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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?

Qmechanic
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theage
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    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. – ACuriousMind Oct 21 '14 at 22:01
  • related: http://physics.stackexchange.com/q/17651/ –  Oct 21 '14 at 22:01
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    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. –  Oct 21 '14 at 22:03
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    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? – CuriousOne Oct 21 '14 at 22:22
  • 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. – Qmechanic Oct 21 '14 at 22:58
  • Following the hints of Ben and CuriousOne, divide by c and see if it is a huge frequency: 3 10^14 Hz, or 300 TeraHertz. Yum, sounds huge, but visible light is the band 400-800 Thz. So we are just ok; surely we are interested in acceleration differences, but also the electron goes slower than c. – arivero Sep 28 '15 at 23:13

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The enormity of that acceleration shows that the atom will only exist for a fraction of a second, because the electron will radiate in a very short time almost all its energy. The radiated power, according to Larmor Formula is:

$$P = \frac{e^2 a_C^2}{6\pi\varepsilon_0 c^3} \approx 5\cdot 10^{-8}\ \text{J}\cdot{s}^{-1}$$

The potential difference between a typical atomic orbital and the surface of the nucleus is:

$$\Delta V \approx \frac{e^2}{4\pi\varepsilon_0}\left(\frac{1}{r_N} - \frac{1}{r_a} \right) \approx 2\cdot 10^{-14}\ \text{J}$$

This gives an upper bound on the time at which the electron will fall on the proton:

$$t_{fall} < \frac{\Delta V}{P} \approx 5\cdot 10^{-6}\ \text{s}$$

Davius
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