Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Energy differences between sub shells

  1. Jul 10, 2012 #1

    CAF123

    User Avatar
    Gold Member

    Just a quick question: what has the smaller energy gap and why?
    A transition from 2s to 2p or 2p to 3s?

    My thinking is the former since this is part of the same principal shell, but I'm not entirely sure.
    Thanks.
     
  2. jcsd
  3. Jul 10, 2012 #2

    Simon Bridge

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    2016 Award

    Why not just calculate it?
     
  4. Jul 10, 2012 #3

    CAF123

    User Avatar
    Gold Member

    How would you calculate it?
    Apologies if I am missing something rather obvious!

    What I ultimately want to find is the spectroscopic notation for an atom in its 1st excited state, which i believe corresponds to the smallest energy gap of all the transitions possible.
     
    Last edited: Jul 10, 2012
  5. Jul 10, 2012 #4

    Simon Bridge

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    2016 Award

  6. Jul 10, 2012 #5

    CAF123

    User Avatar
    Gold Member

    I am doing it for the 1st excited state of nitrogen.
    It seems from the graphs in the link provided that 2p is only slightly greater in energy than 2s because it is more shielded from the nuclear charge.
    However, I still don't know how to verify this by calculation. The formula provided on the link I am well aware of, but this gives energy gaps for only the principal quantum number (n) and is valid only for hydrogen.
    Is there an equation somewhere that can be used to calculate energy differences between sub shells in multielectron atoms?
     
  7. Jul 10, 2012 #6

    Simon Bridge

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    2016 Award

    Excuse me I didn't intend to point you at the whole solution for you - more to an understanding of what you are trying to do.

    It's not straight forward...
    http://electron6.phys.utk.edu/phys250/modules/module 3/Multi-electron atoms.htm

    Nitrogen atoms have 5 valence electrons so the Schodinger equation is a total pain to solve. You end up treating it as hydrogen, with bigger Z ... you'll get screening from the closed s shell.

    (I'm amazed I can't find an example ... I'll keep looking but it's 3am down here.)
    The principle is that the lower electrons in the 2s shell slightly screen the nucleus.
     
  8. Jul 10, 2012 #7

    CAF123

    User Avatar
    Gold Member

    Many thanks for the link and your input.
    Do you have any idea on what the 1st excited state spectroscopic notation would be for nitrogen?
    This question was given as first year university material, so the reasoning must be along the lines of knowing that the 1st excited state corresponds to that of lowest energy gap. Looking at a graph of the energies of subshells, you see that 2p is greater in energy than 2s by only a small amount so this would make me think that promotion of an electron from 2s to 2p.
    However, this gives two questions :
    1) promotion of electron from 2s to 2p requires disturbing full 2s subshell, so means more energy than normal? I understand now there may be a way to calculate, but as you mentioned, things get complicated and this question is posed as first year material, so there mut be some simple reasoning rather than some mathematical calculation.
    2) I remember my lecturer telling me that only valence (or outside) electrons can be promoted?
    Will this mean that electron in 2p gets promoted to 3s? This I what my lecturer thinks, but how does this reconcile with energy gaps etc..?

    Sorry for all my questions, but this question has been bothering me for quite a while. Any input is greatly appreciated. Thanks.
     
  9. Jul 11, 2012 #8

    Simon Bridge

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    2016 Award

    N has 5 valence electrons, 2 in 2s and 3 in 2p
    You could presumably get 2s -> 2p promotion ... there's an empty spot for it.

    In a basic course you'd only be expected to use the principle states ... so you'd promote one electron to the 3s state as you said. Since those are easier to calculate.

    This is annoying me now - I used to know this off the top of my head :(
     
  10. Jul 11, 2012 #9
    I must be missing something. Aren't the 2p/2s states degenerate?
     
  11. Jul 11, 2012 #10
    For an hydrogen atom the energy depends only on the principle quantum number, n. Therefore there's no energy change between 2s and 2p. Therefore that's the smaller energy gap since its zero.
     
  12. Jul 11, 2012 #11

    Simon Bridge

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    2016 Award

    2p and 2s states are degenerate states - there are two 2s states and 6 2p states ... they have the same principle quantum number so for hydrogen, as Boston_guy says, these states all have the same energy level. However, form multi-electron atoms, tightly bound shells (like the s ones) shield the nuclear charge, leading to a higher energy for the p shells. Similarly the p shell screens the nucleus for the higher shells.

    And then there's spin-orbit coupling, and the Lamb effect...

    I keep thinking there is something very important about this that I'm forgetting...
     
  13. Jul 11, 2012 #12
    Ah. I haven't been doing the heavier elements.
     
  14. Jul 18, 2012 #13

    Simon Bridge

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    2016 Award

    Sorry for being away so long ... how did you get on in the end?

    The way I'd suggest is as follows:
    Start with the standard formula for the energy levels for 1-electron and atomic number Z.

    The orbital energy levels are determined by the principle quantum umber n, but Z changes according to the shielding effect of the more tightly distributed, occupied, sub-shells.

    the 1s use the unmodified formula.
    for the 2s shell, reduce Z by 2
    for the 2p shell, reduce Z by 4
    ... that should give you something close, certainly E(2s)<E(2p). I have a feeling that the lower shell electrons do not completely shield "their" protons though.
     
  15. Jul 19, 2012 #14

    CAF123

    User Avatar
    Gold Member

    In the end, I emailed the lecturer himself to clarify things. I have not received any reply yet, i think, because he may be on holiday.

    What 'standard formula' do you refer to in your last post?
    (the only one electron system i have considered is the energy levels describing hydrogen i.e E = -13.6/n^2 eV and there is no Z present).

    What do you think of the resulting spectroscopic notation for the 1st excited state? Will it still be that corresponding to the lowest energy difference between the states? If so, how does reconcile with the fact that disturbing a full subshell (ie 2s to 2p) requires more energy? Just wondering because (possibly) this fact does not come across in the equation.?

    Many thanks.
     
  16. Jul 19, 2012 #15

    Simon Bridge

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    2016 Award

    En = -Z2*13.6 eV/n2
    It's just like the hydrogen one, but with a Z in it, in the links I showed you above ... like this:
    [tex] E_n = (-13.6eV)\frac{Z_{eff}^2}{n^2}[/text]

    http://electron6.phys.utk.edu/phys250/modules/module 3/Multi-electron atoms.htm
    ... discusses screening factors - scroll down - you shouldn't really screen the entire charge like I suggested so you may have notes for that. OR it may be you are supposed to realize that the model does not work well for nitrogen :) or maybe you are just expected to use the principle quantum numbers. However, nitrogen is the last of the "well behaved" atoms in terms of the approximate model.

    disturbing a full subshell "requires more energy" than what?
     
    Last edited: Jul 19, 2012
  17. Jul 19, 2012 #16

    CAF123

    User Avatar
    Gold Member

    ok, thanks for clarifying that.
    I know that if a subshell is full, then removal of an electron from that shell requires significantly more energy than if it was half full or partly full. This is used as a basis for explaining the ionisation energies of atoms.

    Using the equation suggested, I can approximate the energy gaps by taking into account the shielding effect, right? My question is, in using the equation all you do is change Z and n. So I am considering 2s to 2p and 2p to 3s. Now it is clear that 2s is only slightly lower in energy than 2p but 2p moving to 3s involves changing principle quantum number, so probably greater in energy (which I will verify by the suggested approximate calculation). I am therefore led to think that the transition 2s to 2p would be involved in the 1st excited state of nitrogen since this is the lowest energy.

    However, removal of an electron from 2s to 2p requires way more energy than from 2p to 3s, simply because 2s is filled. So how do we take this into account?

    Thanks again
     
  18. Jul 20, 2012 #17

    Simon Bridge

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    2016 Award

    May need clarification - you are saying that it takes more energy to singly ionize neutral Helium than it does to doubly ionize He+? Or that it takes more energy to ionize neutral He than it does to ionize hydrogen?

    That link I gave you explains the ionization energies (binding energy of the last electron anyway) up to nitrogen entirely by the approximate model provided.

    Additional effects are, off the top of my head: Lamb shift, spin-orbit coupling, and spin-spin coupling.
     
  19. Jul 20, 2012 #18

    CAF123

    User Avatar
    Gold Member

    In your example, I meant to say something like it takes more energy to ionise neutral helium than hydrogen.

    Is it possible to eject a 2s electron before the 3 2p electrons are ejected ?
     
  20. Jul 20, 2012 #19

    Simon Bridge

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    2016 Award

    That's accounted for in the approximate model provided.

    That would amount to an excited state of singly ionized nitrogen wouldn't it?
     
  21. Jul 20, 2012 #20

    CAF123

    User Avatar
    Gold Member

    I did the calculations and got [itex] E_(2s)= -85eV [/itex] and [itex] E_(2p) = -30.6 eV [/itex], confirming that indeed [itex] E_(2s) < E_(2p) [/itex].

    However, how do you find the energy of the transition [itex] 2p \rightarrow3s [/itex]? I think I can use the same energy above calculated for [itex] 2p [/itex] but I am unsure of how to calculate the energy of the 3s state since effectively all electrons screen the nucleus.
    What I did was try to imagine the electron already there and so only 6 electrons would shield the nucleus. Using this, I get [itex] E_(3s) = -1.51 eV [/itex].

    However, now I have another problem. The transition [itex] 2s \rightarrow2p [/itex] has energy gap [itex] 54.4 eV [/itex] while that for [itex] 2p \rightarrow3s [/itex]is [itex] 1.51 eV [/itex] so this leads to me to think that transition 2p to 3s is the lowest energy transition. ( in case of any confusion I got these values by subtracting the energies of the two states concerned and this gave me the difference). This results goes against my prediction and the chart showing the progressions in energy of the states I have seen in my notes.

    Can you assist ? Many thanks again.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook