- #1

JD_PM

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- Homework Statement
- Compute the magnetic moment of 57-Ni and compare it with its experimental value of ##-0.8 \mu_N##. Note that I have computed two extra magnetic moments for odd-even nucleus since I wanted to understand why I was getting a wrong answer (I also got those wrong). This is extra work, since the original question is about ##\mu## of 57-Ni. I want to understand what I am doing wrong.

- Relevant Equations
- ##\mu_J=g_J\times j\times \mu_N##

I am having difficulties computing the magnetic moment for an even-odd (proton-neutron) nucleus.

The formula is:

$$\mu_J=g_J\times j\times \mu_N$$

I checked this helpful post: https://physics.stackexchange.com/q...ic-moments-how-does-one-decide-between-l-frac

Based on the shell model we see that only the unpaired neutron contributes to a non-zero magnetic moment; all neutron shells can be filled up to ##2p_{3/2}##. Thus the quantum numbers are:

$$j=\frac{3}{2}, s=\frac{1}{2}, l=1.$$

Thus we know that ##j=l+\frac{1}{2}##, so to calculate ##g_J##:

$$g_J=\Big(1+\frac{1}{2j}\Big)g_l+\frac{1}{2j}g_s$$

For neutrons: g_l = 0 and g_s =-3.8260837.

Knowing that we get for 57-Ni case:

$$g_J = -1.275$$

Thus:

$$\mu_J = -1.913 \mu_{N}$$

I checked more similar examples:

a) 87-Sr (38 protons). ##J^\pi=\frac{9}{2}^+## and ##j=l+\frac{1}{2}## Applying the same method I get:

$$\mu_J = -1.913 \mu_{N}$$

Experimental value: ##\mu_J = -1.093 \mu_{N}##.

Again a significant difference; something is wrong.

b) 91-Zr (40 protons). ##J^\pi=\frac{5}{2}^+## and ##j=l+\frac{1}{2}## Applying the same method I get:

$$\mu_J = -1.913 \mu_{N}$$

Experimental value: ##\mu_J = -1.304 \mu_{N}##.

Again a significant difference; something is wrong.

1) Get ##j## based on the shell model (note that in some cases there are exceptions, but we are not concerned with that in this post).

2) Get ##g_J##.

2.1)If ##j=l+\frac{1}{2}## meets your case then use:

$$g_J=\Big(1+\frac{1}{2j}\Big)g_l+\frac{1}{2j}g_s$$

Finally calculate the magnetic moment:

$$\mu_J=g_J\times j\times \mu_N$$

2.2)If ##j=l-\frac{1}{2}## meets your case then use:

$$g_J=\frac{1}{j+1}\Big[\Big(j+\frac{3}{2}\Big)g_l-\frac{1}{2}g_s\Big]$$

Finally calculate the magnetic moment:

$$\mu_J=g_J\times j\times \mu_N$$

What's wrong with it?

Thanks.

PS: Let me know if something needs to be added and I will do it.

I've asked this question at PSE as well, but it's not catching a lot of attention.

https://physics.stackexchange.com/questions/496458/method-for-calculating-nuclear-magnetic-moments

The formula is:

$$\mu_J=g_J\times j\times \mu_N$$

I checked this helpful post: https://physics.stackexchange.com/q...ic-moments-how-does-one-decide-between-l-frac

**I worked out the magnetic moment for 57-Ni.**Based on the shell model we see that only the unpaired neutron contributes to a non-zero magnetic moment; all neutron shells can be filled up to ##2p_{3/2}##. Thus the quantum numbers are:

$$j=\frac{3}{2}, s=\frac{1}{2}, l=1.$$

Thus we know that ##j=l+\frac{1}{2}##, so to calculate ##g_J##:

$$g_J=\Big(1+\frac{1}{2j}\Big)g_l+\frac{1}{2j}g_s$$

For neutrons: g_l = 0 and g_s =-3.8260837.

Knowing that we get for 57-Ni case:

$$g_J = -1.275$$

Thus:

$$\mu_J = -1.913 \mu_{N}$$

**This result is wrong**. It is far from its experimental value; ##\mu_J = -0.8 \mu_{N}##. I have read (Krane page 126-127) that it is acceptable to get a theoretical value slightly different from the experimental one. That is not the case here of course.I checked more similar examples:

a) 87-Sr (38 protons). ##J^\pi=\frac{9}{2}^+## and ##j=l+\frac{1}{2}## Applying the same method I get:

$$\mu_J = -1.913 \mu_{N}$$

Experimental value: ##\mu_J = -1.093 \mu_{N}##.

Again a significant difference; something is wrong.

b) 91-Zr (40 protons). ##J^\pi=\frac{5}{2}^+## and ##j=l+\frac{1}{2}## Applying the same method I get:

$$\mu_J = -1.913 \mu_{N}$$

Experimental value: ##\mu_J = -1.304 \mu_{N}##.

Again a significant difference; something is wrong.

**Note that in all three cases I get the same mistaken theoretical value. There must be something I am missing.**

To recap, this is the method that I've used:To recap, this is the method that I've used:

1) Get ##j## based on the shell model (note that in some cases there are exceptions, but we are not concerned with that in this post).

2) Get ##g_J##.

2.1)If ##j=l+\frac{1}{2}## meets your case then use:

$$g_J=\Big(1+\frac{1}{2j}\Big)g_l+\frac{1}{2j}g_s$$

Finally calculate the magnetic moment:

$$\mu_J=g_J\times j\times \mu_N$$

2.2)If ##j=l-\frac{1}{2}## meets your case then use:

$$g_J=\frac{1}{j+1}\Big[\Big(j+\frac{3}{2}\Big)g_l-\frac{1}{2}g_s\Big]$$

Finally calculate the magnetic moment:

$$\mu_J=g_J\times j\times \mu_N$$

What's wrong with it?

Thanks.

PS: Let me know if something needs to be added and I will do it.

I've asked this question at PSE as well, but it's not catching a lot of attention.

https://physics.stackexchange.com/questions/496458/method-for-calculating-nuclear-magnetic-moments

Last edited: