What is the spin quantum number of an atom in 4P state?

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SUMMARY

In a 4P state, the spin quantum number is definitively s=3/2, derived from the term symbol notation in Russell-Saunders (LS) coupling, where 2S+1=4 indicates S=3/2. The angular momentum quantum number l is established as 1, corresponding to the P state. The term symbol construction follows the format 2S+1LJ, with L representing the total orbital angular momentum. The notation used is a legacy from early atomic physics, aligning with the nomenclature for orbitals.

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shedrick94
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If an atom described by spin-orbit coupling is in a 4P state, why is the spin quantum number s=3/2?
 
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I guess you mean 4P. How is a term symbol built up?
 
That's the terminology in the question yes. I don't quite understand what 4P means though? What do you mean by term symbol?
 
The description of the angular momentum of a state is called a term symbol. In Russell-Saunders (LS) coupling, it has the form 2S+1LJ. In your particular case, the value of J has been omitted.
 
So the angular momentum quantum number is l=1 as it's a P state? How is the value of s determined?
 
shedrick94 said:
So the angular momentum quantum number is l=1 as it's a P state? How is the value of s determined?
The exponent is 2S+1, so for a quadruplet, 2S+1 = 4 means S=3/2.
 
Sorry I've never seen this notation before so I wasn't aware of the relationship. So why do we have L=P then?
 
shedrick94 said:
Sorry I've never seen this notation before so I wasn't aware of the relationship. So why do we have L=P then?
It's a legacy from the earlier days of atomic physics. It follows the same nomenclature as orbitals: s, p, d, f, etc., but with uppercase letters (just as the total orbital angular momentum is L, instead of l for the orbital angular momentum of a single electron).
 

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