Chemical Forums
Specialty Chemistry Forums => Nuclear Chemistry and Radiochemistry Forum => Topic started by: tisu85 on October 04, 2006, 03:08:12 AM
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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??
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Use the shell model. Does 37Cl have an odd proton or an odd neutron.
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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?
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If you use gl = 1 u0 (protons), you seem to get the right answer so try that.
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But gl = l u0 , as in "l" the alphabet not "1" Wouldn't that mean we should use the angular momentum value: 4???
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Just try it, see what you get for the values.
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so , is it gl always equal to 1 times u0 (nuclear magneton)
The book says it is l times u0
i m confused
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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?
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so , is it gl always equal to 1 times u0 (nuclear magneton)
The book says it is l times u0
i m confused
Does it work?
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I've tried setting gl = l as well as gl = 1. The latter seems to work better when you compare your answer to Fig. 6.6 in the text.
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Yeah, there should be no overall problem; the answers are literally in Figure 6.6
Ask Cerny tomorrow in class.