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Suppose we want to analyze a hydrogen atom using purely classical mechanics. This obviously is not exactly how things work - quantum mechanics plays a huge role and probability distributions are involved - but there is an interesting catch here.

Suppose naively that an electron orbits the nucleus to create a hydrogen atom (Bohr radius $r=5.3\times 10^{-11}\,m$. Then the nucleus (having one proton of charge $+1e$) itself has charge $+1e$, and the electron has charge $-1e$. Plugging this into Coulomb's law (for the magnitude of the force), we get

$$F_C = k\frac{|q_1q_2|}{r^2} = k\frac{|(-1e)(1e)|}{2.809\times10^{-21}\,m^2} = 8.2\times 10^{-8}\,N.$$

Next, using $F_C = ma_C$ (or $a_C = \frac{F_C}{m}$), we plug in the mass of an electron to get:

$$a_C = \frac{8.2\times 10^{-8}\,N}{9.1\times 10^{-31}\,kg} = 9\times 10^{22}\,\frac{m}{s^2}.$$

This is a shockingly large number for the centripetal acceleration of the particle (centripetal because the radius vector is between the nucleus and the electron). In the microscopic atomic world, is there any meaning to this classical analysis and if so, why is $a_C$ so large?

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    $\begingroup$ Have you also calculated the centripetal force that would be required to stay on an orbit with the Bohr radius? Does it match with the Coulomb force? If not, you can throw this whole analysis away without further thought. $\endgroup$ – ACuriousMind Oct 21 '14 at 22:01
  • $\begingroup$ related: physics.stackexchange.com/q/17651 $\endgroup$ – Ben Crowell Oct 21 '14 at 22:01
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    $\begingroup$ We expect accelerations at the atomic level to be huge, simply because subatomic particles are traveling very fast (the electron in hydrogen at ~1% of the speed of light), and are confined to very small regions of space. $\endgroup$ – Ben Crowell Oct 21 '14 at 22:03
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    $\begingroup$ Electrons don't seem to care about the scale of this semi-classical approximation, so why are you upset about it? What makes one number "shocking" and another number (like $10m/s^2$) normal? $\endgroup$ – CuriousOne Oct 21 '14 at 22:22
  • $\begingroup$ Hi @BMS: I took out the word QM in the question formulation because it seems OP would like to try to understand the hydrogen atom without using QM. $\endgroup$ – Qmechanic Oct 21 '14 at 22:58

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