Using nuclear shell model to determine parity and spin

[Flucky]

Homework Statement

Use the shell model to determine the parity and spin assignments for all the stable isotopes of calcium.

Homework Equations

n/a

The Attempt at a Solution

The stable isotopes seem to be ⁴⁰Ca, ⁴²Ca, ⁴³Ca, ⁴⁴Ca, ⁴⁶Ca, and ⁴⁸Ca

I believe all of those isotopes except ⁴³Ca are even-even which means that they have spin=0 and parity=even, however I'm a bit stuck on how to work out the spin and parity of ⁴³Ca.

There are 23 neutrons, so 20 of those fill up the shells to the magic number (20) and we have 3 left over.

How do I work out the orbital angular momentum l and j-value of those?

With these values I could work out the spin of the unpaired nucleon (which would be the j-value) and its parity (-1)^l

 

[e.bar.goum]

I believe all of those isotopes except ⁴³Ca are even-even which means that they have spin=0 and parity=even

Yes.

There are 23 neutrons, so 20 of those fill up the shells to the magic number (20) and we have 3 left over.

How do I work out the orbital angular momentum l and j-value of those?

With these values I could work out the spin of the unpaired nucleon (which would be the j-value) and its parity (-1)^l

You're almost there - you clearly know that paired nucleons will couple to 0⁺ in the lowest energy, since you know what the GS spin and parity is for even-even nuclei. Then, you know that you only have one neutron left over that you need to worry about. Now, to work out the spin and parity, you need to know what shell it will sit in.

I don't know where in learning about the shell model you are. Have you come across this http://upload.wikimedia.org/wikipedia/commons/thumb/e/e5/Shells.png/220px-Shells.png sort of diagram? Or have you gotten to the Nilsson diagram? http://ie.lbl.gov/toipdf/nilsson.pdf. As 43Ca will be very close to spherical, either will work to answer the question.

 

[Flucky]

I don't know where in learning about the shell model you are. Have you come across this http://upload.wikimedia.org/wikipedia/commons/thumb/e/e5/Shells.png/220px-Shells.png sort of diagram? Or have you gotten to the Nilsson diagram? http://ie.lbl.gov/toipdf/nilsson.pdf. As 43Ca will be very close to spherical, either will work to answer the question.

I'm pretty sure we'd only be working with the first one, the Nilsson diagram looks more advanced.

Going off the first diagram it looks like our lone neutron would be in the 1f_\frac{7}{2} shell where nuclear spin j = \frac{7}{2} and parity = (-1)³ [to the power 3 because it is in the f shell] = -1 = odd?

 

[e.bar.goum]

I'm pretty sure we'd only be working with the first one, the Nilsson diagram looks more advanced.

The Nilsson diagram becomes important if you want to talk about deformed nuclei. If you look at the center line of the Nilsson diagram, you'll see that it's the same as the first diagram.

Going off the first diagram it looks like our lone neutron would be in the 1f_\frac{7}{2} shell where nuclear spin j = \frac{7}{2} and parity = (-1)³ [to the power 3 because it is in the f shell] = -1 = odd?

$\frac{7}{2}^-$ -- You've got it! You can always check your answer by looking at the data for nuclei. Here's a good place to look. http://www.nndc.bnl.gov/chart/chartNuc.jsp

 

[Flucky]

$\frac{7}{2}^-$ -- You've got it! You can always check your answer by looking at the data for nuclei. Here's a good place to look. http://www.nndc.bnl.gov/chart/chartNuc.jsp

Great, thanks for the help.