When we apply magnetic field to a hydrogen atom such that the Magnetic Lorentz Force acts inwards on the electron, will its radius shrink to a value lesser than the Bohr's radius, because according to Bohr's quantization radius of electron orbit can never be less than Bohr's radius?
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10$\begingroup$ The basic problem with this question is that the electrons aren't in orbit. Yes, that is the model Bohr used, but no it is not correct. Not even vaguely correct. The word used for the real quantum description is "orbital", and the ground state orbital has a higher chance for the electron to be inside the nucleus than in any other comparably sized volume. $\endgroup$– dmckee --- ex-moderator kittenCommented Feb 20, 2017 at 4:36
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In quantum mechanics the electron does not have orbits, only orbitals, probability loci for where the electrons will be if a measurement is attempted. The Bohr model is an averaging out of the underlying quantum mechanical behavior. An average radius can be calculated and that is what may change if a magnetic field is imposed.
There are measurements of atomic orbitals and orbitals that are important in chemical studies. There are calculations of how orbitals change where a change in average radius is taken into account , so yes, the average radius can change by external magnetic fields.
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1$\begingroup$ anna, only for my autodidactic understanding: Bohr developed a model but the reality is QM? I see this many times that the reality is confused with QM. Wouldn't it be better to talk about the best fitted model of the reality on atomic level? Ok this is to long. How about modern atomic physics is quantum mechanical? $\endgroup$ Commented Feb 20, 2017 at 5:55
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$\begingroup$ @HolgerFiedler Yes, Bohr's model is an approximation of the underlying quantum mechanical probabilistic level. Atomic physics too assumes the underlying probabilistic quantum mechanical level. We say that QM is the underlying framework because the assumption fits all the data. One does not have to use bohr type approximations all over the place, but develop the quantum mechanical model and project it to the macroscopic measurements. All macroscopic numbers emerge from underlying quantum mechanical behavior. $\endgroup$– anna vCommented Feb 20, 2017 at 6:09
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2$\begingroup$ @HolgerFiedler QM is the best description we have right now. This may of course change in the future, but right now, we can't tell the difference between QM and reality, so we consider QM (or some form combining QM and GR, like string theory) real. But it's still just a model - we will never know what's "real reality" (that's impossible in principle, and really more philosophical than physical), we just keep making more accurate models. Bohr's model was limited - it explained the emission spectra of hydrogen, but that's about it; QM explains everything we know, and predicted lots we didn't. $\endgroup$– LuaanCommented Feb 20, 2017 at 9:25
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2$\begingroup$ @Luaan Not my mother tongue. The only thing I wanted to say with my replay was that QM is a description of reality and not reality by itself. $\endgroup$ Commented Feb 20, 2017 at 11:51
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$\begingroup$ For confirming. Does the orbit of an electron reduce (in older QM) in the presence of an external strong magnetic field. Or In new QM does the probability of finding it near the nucleus increase in the above case. $\endgroup$ Commented Apr 4, 2017 at 10:05