Nuclear Magnetic Moments - Oxygen 15

BOAS
Messages
546
Reaction score
19

Homework Statement


I am trying to calculate the nuclear magnetic moment for Oxygen 15 but my answer differs strongly from the value quoted in literature.

Homework Equations

The Attempt at a Solution


[/B]
##\mu_J = g_J \mu_N J##

This isotope of Oxygen has an unpaired Neutron with ##J = \frac{1}{2}##.

Calculating the nuclear g-factor for ##J = l + \frac{1}{2}##

##g_J = g_l (1 - \frac{1}{2J}) + g_s(\frac{1}{2J})##

For a neutron ##g_l = 0## and so in this case ##g_J = g_s = -3.82##

I therefore find that ##\mu_J = -1.91 \mu_N## which is significantly different from the value I find in the NIST table of ##\pm 0.7189 \mu_N##

http://nist.gov/data/PDFfiles/jpcrd85.pdf

Is my treatment too simplistic to get a decent prediction, or am I making a fundamental error?

Thanks for reading!
 
Physics news on Phys.org
BOAS said:

Homework Statement


I am trying to calculate the nuclear magnetic moment for Oxygen 15 but my answer differs strongly from the value quoted in literature.

Homework Equations

The Attempt at a Solution


[/B]
##\mu_J = g_J \mu_N J##

This isotope of Oxygen has an unpaired Neutron with ##J = \frac{1}{2}##.

Calculating the nuclear g-factor for ##J = l + \frac{1}{2}##

##g_J = g_l (1 - \frac{1}{2J}) + g_s(\frac{1}{2J})##

For a neutron ##g_l = 0## and so in this case ##g_J = g_s = -3.82##

I therefore find that ##\mu_J = -1.91 \mu_N## which is significantly different from the value I find in the NIST table of ##\pm 0.7189 \mu_N##

http://nist.gov/data/PDFfiles/jpcrd85.pdf

Is my treatment too simplistic to get a decent prediction, or am I making a fundamental error?

Thanks for reading!

I think I have found my mistake.

My case is for J = L - S, with L = 1.

My g-factor is therefore ##g_J = g_l (1 + \frac{1}{2l + 1}) - g_s (\frac{1}{2l + 1})##, with ##g_l = 0##

I find that ##g_J = -0.63 \mu_N## which is much closer to my expected value.

Sorry for posting prematurely.
 
Thread 'Need help understanding this figure on energy levels'
This figure is from "Introduction to Quantum Mechanics" by Griffiths (3rd edition). It is available to download. It is from page 142. I am hoping the usual people on this site will give me a hand understanding what is going on in the figure. After the equation (4.50) it says "It is customary to introduce the principal quantum number, ##n##, which simply orders the allowed energies, starting with 1 for the ground state. (see the figure)" I still don't understand the figure :( Here is...
Thread 'Understanding how to "tack on" the time wiggle factor'
The last problem I posted on QM made it into advanced homework help, that is why I am putting it here. I am sorry for any hassle imposed on the moderators by myself. Part (a) is quite easy. We get $$\sigma_1 = 2\lambda, \mathbf{v}_1 = \begin{pmatrix} 0 \\ 0 \\ 1 \end{pmatrix} \sigma_2 = \lambda, \mathbf{v}_2 = \begin{pmatrix} 1/\sqrt{2} \\ 1/\sqrt{2} \\ 0 \end{pmatrix} \sigma_3 = -\lambda, \mathbf{v}_3 = \begin{pmatrix} 1/\sqrt{2} \\ -1/\sqrt{2} \\ 0 \end{pmatrix} $$ There are two ways...
Back
Top