What are the quantum numbers (n, L, J)?

[unscientific]

Homework Statement

[/B]
I'm supposed to find the quantum numbers (n, L, J) for the first 3 energy levels in Iridium (Z=77), given that the first 4 ionization energies are $76.1, 13.4, 12.8, 11.2 keV$.

Homework Equations

The Attempt at a Solution

I know that the electronic configuration is $[Xe] 4f^{14} 5d^7 6s^2$.
I suppose lowest level is $n=6, l=0, S=0,J=0$.
Is the next level $6s^1 6p^1$?

 

[ArmanCham]

I want to help you but I didnt understand your question.Example you said "first 3 energy levels" you know first energy level is n=1 l=0 and J=0 so can you be more spesific.I might be understand your question false too.

 

[unscientific]

I want to help you but I didnt understand your question.Example you said "first 3 energy levels" you know first energy level is n=1 l=0 and J=0 so can you be more spesific.I might be understand your question false too.

How can lowest energy level be 1 when the outermost orbital is at $6s^2$?

 

[ArmanCham]

First three energy means n=1 n=2 and n=3 but lowest energy is $6s^2$

 

[unscientific]

First three energy means n=1 n=2 and n=3 but lowest energy is $6s^2$

Ok, so n = 1,2,3,4. What are the L and J for each?

 

[ArmanCham]

Here you can find everything

 

[unscientific]

Here you can find everything

Thanks for your input, but you just said first 3 levels are n=1,2,3 and that directly contradicts what's in the picture.

 

[ArmanCham]

Yes I didnt understand your question and I assumed this

 

[blue_leaf77]

I think you should check again the problem, you said first 4 ionization energies but the values you gave stands in keV range which doesn't make sense for large atom like Ir, plus they are in the order of lowering values, which should be the other way around. There is a good discussion about many electron atoms in Physics of Atoms and Molecules by Bransden and Joachain, you might want to check this out.

 

[unscientific]

I think you should check again the problem, you said first 4 ionization energies but the values you gave stands in keV range which doesn't make sense for large atom like Ir, plus they are in the order of lowering values, which should be the other way around. There is a good discussion about many electron atoms in Physics of Atoms and Molecules by Bransden and Joachain, you might want to check this out.

The question is here:

 

[unscientific]

bumpp

 

[ArmanCham]

I want to help.Is there any equation between energy level and n I found this

but I am not sure

 

[blue_leaf77]

I have found the answer to your question in Google.com.

 

[ArmanCham]

I have found the answer to your question in Google.com.

then post the url

 

[blue_leaf77]

The forum rule forbids helpers to give directly the final answer. Very unfortunate that I can't help with physical arguments.

 

[ArmanCham]

Send me a message I can help him maybe

 

[ArmanCham]

Electron binding energy means electron ionization energy.Now there wrote most tightly,it means we have to give a lot of energy to ionize electron.Think this way If Ir has one electron then we should use this formula

to find binding energy of electron.Put the values and you will find n there.Find it and tell me the answer then I will explain the other energies and you will understand the question.

 

[unscientific]

Electron binding energy means electron ionization energy.Now there wrote most tightly,it means we have to give a lot of energy to ionize electron.Think this way If Ir has one electron then we should use this formula

to find binding energy of electron.Put the values and you will find n there.Find it and tell me the answer then I will explain the other energies and you will understand the question.

We can't assume that, as each level has a shielding effect such that $E_n = -hcR_{\infty}\frac{(Z-\sigma_n)^2}{n^2}$.

 

[unscientific]

Ok, I think I got it. Are the first 4 energy levels $1s^2, 2s^2, 3s^2, 4s^2$?

 

[blue_leaf77]

For such heavy atom, I suspect the relativistic effect which manifests itself in the coupling between orbital and spin angular momenta will be too big to be neglected, due to which the J quantum number may have to appear in certain form in the notation of the states.

 

[unscientific]

For such heavy atom, I suspect the relativistic effect which manifests itself in the coupling between orbital and spin angular momenta will be too big to be neglected, due to which the J quantum number may have to appear in certain form in the notation of the states.

Ok, but $n$ is defined as the number beside $1s,2s,3s$ right? So in that case the first 4 levels are $1s^2, 1s1p, 2s^2, 2s2p, 3s^2$? So energy levels are $n=1,1,2,2,3$.

 

[blue_leaf77]

where does 1p come from?

 

[unscientific]

where does 1p come from?

Oops. Then $1s^2, 2s^2, 2s2p, 3s^2, 3s3p$?

 

[blue_leaf77]

Alright, upon some thorough searches I can finally come up with a conclusion about the commonly used notation in the electronics structure to denote the electron binding energy. I will just give you the hint, the ordering of the electron binding energies follows exactly from one electron atom taking into account fine structure effect. Are you familiar with fine structure of hydrogenic atoms? Unfortunately I cannot figure out why spectroscopists defined the notation this way.