An atom of hydrogen is placed in a very strong magnetic field. The magnetic moment of the orbit of the electron may either align with the external field or may oppose it. What will happen with the electron in each case? In both cases the Lorentz Force is acting on the electron.
Suppose the magnetic moment of the orbit of the electron is aligned with the external magnetic field. What will happen in this case, will the Lorentz Force push the electron away from the nucleus? If the magnetic field is strong enough, will this ionize the atom? If so, at whose expense, who's paying the energy for the ionization? Remember that the electron won't loose its velocity, due to the Bohr's postulate.
Now consider the opposite situation, the magnetic moment of the orbit of the electron opposes the external magnetic field. In that case the electron will be pushed towards the nucleus, again due to Lorentz Force. If the external magnetic field is strong enough, will it cause the electron to smash into the nucleus, and thus causing a nuclear reaction?
This situation is particularly interesting to me. Magnetars are neutron stars and have the strongest magnetic field in the universe. Is it due their extremely powerful magnetic field that their neutrons are unable to decay and emit beta radiation?
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$\begingroup$ @TLDR Consider an electron moving in a circular path due to the attraction to the positively charged nucleus. Now consider an external magnetic field is applied to the electron. If the B vector is coming out of the screen and the electron's orbit is counter-clockwise, then the Lorentz Force will pull the electron away from the nucleus. And if the orbit is clockwise, the Lorentz Force will push the electron towards the nucleus. $\endgroup$– question-askerCommented May 22 at 3:53
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$\begingroup$ @TLDR Lithium has zero orbital angular momentum in its ground state. $\endgroup$– my2ctsCommented May 22 at 4:22
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1$\begingroup$ @my2cts Well, that's embarrassing. Anyways, you know what I meant. Lithium may have zero orbital angular momentum... but boron doesn't! $\endgroup$– TLDRCommented May 22 at 4:24
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1$\begingroup$ @JosBergervoet the Zeeman effect is certainly related to this discussion, but I think the standard treatment ignores the term proportional to $B^2$ (which becomes relevant when $|B|$ is large.) $\endgroup$– TLDRCommented May 22 at 5:44
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1$\begingroup$ There is a literature on this subject that I found by searching for 'hydrogen strong magnetic field' :-). This may be the most clearly written paper: arxiv.org/abs/2203.02730. Any answer here should involve a basic understanding of the literature. $\endgroup$– my2ctsCommented May 22 at 8:46
1 Answer
The essential problem in this question is whether the electron orbit is squeezed to the center or whether it is pushed outward even to the point of becoming unbound. To answer that we can look at the case where there is only a very strong $B$-field so the attraction by the nucleus is ignored.
This leads to the standard solution with Landau levels and associated wave functions. The wave function in the $xy$-plane (perpendicular to $\bf B$) can be found for instance here [Cambridge] in Eq. (6.20) at p. 174, where we see that the lowest state is indeed squeezed to the "magnetic length" as defined on p. 171, which decreases with the square root of the field strength: $$\ell_B=\sqrt{\frac{\hbar}{q\,B}}$$
The wave function in this treatment is not confined in the $z$-direction, but it should be clear that if we do keep the attraction of the nucleus it would solve that problem, and we would have a cigar-shaped solution, strongly confined in the $xy$-plane and weakly in the $z$-direction. Solving this mathematically would make things unnecessarily complicated.
We can also see that we don't get solutions of lower energy with the electron "pushed out" if it is orbiting aligned with the field. This is perhaps not very surprising, as it is well-known that the magnetic field does no work, since its force is perpendicular to the motion, so any solutions with the electron orbiting at higher speed will just be stationary solutions with higher energy.
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$\begingroup$ It might also be interesting to explore the behavior of solutions in a neighborhood of the origin. Normally, the set of Landau orbitals has a continuous translation symmetry that one might expect is preserved (approximately) away from the origin, but which could be broken at or near $r=0$. $\endgroup$– TLDRCommented Jun 9 at 19:06