## Saturday, July 24, 2010

### More Quantum Problems  We left off in the previous blog with a problem on multiple L-S coupling. The solution is straightforward.

s' = s1 + s2 = 1/2 + 1/2 = 1

l' = l1 + l2 = 2 + 3 = 5

Then: j' = l' + s' = 1 + 5 = 6

So: s' = 1, l' = 5 and j' = 6

This is the maximum for the combination of letters. The minimum is then calculated from:

s' = 1/2 - 1/2 = 0

and:

l' = 3 - 2 = 1

so: j' = l' + s' = 1 + 0 = 1

Then: s = 0, l' = 1 and j' = 1

all the other intermediary values can easily be computed in the same way and these are shown in the table of Fig. 1.

Now, for a more complex problem which gets into the splitting of spectral lines, according to L-S coupling separations. We want to look at the 4s 3d atomic configuration and use that information to do a spectral configuration diagram.

For 4s 3d we have:

l1 = 2, s1 = 1/2, l1 = 0 and s1 = 1/2. Then for the maximum value: l1 + l2 = 2 + 0 = 2

and: s= = s1 + s2 = 1/2 + 1/2 = 1

The LOWEST energy level is then: 4s 3d (3D1) (Since 2s' + 1 = 3, leading to minimum j' = [s' -l'] = 1. )

Using the assorted combinations, for l'= 0 and l' = 2, to get the respective j' values (in combination with s' = 0) and then further for s' = 1, we arrive at the configuration diagram shown in Fig. 2.)

Readers should be able - if they've followed all the quantum blogs so far- to work through all the combinations and verify this result!)

For next time:

Draw a schematic energy diagram for the 2p 3s configuration for the carbon atom (12C 6) and lable each level with spectroscopic notation.