Finding components of 2s2p configuration using LS coupling

  • Thread starter omicgavp
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  • #1
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Homework Statement


The lowest excited states of the Be atom correspond to a 2s2p configuration of optically active electrons. What 2s+1Lj components originate in this configuration? Assume that Hund's rules hold for the multiplet and deduce the ordering of the energy levels. Refer to a table of beryllium levels to see if the ordering occurs as predicted.


Homework Equations


l=|l1-l2|,|l1-l2|+1,...,l1+l2-1,l1+l2


j=|l-s|,|l-s|+1,...,l+s-1,l+s

The Attempt at a Solution


spin-orbital assignments of 2s2p configuration:
(l1=0,s1=1/2) and (l2=1,s2=1/2)

So that,
total spin states: s=0 and 1
and
l=0,1, and 2
...............................???
I don't know what to do next.

Answer at the back of the book:
3P0,1,2,1P1
 

Answers and Replies

  • #2
30
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You pretty much have it. Except, I don't believe you calculated the possible l values properly.

|1-0|=1, so only l=1 works.

Now, all you have to do is use Hund's rules to find the order.

This will lead to the answer in the book.
 
Last edited:

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