Energy differences between sub shells

[CAF123]

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.

 

[Simon Bridge]

Why not just calculate it?

 

[CAF123]

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.

 

[Simon Bridge]

http://hyperphysics.phy-astr.gsu.edu/hbase/quantum/qnenergy.html

Calculating the energy gap is a matter of working out the n-state energies and applying the level-splitting. With multiple electrons it can get quite complex. What level are you trying to do this at?

 

[CAF123]

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?

 

[Simon Bridge]

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 Schrödinger 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.

 

[CAF123]

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.

 

[Simon Bridge]

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 :(

 

[mikeph]

I must be missing something. Aren't the 2p/2s states degenerate?

 

[Boston_Guy]

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.

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.

 

[Simon Bridge]

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...

 

[mikeph]

Ah. I haven't been doing the heavier elements.

 

[Simon Bridge]

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.

 

[CAF123]

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.

 

[Simon Bridge]

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:
$$E_n = (-13.6eV)\frac{Z_{eff}^2}{n^2}[/text]<br /> <br /> <a href="http://electron6.phys.utk.edu/phys250/modules/module%203/Multi-electron%20atoms.htm" target="_blank" class="link link--external" rel="nofollow ugc noopener">http://electron6.phys.utk.edu/phys250/modules/module 3/Multi-electron atoms.htm</a><br /> ... discusses screening factors - scroll down - you shouldn't really screen the entire charge like I suggested so you may have notes for that. OR it may be you are supposed to realize that the model does not work well for nitrogen :) or maybe you are just expected to use the principle quantum numbers. However, nitrogen is the last of the "well behaved" atoms in terms of the approximate model.<br /> <br /> disturbing a full subshell "requires more energy" than what?$$

 

[CAF123]

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

 

[Simon Bridge]

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.

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?

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?

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.

 

[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]

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

That's accounted for in the approximate model provided.

Is it possible to eject a 2s electron before the 3 2p electrons are ejected ?

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

 

[CAF123]

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

However, how do you find the energy of the transition $2p \rightarrow3s$? I think I can use the same energy above calculated for $2p$ 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 $E_(3s) = -1.51 eV$.

However, now I have another problem. The transition $2s \rightarrow2p$ has energy gap $54.4 eV$ while that for $2p \rightarrow3s$is $1.51 eV$ 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 $2p →3s$ transition corresponding to the lowest energy?
Is it simply that removal of an electron from $2s →2p$ 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 $2s$ electron from an atom before you remove a $2p$ electron, right? And once this is done, a $2p$ 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 $3s$ state. Doing it so only full subshells screen the nucleus results in this having the same energy as $2p$, 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.