[tisu85]

i have a homework problem that asks me to calculate the magnetic dipole moment for 37Cl with J= l - 1/2
I know that in order to find the moment, i need to find J
However , i don't know how to find l for a certain nucleus
Can you guys help me with this??

[Mitch]

Use the shell model. Does ³⁷Cl have an odd proton or an odd neutron.

[ethylenediamine]

I tried the calculation for 137-Cesium, but I used this equation:

[J/(J+1)] [(l+1)g_l -0.5g_s],                     where g_l = l u_0, and g_s = 5.5845 u_0 (protons)

I find that l=4, because the last proton is in the g 7/2 state.

I would then get, [(7/2)/(9/2)][5*4-0.5*5.5845]u_0 = 13.3 u_0, but I thought the answer was 1.8 u_0. Am I wrong.  How come when I look online, at:
http://www.webelements.com/webelements/elements/text/Cs/radio.html

I see 2.84 as the listed value? Are they assuming J = l+1/2, even though I am calculating for J=l-1/2?

[Mitch]

If you use g_l = 1 u₀ (protons), you seem to get the right answer so try that.

[ethylenediamine]

But g_l = l u₀ , as in "l" the alphabet not "1" Wouldn't that mean we should use the angular momentum value: 4???

[Mitch]

Just try it, see what you get for the values.

[tisu85]

so , is it g_l always equal to 1 times u₀ (nuclear magneton)
The book says it is l times u₀
i m confused

[ethylenediamine]

How would the quadrupole moment be found? Does it have anything to do with j? Or does it have to do with the magic numbers?

[Mitch]

so , is it g_l always equal to 1 times u₀ (nuclear magneton)
The book says it is l times u₀
i m confused

Does it work?

[toluene]

I've tried setting g_l = l as well as g_l = 1. The latter seems to work better when you compare your answer to Fig. 6.6 in the text.

[Mitch]

Yeah, there should be no overall problem; the answers are literally in Figure 6.6

Ask Cerny tomorrow in class.