How would you find the magnetic field of a spinning nucleus?


by Crashwinder
Tags: field, magnetic, nucleus, spinning
Crashwinder
Crashwinder is offline
#1
Jan26-05, 11:21 PM
P: 5
I'm not sure if these are very clear questions, but here goes:

1. If a fairly large nucleus is set spinning, then it should generate a small magnetic dipole, right? I'm wondering, how would you calculate it's magnetic moment, using the properties of that particular nuclide (e.g. mass, charge, etc), the speed at which it's spinning, etc?

2. Same as above, except with a cation that has a given number of electrons.
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marlon
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#2
Jan27-05, 03:04 AM
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Quote Quote by Crashwinder
I'm not sure if these are very clear questions, but here goes:

1. If a fairly large nucleus is set spinning, then it should generate a small magnetic dipole, right? I'm wondering, how would you calculate it's magnetic moment, using the properties of that particular nuclide (e.g. mass, charge, etc), the speed at which it's spinning, etc?

2. Same as above, except with a cation that has a given number of electrons.
Very good question.

Quantummechanics learns us that the magnetization M is equal to [tex]\vec{M} = \frac{- \mu_{B}( \vec{L} + g_{s} \vec{S})}{\hbar}[/tex] The mu represents the Bohr-magneton, L is angular momentum and S is the spin. The g represents the gyromagnetic ratio. This formula was check experimentally via the Stern and Gerlach experiment and the Zeemann-effect for the spin quantumnumber. Now, for a given atom we can derive the possible L and S values by applying the laws of quantummechanics. This formula is especially valid for the orbiting electrons that indeed are equivalent to a magnetic dipole which yields the magnetization M. The orbit-speed is incorporated in L.
An analoguous formula can be derived for an atomic nucleus. In order to identify an unknown atom , one can excite it and then register it's emission-spectrum coming from the emitted radiation when the atom de-excites. each atom has a different emission-spectrum so this is like looking at the pass-port of an unknown atom. Many other options (using the above explained theory predicted by QM) are possible. Eg : nuclar magnetic resonance, etc

regards
marlon
Crashwinder
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#3
Jan28-05, 01:23 AM
P: 5
Quote Quote by marlon
Very good question.

Quantummechanics learns us that the magnetization M is equal to [tex]\vec{M} = \frac{- \mu_{B}( \vec{L} + g_{s} \vec{S})}{\hbar}[/tex] The mu represents the Bohr-magneton, L is angular momentum and S is the spin. The g represents the gyromagnetic ratio. This formula was check experimentally via the Stern and Gerlach experiment and the Zeemann-effect for the spin quantumnumber. Now, for a given atom we can derive the possible L and S values by applying the laws of quantummechanics.
Sorry for being ignorant (my understanding of this stuff is first year university level), but how exactly would you derive those?

This formula is especially valid for the orbiting electrons that indeed are equivalent to a magnetic dipole which yields the magnetization M. The orbit-speed is incorporated in L.
An analoguous formula can be derived for an atomic nucleus. In order to identify an unknown atom , one can excite it and then register it's emission-spectrum coming from the emitted radiation when the atom de-excites. each atom has a different emission-spectrum so this is like looking at the pass-port of an unknown atom. Many other options (using the above explained theory predicted by QM) are possible. Eg : nuclar magnetic resonance, etc
I'm afraid i don't understand how i should apply this information, i would greatly appreciate an example or two.


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