1

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
  • 220,844
theage
  • 586

1 Answers1

2

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
  • 1,700