# Homework Help: How do you calculate the magnetic moment of the ground state of Calciu

1. Dec 5, 2013

### zhillyz

1. The problem statement, all variables and given/known data

Calculate the magnetic moment of the ground state of $\,_{20}^{39}Ca$. Compare to the experimental value in table 1.

2. Relevant equations

Nuclear Shell Model knowledge

3. The attempt at a solution

Well firstly the magnetic moment of the nucleus similar to the spin is based only on the unpaired nucleon. With the above having one unpaired neutron in the 1d3/2 giving it spin 3/2 and parity +.

I thought the formulae for magnetic moment was:

$\mu_{n} = \dfrac{g_{n}\mu_{N}\vec{s}}{\hbar}$

but that would only involve googling the g value and multiplying it by 3/2 I think which would not get me all the marks this question is worth. Please give me a point in the right direction.

Last edited: Dec 5, 2013
2. Dec 7, 2013

### dmacbeath

I thought the subshell it would be in would be 0d3/2

3. Dec 7, 2013

### zhillyz

Yes sorry the subshell is 0d3/2 also its total angular momentum is 3/2 not its spin which I said before. I continued studying on my own when I didnt get an answer and in case anyone is interested the answer (I think) would be to use this formula (derived from the first equation I mentioned taking into account that s and l are not exactly defined in a system where j is, so replacing all terms in quantities related to j using vectors and expectation values);

$\mu_{n} = \left[g_{l}\dfrac{j(j+\dfrac{1}{2})}{(j+1)}-\dfrac{1}{2}\dfrac{1}{j+1}g_{s}\right]\mu_{N}$

4. Dec 8, 2013

### dmacbeath

for the magnetic moment i got μn=3.865x10^-27 J/T
where gl=0 and gs=-3.8261

5. Dec 8, 2013

### zhillyz

That is what I got if you convert to those units. Though the comparison against the experimental value in table 1 is unitless i.e. magnetic moment/nuclear magneton and the equation above gives you an answer in units of nuclear magneton(if you dont put in the number of mu_N and just leave it as mu_N). So its just that number(0.76526)