Where the electron is

An orbital is a function rather than a place. Ask where the electron in a hydrogen atom is and there are at least three defensible answers, all computable, all different.

The probability density is greatest at the nucleus. The most probable radius is one bohr. The ninety-per-cent contour is at 2.661 bohr. None of these contradicts the others, and confusing them is one of the more common ways to go wrong about the shape of an atom.

The density and the distribution #

$|\psi|^2$ is a probability density: probability per unit volume. For a 1s orbital it is largest at the nucleus and falls off from there monotonically.

The radial distribution is a different quantity: the probability of finding the electron anywhere in a thin shell at radius $r$. A shell of thickness $dr$ has volume $4\pi r^2 dr$, so the radial distribution is

$$P(r) = 4\pi r^2 |R(r)|^2.$$

The factor of $r^2$ is the whole story. Close to the nucleus the density is high and the volume is tiny; far out the volume is large and the density is negligible. The product peaks in between, and for hydrogen’s 1s orbital it peaks at exactly one bohr — the Bohr radius, which is where the name comes from and one of the few places the old model gives the right answer.

Why the density peaks at the nucleus and the electron does not #

This is the point that reliably surprises people, and it has an everyday analogue.

The most likely single point at which to find a raindrop landing on a circular target may be the exact centre. The most likely distance from the centre is not zero, because there is almost no target at radius zero and a great deal of it further out. Nothing is inconsistent; the two questions differ by the area available.

The same happens in three dimensions with the volume of a shell, and the $r^2$ is the same factor. So it is entirely correct to say the electron is most likely to be found near the nucleus per unit volume, and equally correct that it is most likely to be found at about one bohr from it.

The third answer, and why it is larger #

The contour that encloses ninety per cent of the density sits at 2.661 bohr, a good deal further out than the peak of the distribution.

That is not a contradiction either. The distribution peaks at one bohr and then falls away slowly; getting to ninety per cent of the total requires going well past the peak. Half the density is inside 1.337 bohr, ninety per cent inside 2.661, ninety-nine per cent inside 4.203.

The expectation value, which is a fourth number #

For completeness there is one more, and it is the one a physicist usually means.

The expectation value of the radius, $\langle r \rangle$, is the average distance weighted by the distribution. For hydrogen’s 1s orbital it is 1.5 bohr — larger than the most probable radius, because the distribution is skewed, falling away far more gradually on the outer side than on the inner.

So: 0 for the density maximum, 1.0 for the most probable radius, 1.5 for the mean radius, 2.661 for the ninety-per-cent contour. Four numbers, one orbital, no disagreement. Which one is “the size of a hydrogen atom” depends entirely on what the number is going to be used for.

What changes with n and l #

The pattern across orbitals is worth having because it explains a good deal about periodic trends.

Higher $n$ means further out, roughly as $n^2$. The most probable radius for hydrogen’s 2s is about 5.2 bohr against 1 for the 1s.

Higher $l$ within a shell means less inner density. A 2s orbital has a small peak inside its node, sitting close to the nucleus; a 2p has none. That inner peak is penetration, and it is why the s orbital of a shell lies lower in energy than the p in every atom except hydrogen: the s electron spends part of its time inside the screening of the inner shells and feels a larger effective nuclear charge.

That is a real explanatory chain from a shape to an energy ordering to the structure of the periodic table, and it runs entirely through the radial distribution rather than through the contour picture.

What the contour picture cannot tell #

Which brings the argument back to the drawings.

An orbital contour is a statement about a proportion of the density. It says nothing about where within that region the electron is likely to be, and in particular the region is not uniformly occupied — a 1s orbital’s ninety-per-cent sphere has far more density near its centre than near its surface.

So a contour picture answers “where is most of it” and cannot answer “where is it most likely to be”. The contour is a choice and the distribution is not, which is a good reason to show both.

Both are drawn here for exactly that reason, and this essay exists because the two are routinely conflated — including in the common claim that a p orbital’s lobes are “where the electron spends its time”, which is not what a contour means.

Nodes in the distribution #

One consequence for reading these plots.

The radial distribution vanishes at a radial node, because the wavefunction does. It also vanishes at $r=0$ for every orbital, because the $r^2$ factor does — including for s orbitals, whose wavefunction is maximal there.

That second zero is not a node. Nothing changes sign, and the wavefunction is perfectly finite; the distribution goes to zero only because the shell has no volume. Counting it as a node is a common slip and it makes every count come out one too high. The node counts here are made on $R(r)$, not on the distribution, for exactly this reason.

What it costs to have four numbers rather than one #

Everything above turns on the difference between quantities that a single phrase — “where the electron is” — runs together, and getting all four takes more work than getting any one.

The most probable radius is a maximum of the radial distribution, found by sampling the computed function on a fine grid and taking the largest value. That is a few hundred evaluations, and the answer for 1s comes out at one bohr to the precision of the grid, which is the check: the exact answer is $a_0$ and the search is never told it.

The mean radius is an integral, $\int r\,|\psi|^2\,dV$, and costs a quadrature. It is larger than the most probable radius for every orbital, because the distribution has a long tail on the outside and none on the inside, and the gap between them is exactly the asymmetry that makes “the size of an atom” ambiguous.

The contour radius is a root-find on an integral: the level enclosing a stated fraction is bisected for, and each step of the bisection is another quadrature. That is the expensive one, and it is the number this site is built around.

The maximum of $|\psi|$ is at the nucleus for an s orbital, costs nothing, and is the one most likely to be quoted by accident.

So the bill runs from nothing to a few thousand function evaluations depending on which question is asked — and the whole point of this essay is that they are different questions with different answers, spanning a factor of nearly three for a hydrogen 1s electron.

What is worth paying for is that each is computed rather than recalled, because the failure mode here is not arithmetic. It is answering the wrong one. A source that reports 2.66 bohr as the size of a hydrogen atom and one that reports 1.0 are not in disagreement; they have computed different quantities and neither has said which.

Why the distribution has a maximum at all #

The peak in the radial distribution is the point of the whole essay, and it is worth saying where it comes from, because it is the product of two competing effects and neither on its own would produce it.

The density $|\psi|^2$ falls monotonically with radius for a 1s orbital — it is largest at the nucleus and decays exponentially. The volume of a spherical shell of thickness $dr$ grows as $4\pi r^2$. The radial distribution is the product, and a decaying function times a growing one has a maximum where the two rates match.

That is the whole of it, and the arithmetic is short enough to do: differentiating $r^2 e^{-2r}$ gives a maximum at $r = 1$ in atomic units, which is the Bohr radius exactly. The coincidence is not a coincidence — Bohr’s orbit radius and the maximum of the quantum-mechanical radial distribution agree for the ground state, and disagree for every excited state, which is a fair summary of how much the old quantum theory got right.

The same reasoning explains why the distribution vanishes at the nucleus while the density does not. At $r = 0$ there is no shell to be in: the volume factor is zero however large the density is. A statement about probability per unit volume and a statement about probability per unit radius differ by a factor that goes to zero, and running them together is how the two most commonly confused answers in this essay get confused.

Where the model stops #

Three limits, all standing.

One electron. Every number here is for a hydrogenic orbital. In a many-electron atom the radial distributions are contracted by the increased nuclear charge and distorted by the other electrons, and the orbitals themselves are an approximation.

A distribution is not a trajectory, and nor is the contour a boundary. Saying the electron is most likely to be found at one bohr does not mean it travels at that radius, or travels at all. The distribution describes the outcome of a measurement, not a path between measurements.

Every number here is spherically averaged, and for most orbitals that discards information. The radial distribution of a 2p orbital is the same whichever of $p_x$, $p_y$ and $p_z$ is meant, because the angular part integrates away — so the radial account cannot distinguish orbitals that are pointing in different directions, and the directionality is most of what makes a p orbital useful in a molecule. The radial and angular questions separate cleanly for a hydrogenic function, which is what makes this essay possible; it is also what makes it partial.

The one place the distinction changes an answer #

It would be fair to ask whether any of this matters outside a pedantic footnote, and there is a case where it does.

The periodic table’s shape depends on which orbital an electron enters, and that depends on comparing the energies of, say, a 4s and a 3d orbital in an atom with several electrons. The comparison is usually explained by penetration: a 4s orbital has a small inner lobe that reaches close to the nucleus, feels the unscreened charge there, and is stabilised relative to a 3d orbital that has no such lobe.

That argument is about the radial distribution and it cannot be made with any of the other three numbers. The 4s orbital’s most probable radius is much larger than the 3d’s; its mean radius is larger; its ninety-per-cent contour is larger. By every measure of “where the electron is” except one, 4s is further out — and the one exception, the small amount of density it places very close in, is what decides the order.

So the four answers are not interchangeable even in sign. The quantity that settles the filling order is the one that is smallest, not the one that is largest, and reaching for the wrong one gives the wrong answer with complete confidence.

The figure also shows why the argument is a hydrogenic one being borrowed. In hydrogen all three of 3s, 3p and 3d have the same energy exactly, so penetration decides nothing; it starts deciding only when a second electron is present to screen the nucleus, and the shape of the distribution is then read off the one-electron function and applied to a situation it does not describe. That is a reasonable and very widely used move, and it is worth being clear that it is a move.

The value at the nucleus is measurable, and it decays #

The wavefunction being largest at the nucleus is the least intuitive of the three numbers and the one usually passed over as a curiosity of where a maximum sits. It is not a curiosity: it is a quantity nuclear physics depends on, and it can be changed by chemistry.

Some nuclei decay by capturing one of their own atom’s electrons. The rate at which they do so is proportional to the density of electrons at the nucleus — to $|\psi(0)|^2$ — because the process requires the electron to be there, and a probability of being at a point is what the wavefunction squared supplies.

That immediately explains a selection rule. Only s orbitals have any amplitude at the nucleus; every orbital with angular momentum has a node there. So electron capture takes s electrons, overwhelmingly the 1s, and the rate is set by the value this essay says is the maximum of the wavefunction.

The measurable consequence is the striking part. If the rate depends on the electron density at the nucleus, and chemistry rearranges the outer electrons, then the half-life of such a nucleus should depend on its chemical environment. It does. Beryllium-7 decays by electron capture, and its decay rate differs measurably between beryllium metal, beryllium oxide and beryllium trapped inside a fullerene cage — by a fraction of a per cent, small but far outside the measurement’s error, and reproducible.

A nuclear half-life is the standard example of a quantity chemistry cannot touch. This is the exception, it is small, and it exists because a wavefunction has its maximum at a point that a chemical bond can very slightly rearrange.

Where the number came from #

The Bohr radius is 0.529 ångström, and it is worth noticing what it is doing in a theory that abandoned Bohr’s model.

Bohr’s 1913 model put the ground-state electron in a circular orbit of exactly that radius. The model is wrong in almost every respect — it has trajectories, it has no explanation for why the orbits are stable, and it fails completely for helium — and this one number survives, as the most probable radius of the 1s distribution and as the natural unit of length in atomic physics.

That is a reasonably common pattern in physics: a superseded model leaves behind a scale that turns out to be right, because the scale was fixed by dimensional analysis rather than by the wrong dynamics.

Comparing the shells #

Setting several distributions side by side makes the trends visible that individual plots do not.

Higher n moves the peak out, roughly as n². Higher l within a shell removes inner structure, because the centrifugal term keeps a high-angular-momentum electron away from the nucleus.

The second trend is the one that shapes the periodic table. Penetration is what breaks the hydrogenic degeneracy, and once broken the filling order becomes 4s before 3d — which is the reason the first transition series exists where it does, and the reason its chemistry is what it is.

Contours, nodes, and one-electron functions #

The contour question is what the surface encloses, which is the discipline every orbital picture needs.

The structural feature behind the two-peaked distributions is the node.

And the caution that applies to every number on this page is that these are one-electron functions.

What the pictures here cannot show. A radial distribution is an average over all directions, so it says nothing about shape — a 2p and a 2s of the same $n$ have quite different distributions and a $2p_x$ and a $2p_z$ have identical ones. Direction is precisely what these plots integrate away.