Zeeman Effect for a Muon: Differences Compared to Ordinary Hydrogen Atoms

AI Thread Summary
The discussion centers on the differences in the Zeeman Effect between muon atoms and ordinary hydrogen atoms. The energy change for hydrogen is given by delta E = 2*(bohr magneton)*(magnetic field), but for muons, the energy difference is affected by their greater mass, leading to a modified equation. The proposed equation for the muon is Energy difference = 2*[ (e*h-bar)/(2*207*mass of an electron)]*(magnetic field), which is correct as the muon's spin does not need additional consideration. It is concluded that the energy difference for the muon will be smaller than that for hydrogen, resulting in greater spectral line splitting for the muon. The discussion emphasizes the importance of mass and spin in determining the Zeeman Effect in different atomic systems.
pka
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Homework Statement


Consider an atom composed of a proton nucleus and an orbiting negatively charged muon (which has a mass of 207*(mass of an electron)). What difference, if any, would you expect between the Zeeman Effect in such atoms and in ordinary hydrogen atoms? (Hint: the muon has spin -1/2, but a greater mass than the electron).


What I've done:
The change in energy for a hydrogen atom according to the Zeeman effect (or because of it) is: delta E = 2*(bohr magneton, Mb)*(magnetic field).

This is where I am getting a little confused...For the muon, the energy difference will not be the same. What I'm having trouble with is the equation for which to use in order to determine the energy difference for the muon. What I've come up with is this:

Energy difference = 2*[ (e*h-bar)/ (2*207*mass of an electron) ]*(magnetic field)
Is this correct? What makes me unsure is the fact that the muon now has spin (not just a greater mass) so shouldn't my energy difference also be multiplied by the spin of the muon?

Assuming that both the hydrogen atom and the muon are in a uniform magnetic field of 1 Tesla and Mb = [(e*h-bar)/ (2*mass of an electron)] = 9.274 x 10^-24 J/T.

My conclusion is that the energy difference for the muon will be smaller than that for the hydrogen atom because splitting is greater for the atom with spin. Also, the muon has greater mass so the energy difference will end up being closer to zero than the hydrogen atom and thus being smaller and leading to a greater splits in the spectra when you look at the little lines for the muon. On the other hand, the hydrogen atom will simply have three splits on its spectra.

Any help or advice anyone can offer would be fantastic! If I'm thinking about this as I shouldn't be I'd greatly appreciate it with anyone could point me in the right direction. Many thanks in advance.
 
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pka said:

Homework Statement


Consider an atom composed of a proton nucleus and an orbiting negatively charged muon (which has a mass of 207*(mass of an electron)). What difference, if any, would you expect between the Zeeman Effect in such atoms and in ordinary hydrogen atoms? (Hint: the muon has spin -1/2, but a greater mass than the electron).


What I've done:
The change in energy for a hydrogen atom according to the Zeeman effect (or because of it) is: delta E = 2*(bohr magneton, Mb)*(magnetic field).

This is where I am getting a little confused...For the muon, the energy difference will not be the same. What I'm having trouble with is the equation for which to use in order to determine the energy difference for the muon. What I've come up with is this:

Energy difference = 2*[ (e*h-bar)/ (2*207*mass of an electron) ]*(magnetic field)
Is this correct? What makes me unsure is the fact that the muon now has spin (not just a greater mass) so shouldn't my energy difference also be multiplied by the spin of the muon?
No, you don't have to worry about the spin. The muon has the same spin as an electron.
 
Alrighty then! Sounds good to me. So I've got the right idea about the rest of the problem? Thanks so much. :D Fantastic!
 
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