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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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1 Answers1

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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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