Magnetic moment of proton and electron of H-atom in ground state L=0

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SUMMARY

The discussion centers on calculating the magnetic moments of the proton and electron in a hydrogen atom under a magnetic field of 2T while in the ground state (L=0). The magnetic moment formula μ = (q/2m)L yields μ=0 for both particles due to L=0, which raises concerns about the validity of this result. The correct approach involves using the spin magnetic moments, expressed as μspin = -2 (eħ/2me)ms for the electron and μ = -2.79 (eħ/2mp)ms for the proton. The expectation value of the magnetic moment operator in the fundamental state (1s) is crucial for accurate calculations.

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Homework Statement



Consider the Hydrogen atom in a magnetic field of 2T. If the atom is in its ground state (orbital angular momentum L=0):
(a) Write down the magnetic moments of the proton and the spinning electron. [1]

Homework Equations



I am aware that the magnetic moment μ = (q/2m)L

The Attempt at a Solution

but because L=0 this means that μ=0 for both the proton and the electron, which doesn't seem to me to be a valid answer.

Is this the correct way to be attacking this problem or am i missing something?

Any help with pointing me in the right direction would be much appreciated.

Thanks.
 
Last edited:
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I think it asks for the expectation value of the magnetic moment operator in the fundamental state (1s).
 
Thanks for the help, but I've looked at it some more alongside my lecture notes and I think what they want is:

\mu^{spin} = -2 \frac{e\hbar}{2m_{e}}m_{s}

for the electron, and for the proton:

\mu = -2.79 \frac{e\hbar}{2m_{p}}m_{s}


Do you think that seems reasonable? I am going to ask the lecturer tomorrow if it is the expectation value he wants or not.

Thanks again for the feedback!
 

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