Energy differences between sub shells

CAF123

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 ?

Simon Bridge

That's accounted for in the approximate model provided.

That would amount to an excited state of singly ionized nitrogen wouldn't it?

CAF123

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.

Simon Bridge

Yeah - the usual approach is to treat single electron excitations like hydrogen - screen all but one proton. It's not great - nitrogen is nowhere near hydrogenic. Also - IRL you don't get total screening like it sounds like you've been doing. So look closer: in what way does the progression differ from the one you expected?

You could try working out the energy levels assuming only the full subshells screen the nucleus. (Only the full shells have the spherical symmetry.) Should give you tighter binding... which should reinforce your result that the 2p -> 3s transition has lower energy. (It will also give you the 3p states with same energy as the 3s states - this is because the splitting is due to filled s states and they are empty.)

Another approach is to follow the link's examples and work out the actual screening factor from the ionization energy and apply that for the n > 2 energy levels.

However - if you google for "electron configuration of 1st excited state of nitrogen" you get a whole lot of people agreeing that it is 1s², 2s², 2p², 3s¹ ...

So I think at this stage we need to focus on the specifics of the problem at hand: is this something you are doing for your own interest or is it a set question?

CAF123

Many thanks again for your reply,
I am on holiday right now but will try your suggested changes when I return.

For the moment though, I notice you say the 2p to 3s level corresponds to the lowest energy. (As I also got when I did my calculations)
How does this fit in with the chart which shows progressive energies of subshells?
(I have seen this chart on my notes but can´t find it on wikipedia yet - what it shows is the 2p level only slightly greater in energy than the 2s, whereas the gap between 2p and 3s is a lot bigger).

Simon Bridge

The charts are usually generalized sketches. You'd have to find one which specifically plots the energy levels for Nitrogen for a good comparison. The heavier an element is the less orderly the sequence of subshells ... look at the valence shells for platinum.

CAF123

It turns out the calculations I did did only take into account full subshells that screen the nucleus.
Can I ask if my numbers are right in my calculations?

What is the physical reasoning for the [itex] 2p →3s [/itex] transition corresponding to the lowest energy?
Is it simply that removal of an electron from [itex] 2s →2p [/itex] requires disturbing a full subshell?

One more question: I asked you this earlier in this thread, but just to make things clearer.
It is possible to remove a [itex] 2s [/itex] electron from an atom before you remove a [itex] 2p [/itex] electron, right? And once this is done, a [itex] 2p [/itex] electron moves down to fill the 'gap' - which corresponds to some excited state of the atom?

CAF123

Any ideas at all?

Also, I am a little bit unsure of how to calculate the energy of the [itex] 3s [/itex] state. Doing it so only full subshells screen the nucleus results in this having the same energy as [itex] 2p[/itex], which when we find out the energy difference is 0? Clearly, I have done something wrong here, but I am unsure of what to try next?

By the way, this question stemmed purely from my interest and is not a set question or anything, so any ideas/ rough approximations to yield an estimated answer are fine.
Thanks.

Simon Bridge

I think if you ionized an atom by removing a non-valence electron you'd see an absorption line of corresponding energy and you don't (afaik). But the configuration that leaves a hole in the 2s shell is that of an excited state of the ion.

For the calculation - you should have another look at the link earlier.