How do you calculate the magnetic moment of the ground state of Calciu

AI Thread Summary
The magnetic moment of the ground state of calcium-39 is primarily determined by its unpaired neutron in the 1d3/2 subshell, which has a total angular momentum of 3/2. The formula for calculating the magnetic moment involves using the g factors for the orbital and spin contributions, leading to a calculated value of μn = 3.865 x 10^-27 J/T. This value can be converted to a unitless form for comparison with experimental data, resulting in a dimensionless magnetic moment of approximately 0.76526 when expressed in terms of the nuclear magneton. The discussion emphasizes the importance of correctly applying the nuclear shell model and understanding the contributions of angular momentum and spin. Overall, the calculated magnetic moment aligns closely with the experimental value, confirming the theoretical approach.
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Homework Statement



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

Homework Equations



Nuclear Shell Model knowledge

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.
 
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I thought the subshell it would be in would be 0d3/2
 
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}
 
for the magnetic moment i got μn=3.865x10^-27 J/T
where gl=0 and gs=-3.8261
 
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 don't put in the number of mu_N and just leave it as mu_N). So its just that number(0.76526)
 
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