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It is well known that the Rutherford model of atom was not satisfactory since it contradicted the energy conservation law and the Maxwell equations. Indeed according to the latter the electron moving around the nucleus has to emit electromagnetic waves and hence loose energy. It has to loose all of its energy and fall onto the nucleus.

I am looking for a reference with a quantitative version of the above argument. I am wondering what is the estimated time when the electron falls onto the nucleus.

MKO
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  • "it contradicted the energy conservation law and the Maxwell equations" I find that very hard to believe. – my2cts Aug 29 '20 at 13:42

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I've given this problem as part of assignments in the past, so I don't feel comfortable just doing the entire calculation for you, but it's really not hard so I'd urge you to calculate it yourself. Here's a sketch of the argument:

  1. An accelerated charge radiates electromagnetic radiation. It can be shown (see here) that the total power radiated by such a charge is given by the Larmor formula: $$P = \frac{1}{4\pi\epsilon_0}\frac{2 e^2}{3 c^3}a^2.$$

  2. Classical electrons are assumed to be in circular trajectories. Their energy is thus given by $$E = - \frac{1}{4\pi \epsilon_0} \frac{e^2}{2 r}$$

  3. Now as the electron is assumed to be accelerating around the nucleus, it radiates power thus decreasing its energy and "spiraling in". Using the above expression for energy, you can calculate its rate of change and equate it to $P$ given above, i.e.: $$\frac{\text{d}E}{\text{d}t} = P$$

  4. If you do this, you should get a non-linear differential equation involving $\dot{r}$ and $\ddot{r}^2$. The trick to solving the equation is to relate $\ddot{r}$ (the acceleration) to the Coulomb force, i.e: $$m\,\ddot{r} = -\frac{1}{4\pi\epsilon_0}\frac{e^2}{r^2}.$$

  5. If you've done everything right, the non-linear differential equation above should reduce to a simpler one that looks like $$\frac{\text{d}r}{\text{d}t} \propto \frac{1}{r^2}.$$

  6. You can solve this equation easily by integrating both sides: $$\int_0^T \text{d}t \propto \int_{a_0}^0 r^2 \text{d}r,$$ where $T$ is time you want to find, and $a_0$ is the Bohr Radius of the atom.

  7. If you plug in all the constants (and nothing's gone wrong), you should find that $$T \sim 10^{-11}\text{s},$$ which I hope you'll agree would have been noticed!

Philip
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    Or you can just divide the KE of the electron by the Larmor power for the same order of magnitude estimate. $(e^2/2r) /(2e^2 a^2/3c^3) $, where $a = e^2/4\pi \epsilon_0 r^2 m_e$ – ProfRob Aug 29 '20 at 13:33
  • @RobJeffries That's true. This entire argument is based on many approximations, so an order of magnitude estimate would work just as well, I suppose. Though it is a good undergraduate exercise :P – Philip Aug 29 '20 at 13:36
  • @Philip: Thank you very much. This is very helpful. I am wondering if there is a more rigorous treatment which would cover also 1) elliptic orbits; 2) more general expression for energy $E$ taking into account the fact that the trajectories are not exact circles or ellipses but spirals. – MKO Aug 30 '20 at 10:35
  • I'm sure you could, but I don't see the point of it. You could get an expression for the energy of an elliptical orbit and use it, the rest of the method should be similar. But I don't see the point of it: frankly, Rob's order of magnitude analysis is just as valid as my more "rigorous" treatment: atom's aren't classical, so speaking of such trajectories is inherently flawed. It's like trying to solve the problem of angels dancing on the head of a pin by including the actual dance steps. ;) – Philip Aug 30 '20 at 10:45