# Why are atoms not destroyed when dropped?

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.

• "because obviously the ring does not take any damage when dropped" Are you sure about that? Have you checked with an electron microscope? – Jahan Claes May 21 at 20:11
• Related if not duplicate: Why doesn't matter pass through other matter if atoms are 99.999% empty space? – Ruslan May 21 at 20:13
• Re, "It is like trying to stop a Mercedes by pushing against the star mounted on the bonnet." Perhaps you underestimate how strongly the "star" is attached. A gold atom is not a teeny-tiny Mercedes Benz, and the forces that hold its component parts together operate on a much different scale from the forces with which you are familiar on a human scale. – Solomon Slow May 21 at 20:16
• If the electromagnetic force is stopping the atom it can only act on the shell, and therefore the nucleus is simply continuing to follow his trajectory because of the law of inertia. No. The Coulombic forces act on the electrons AND the nucleus. It's a reciprocal force, so the nucleus is not free to fall AT ALL. – Gert May 21 at 20:53
• @Gert Thanks for pointing this wording imprecision out. Of course the nucleus will at one time be exposed to the EM force but when he does significantly the shell is already under a very strong force due to the nature of the 1/r**2 of the EM force. – Johannes Maria Frank May 22 at 0:30

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.