Why are atoms not destroyed when dropped?

Question

I made the following thought experiment: Dropping a gold ring on a wooden table. It drops, hits the table, bounces off, hits again with less velocity and so on until it finally rests.

Now consider an gold atom inside the ring. It will of course be accelerated and there is no problem with the nucleus and shell having a huge mass difference as the gravitational acceleration is independent of the mass.

Assume it is a carbon atom that is hit when the ring hits the table. This is reasonable as there are a lot of carbon atoms in wood.

Now the only force that can stop our ring is the electromagnetic force, since we only have four forces, there is no anti-gravity and the weak and the strong force do not extend to the outer shell. From the geometry of the two atoms the one shell interacts therefore with the other shell first.

The gold atom is a lot heavier than the carbon atom so the carbon atom will start moving and will in turn move other atoms which distributes the force so the counter force starts to increase and in the end will balance forces which brings the gold atom to a halt and then even pushes back so the ring bounces back. The rules are governed by Hooke’s law, the table acts like a spring.

But the atom is not a solid sphere, it is like a solar system with all the mass centered in the center. And here I am not understanding how this can actually work.

If the electromagnetic force is stopping the atom it can only act on the shell first (because of the speed of light being finite) and therefore the nucleus is simply continuing to follow his trajectory because of the law of inertia. It is thus suddenly pushed out of the center of the atom and even if I ignore that now one side of the shell is pulling harder on the nucleus than the other, the shear difference in mass must just lead to the nucleus crushing through the shells of several atoms.

It is like trying to stop a Mercedes by pushing against the star mounted on the bonnet.

So what is preventing the atom from being destroyed? How is the force that stops the shell actually put on the nucleus, because obviously the ring does not take any damage when dropped.

Answer 1

The appropriate intuition here is that small objects operate on faster timescales. You might not be able to stop a Mercedes by pushing on the star instantly, but you certainly can if you gradually push on it for a couple of centuries.

In the case of an atom, the appropriate timescales are given by the de Broglie relation $E = \hbar \omega$, so $$t \sim \frac{1}{\omega} \sim \frac{\hbar}{E} \sim \frac{10^{-34} \, \text{J s}}{1 \, \text{eV}} \sim 10^{-15} \, \text{s}.$$ If an impact takes a few milliseconds, then in a classical picture, during the collision the electron can go around the nucleus a trillion times. In our solar system, the equivalent timescale for the Sun and the Earth would be a trillion years.

The same intuition holds in the quantum case. The collision is not sudden at all, and there's no reason that impulse can't be gradually transferred from the electron to the nucleus. In fact, for both the classical and quantum cases, this intuition can be formalized by the adiabatic theorem, whose conditions are satisfied extremely well here.