The Stability of Atoms

Section 3.4 The Stability of Atoms

Now we are ready to address the second great failure of classical physics. Maxwell's equations make a crystal clear prediction: accelerating charges send out energy in the form of EM waves. This is a big problem for the classical physics model of the atom. If the electrons orbit the nucleus like the planets orbit the sun, then they definitely have acceleration (equal to \(v^2/R\)), and so they are radiating away their energy.

But where would that energy come from? It comes from the potential energy of the electron's electric force interaction with the proton. To lower its potential energy, the electron must come closer to the proton, just like an object coming closer to the Sun lowers its gravitational potential energy.¹ So that is the classical solution: the electron must radiate away energy by coming ever nearer to the nucleus.

Working out the numbers involved for a hydrogen atom, we find that the electron should spiral in and essentially crash into the nucleus after a time of about \(10^{-12}\Xunits{s}\text{.}\) That is, classical mechanics and electromagnetism predict that atoms aren't stable and shouldn't last long enough for them (and us) to still be around. All the matter in our universe should be simply electrons sitting in the nucleus with the protons and neutrons.

Heisenberg's uncertainty principle resolves this dilemma. An electron cannot sit on top of the nucleus because that would be a very precise position and a very precise momentum (namely, at rest), which violates the uncertainty relation, so the best the electron can do is some compromise. It accepts the minimal spread in position and momentum that it can and then is stuck there. Since it cannot lower its energy further, it quits radiating and is stable.²

Put another way, whenever a particle is confined — i.e., if \(\sigma_x\) is less than \(\infty\) — then there is a minimum spread \(\sigma_p \ne 0\) in the momentum. Assume the product \(\sigma_x\sigma_p\) in the Heisenberg uncertainty relation, (3.14), to be as small as possible, namely, \(\sigma_x\sigma_p\approx \hbar/2\text{.}\) This gives us the inverse relation

\begin{equation} \sigma_p \approx\frac{\hbar}{2\sigma_x}\text{.}\label{eq_sigmap_from_sigmax}\tag{3.15} \end{equation}

If \(\sigma_p \ne 0\text{,}\) there is a non-zero kinetic energy. The average momentum must again be zero by symmetry, \(\langle p\rangle=0\text{,}\) so

\begin{equation} \sigma_p^2 = \langle p^2\rangle - \cancelto{0}{\langle p\rangle^2} = \langle p^2\rangle\text{.}\tag{3.16} \end{equation}

The average kinetic energy can be related to the spread in momentum:

\begin{equation} \langle K\rangle = \frac{1}{2}m\langle v^2\rangle = \frac{1}{2m}\langle p^2\rangle = \frac{\sigma_p^2}{2m}\text{.}\label{eq_ke_from_spread}\tag{3.17} \end{equation}

Conceptually, the since the electron is confined, it is forced to have some spread in velocity or momentum. The larger this spread, the larger the resulting average kinetic energy.

So, what does this mean? If an electron in an atom were to spiral into the center, the spread in its position \(\sigma_x\) would get smaller and smaller. But smaller \(\sigma_x\) would result in a larger and larger \(\sigma_p\) (by the uncertainty principle) and a larger kinetic energy (larger than the classical increase in \(K\) from spiraling inward). For an electron in an atom, the reduction in its potential energy (by moving closer to the center of the atom) would be countered by an increase in its kinetic energy, which grows toward infinity if the electron spirals all the way in. There is an optimal radial distance where the mechanical energy \(E = \langle K\rangle + \langle U\rangle\) reaches its minimum. So, the classical problem of an electron spiraling in doesn't apply, because eventually the energy increases if the electron gets closer and closer to the center of the atom.

So, the condition for the electron to spiral in all the way is gone. It can't keep radiating radiating EM waves, losing energy, and spiraling in to the nucleus of the atom. Heisenberg's principle indicates that the energy of the electron is not a minimum at the center of the atom but rather at a finite radius. So, an electron would need more energy to fall further into the center of the atom,

This is, of course, a simplification of the quantum explanation for the stability of atoms. As we'll see in the next chapter, the energy of an electron in an atom (and, in fact, the energy of any confined particle) is quantized, having only certain, discrete values.

In the meantime, an important consequence of Heisenberg's Uncertainty is the following:

A confined particle can never have zero kinetic energy.

We will say a lot more about this in the next chapter.

Actually, only half of the potential energy is radiated away. The other half goes into increased kinetic energy. But the total mechanical energy is still decreasing.

Thank goodness!