Possible Atomic Levels: How Can J Be Equal to 1?

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SUMMARY

The discussion centers on determining possible atomic levels for the excited state of Helium with the electronic configuration (1s, 2p). The four identified levels are: 1) S=0, L=1, J=1; 2) S=1, L=1, J=2; 3) S=1, L=1, J=0; and 4) S=1, L=1, J=1. The confusion arises regarding the fourth level, where J can equal 1. This is explained by the addition of angular momenta, where the total angular momentum J can take values from |j2-j1| to j1+j2 in increments of 1, resulting in possible values of J being 0, 1, and 2.

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octol
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Hello all,

I wonder if anyone can help me with a problem I don't understand about atomic levels. Given an excited state of He with electronic config (1s,2p), I need to find all possible atomic levels.

The answer sheet lists 4 levels:

1) S=0, L=1, J=1
2) S=1, L=1, J=1+1=2
3) S=1, L=1, J=1-1=0

These first three I understand, but the fourth:

4) S=1, L=1, J=1 (?)

How come J can be equal to 1? Anyone care to explain or point me in the right direction, please?
 
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octol said:
Hello all,

I wonder if anyone can help me with a problem I don't understand about atomic levels. Given an excited state of He with electronic config (1s,2p), I need to find all possible atomic levels.

The answer sheet lists 4 levels:

1) S=0, L=1, J=1
2) S=1, L=1, J=1+1=2
3) S=1, L=1, J=1-1=0

These first three I understand, but the fourth:

4) S=1, L=1, J=1 (?)

How come J can be equal to 1? Anyone care to explain or point me in the right direction, please?
when adding two angular momenta j_1 and j_2 (which can be spin or orbita), the total angular momentum ranges from |j2-j1| to j_1+j_2 in increments of 1. So if you are adding a spin of 1 to an orbital angular momentum L of 1, the total angular momentum goes from |1-1| up to 1+1 in steps of 1, so the possible J are 0,1 and 2 (all values from 0 to 2 in steps of 1).

If you were adding a spin of 3/2 to an orbital angular momentum of 2, say, the possible J would go from 1/2 to 7/2 in steps of 1, so 1/2, 3/2,5/2 and 7/2.
 
Thanks for the help!
 

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