Schrödinger's Cormorant

5.9 Atoms: An uncertain answer to a big question about small things

The uncertainty principle explains how atoms can exist, without the electrons simply falling into the nucleus. Estimation of the size of a hydrogen atom, based on these ideas.

If the electron is attracted to the nucleus, why doesn’t it just crash down into it? You probably remember the first time you asked this question: it’s one of Nefertiti’s most cherished memories of her days in the nursery. It’s not a trivial question to answer.

Cormorant mutters something about it being like planets in perpetual orbit, in apparent defiance of gravity, but Nefertiti cuts him short. This won’t do: Sir Isaac’s classic explanation of why planets don’t crash into the sun only works because there’s no mechanism for the planets to lose their kinetic energy - at least at anything but a negligible rate. If electrons were orbiting the nucleus in the same sort of way, however, there would be just such a mechanism: it’s well established that when a charged particle is accelerated (and circular motion applies continuous acceleration as the direction changes) it loses energy through electromagnetic radiation. If you have any doubts about this, just pay a visit to your friendly local synchrotron light source. So an orbiting electron would recklessly radiate away its energy, experience orbital decay and spiral in disastrously towards the nucleus. And, in any case ……….

We’ve just spent some considerable effort establishing the uncertainty principle which means that we can't simultaneously know precisely the position and the momentum of a particle. In a planetary model, an electron would be travelling on a precise orbit, at a precise angular velocity, around the nucleus. So its position and momentum would be exactly defined at every instant: it would be breaking the rules.

Moreover – now Werner watches expectantly as our mental cogs begin to turn - we just discovered that the uncertainty principle means that localisation of a particle comes at an energy cost. So:

The electrostatic attraction will tend to pull the electron in towards the nucleus which, in terms of our wave model, corresponds to narrowing the wave packet and hence restricting the volume of space in which it is likely to be found. BUT: there will be a limit as to how small a space it can actually be restricted to because the narrower the wave packet, the higher the localisation energy, as we have seen.

In other words, it is the uncertainty principle, and the constraint it places on the properties of a matter-wave, that is the reason atoms exist. The tendency to minimise potential energy keeps the atom together, preventing electrons from just flying away but the increase in their kinetic energy as they become localised around the nucleus opposes this, and means that there is going to be a finite, compromise size to an atom.

Let’s see if we can quantify this idea and, by doing so, predict the size of a hydrogen atom. This should be a good test of whether we are thinking along the right lines. We can place the (centre of) the nucleus at the origin of our coordinate system and look at what happens along any line that passes through it. We’ll begin by writing down an expression for the contributions to an electron’s energy from electrostatic attraction to the nucleus and from localisation.

There is an issue here: if you’ve got this far in this journey, you’ve probably spent so long in one dimension that you’ve stopped noticing, but a hydrogen atom is distinctly three dimensional. However because the electrostatic potential and the localisation effect are the same in all directions, we’ll just focus on one dimension for the moment.

Coulomb’s law gives the electrostatic potential energy resulting from the interaction of two charged particles, A and B, as:

Where qA and qB are the charges, ε0 is a constant (the permittivity of a vacuum) and r is the distance between the charges. Since in a hydrogen atom we have just one proton and one electron, with equal and opposite charges (±e), we can rewrite this as:

where <r> is the mean value of the electron’s distance from the nucleus, according to its wavefunction.

Now let’s bring in the uncertainty principle to estimate the localisation energy. As we know from question 5-7, this is a kinetic energy term that results from the range of momenta needed to achieve a wave packet.

Question 5-16

Combining the kinetic energy term you've derived in Question 5-15, with the electrostatic potential energy term (equation V-16, above), we get a total energy expression for the electron:

You can see a graph of this function in figure V-ix:

Figure V-ix

You can see what’s happening here by following the energy as the electron starts a journey far from the nucleus and moves in closer: the attractive electrostatic attraction becomes stronger as <r> decreases, so the potential energy decreases. This is opposed, however, by the kinetic energy cost of localising the electron wavefunction. This term also gets bigger as <r> decreases but because it depends on 1/<r>2 rather than 1/<r>, it drops off more rapidly with <r>. Therefore the localisation energy starts to become significant only at shorter distances. Eventually, though, it becomes the dominant term, leading to the strongly positive overall energy as <r> approaches zero.

So if you’re an electron in a hydrogen atom, what distance from the nucleus are you going to favour? It seems reasonable to predict that the electron will favour the distance where its total energy is lowest. To see if this is true, let’s calculate the value of <r> that corresponds to the minimum in the energy curve:

Question 5-17

This exercise predicts a mean electron-nucleus distance of around 4 x 10-11m for a hydrogen atom. Remember, from Question 5-15, that this can - at best - only be expected to give us an approximation to the actual size of a hydrogen atom.

In fact, the result we've obtained is, indeed, of the correct order of magnitude: the experimentally determined atomic radius is 1.2 x 10-10 m. Of course the atomic radius is not the same as the mean distance between the electron and the nucleus: we would expect the mean to be smaller than the overall radius, as the electron could be at any radius within the wave packet. The mean distance as calculated by a full theoretical treatment is 5.3 x 10-11 m. So our very simple calculation – which makes several questionable leaps and does not make any attempt to take account of the actual wave form of the electron – actually gets remarkably close to the correct answer.

We can test this further by using our uncertainty principle model to estimate the ionisation energy of a hydrogen atom: that is, the energy we would need to put in to remove the electron from it stable, bound state in the atom – in effect to infinite distance:

Question 5-18

These encouragingly good approximations lend support to our wave packet model for the electron in an atom and our suggestion that the resistance of the atom to collapse is a consequence of the uncertainty principle. Just how big is the energy barrier that prevents the atom from collapsing?

Question 5-19

The answer you should get for the price of localising an electron within the nucleus is an enormous energy term. For comparison, even the strongest chemical bond is weaker by a factor of over one billion (the bond energy of the N≡N triple bond in gaseous nitrogen is less than 1000 kJmol-1). Indeed it is even bigger than the binding energy that holds the protons and neutrons together in a nucleus. Seriously big, so it’s not going to happen under any normal kind of conditions. This really does support the idea that it’s the localisation energy that stops an atom collapsing in on itself.

The other side of this coin, then, is the nucleus: how does it get to be so small? In the case of our hydrogen nucleus, it is just a proton, which is made up of three quarks bound together. Again, therefore, there must be a competition between an attractive force and an unfavourable localisation energy but the attractive force which holds together the quarks that make up the proton is not electrostatic: it is a different force, called the strong interaction. Cormorant has an idea: a proton has a much bigger mass than an electron: could that account for its greater localisation?

Question 5-20

If you have worked through all of this, it's worth pausing a moment to feel pleased with yourself. You have successfully used principles derived from the concept of wave-particle duality not just to provide a convincing answer to one of the great mysteries of physics - how atoms can exist - but you've managed to put together calculations which provide successful estimates of the size and energetics of such atoms. That's pretty impressive.

So, we seem to have a good idea about why atoms are the way they are: electrostatic interaction pulls an electron in towards the nucleus but it cannot get too close because of the kinetic energy cost of localising its wavefunction. We have not, however, yet learned any more about the detail of the wavefunction or why the constraints on this function lead to quantised energy levels. We’ll continue moving towards this goal in the next chapter.

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