How can we explain why electrons don't spiral into the attracting nucleus?

Jul 17, 2017

That is the "energy" of the electron. There is no opposing force.

Explanation:

Essentially, at the atomic scale, the electron is orbiting in a vacuum. With no "drag" there is nothing to reduce its angular momentum around the nucleus.

Jul 17, 2017

Here's how I would explain it.

Explanation:

This is exactly what classical physics predicts.

However, we now know that electrons are governed by the laws of quantum mechanics.

Electrons don't really orbit a nucleus.

Instead, we must think of the electron as a cloud of electron density.

The electron could be anywhere in a spherical "shell" around the nucleus.

We could think of the atom as consisting of many thin shells with radius $r$ and thickness Δr arranged like the rings of an onion, as in (a) below.

(From Chemistry LibreTexts)

For each shell, the area is A = 4πr^2 and the volume is V = 4πr^2Δr.

The probability density is greatest at $r = 0$ [as in (b) above].

As we move outwards, the area $A$ and the volume $V$ increase as in graph (c) above.

However, at large distances from the nucleus, the probability density is quite small, so the probability $P$ is near zero [graph (b) above].

Also, close to the nucleus, the volume of each shell is small, so the probability $P$ is near zero.

So, the highest probability of finding the electron is going to be where the volume of the shell is big enough that $P$ is high, but the distance from the nucleus isn't too large for $P$to be low.

Thus, a plot of radial probability has a maximum at a certain distance from the nucleus [graph (d) above].

Most important, the probability of finding an electron at the nucleus is zero.