- #1
jeebs
- 325
- 4
Hi,
I have a couple of questions about the spin of an atomic nucleus.
Say I had a carbon-12 nucleus, which is 6 protons and 6 neutrons. The Pauli exclusion principle tells us that these should be arranged in shells determined by their orbital angular momenta and spins.
So, the shell order goes 1s, 1p, 1d, 2s, 1f, 2p... and the s shell has orbital quantum number l=0, the p shell l=1, and so on, like electrons in atoms, right?
In the s-shell, there are 2 possible states, one for spin up, one for spin down.
In the p-shell, there are 2(2l + 1) states = 2(2 + 1) = 2(3) = 6 available states.
Also, the shells have a spin-orbit interaction, ie. j = l + s.
This gives us the possible values of the quantum number j of |l+s| to |l-s|
ie. j = 1+(1/2) to 1-(1/2)
= 3/2 and 1/2
Now in the j=3/2 level, there are 4 mj states available: -3/2, -1/2, +1/2 and +3/2
and in the j=1/2 level, there are 2 mj states: -1/2 and +1/2.
So, in this nucleus we have 6 protons. They fill up the shell levels from the lowest states, so the 1s shell gets filled, and from what I can gather, the 1p3/2 level gets filled up.
My first question is, why does this level get filled up first and not the j=1/2 level? Am I right in thinking that the 3/2 level corresponds to the orbital and spin angular momentum vectors adding in the same direction, anf the 1/2 level corresponds to the spin and orbital angular momentum vectors opposing each other?
If so, shouldn't the 1/2 level be at a lower energy and therefore filled up first, seeing as if there is less total angular momentum the nucleus should be in a less excited state?
My other question is, what is the total spin of this carbon-12 nucleus?
Do the spins of each successive nucleon added to the nucleus go in alternating up, down, up, down order, meaning that for an even number of protons they would contribute no spin angular moment to the overall nuclear spin (with the same being true for neutrons)?
Thanks.
I have a couple of questions about the spin of an atomic nucleus.
Say I had a carbon-12 nucleus, which is 6 protons and 6 neutrons. The Pauli exclusion principle tells us that these should be arranged in shells determined by their orbital angular momenta and spins.
So, the shell order goes 1s, 1p, 1d, 2s, 1f, 2p... and the s shell has orbital quantum number l=0, the p shell l=1, and so on, like electrons in atoms, right?
In the s-shell, there are 2 possible states, one for spin up, one for spin down.
In the p-shell, there are 2(2l + 1) states = 2(2 + 1) = 2(3) = 6 available states.
Also, the shells have a spin-orbit interaction, ie. j = l + s.
This gives us the possible values of the quantum number j of |l+s| to |l-s|
ie. j = 1+(1/2) to 1-(1/2)
= 3/2 and 1/2
Now in the j=3/2 level, there are 4 mj states available: -3/2, -1/2, +1/2 and +3/2
and in the j=1/2 level, there are 2 mj states: -1/2 and +1/2.
So, in this nucleus we have 6 protons. They fill up the shell levels from the lowest states, so the 1s shell gets filled, and from what I can gather, the 1p3/2 level gets filled up.
My first question is, why does this level get filled up first and not the j=1/2 level? Am I right in thinking that the 3/2 level corresponds to the orbital and spin angular momentum vectors adding in the same direction, anf the 1/2 level corresponds to the spin and orbital angular momentum vectors opposing each other?
If so, shouldn't the 1/2 level be at a lower energy and therefore filled up first, seeing as if there is less total angular momentum the nucleus should be in a less excited state?
My other question is, what is the total spin of this carbon-12 nucleus?
Do the spins of each successive nucleon added to the nucleus go in alternating up, down, up, down order, meaning that for an even number of protons they would contribute no spin angular moment to the overall nuclear spin (with the same being true for neutrons)?
Thanks.