explain the preference for the 19 eV transition, where the orbital angular momentum does not change (Δl=0) and the total angular momentum remains constant (ΔJ=0).
  • Another participant points out that there is also a transition corresponding to 16.85 eV that involves a change in angular momentum (Δl=-1), which may be less likely to occur.
  • Concerns are raised about whether Δl=0 contradicts selection rules, with one participant clarifying that these are collisional excitations and not governed by the same selection rules as electromagnetic absorption.

Areas of Agreement / Disagreement

Participants express uncertainty regarding the explanations for the observed excitation energies and the implications of angular momentum conservation. There is no consensus on the correctness of the proposed explanations or the relevance of selection rules in this context.

Contextual Notes

Participants acknowledge that the angular momentum coupling in the excited states of rare gases may not follow simple L-S coupling rules, indicating potential limitations in their reasoning.

popeadam
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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?
 
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popeadam said:
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" .
 
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SpectraCat said:
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?
 
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popeadam said:
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.
 
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