# Is the boundary of an atom well defined?

That question might seem weird, but that's what I have been imagining lately; going through Bohr's theory on the hydrogen atom. He clearly defines the radius of an orbit with an exact mathematical expression. But what do they mean when they say any atom (say $\ce{Na}$) has a radius $r$. Are we to consider atoms as solid, rigid spheres? If that's the case, what about water?

If all the atoms are defined as solid, rigid, spherical balls, how come an aggregation of so many solid objects be a liquid? How can it flow? So, in a nutshell what is the boundary of an atom made up of, or even this: does it have a defined boundary?

This troubles me a lot. I think maybe the modern atomic model has dealt with this problem. I haven't read it in detail, and I would be highly thankful to any references.

• You ask too many questions at once. Stop using Bohr's theory. Atoms are neither solid nor rigid, and don't have any boundary at all. Then again, a collection of solid spheres may pretty well act as a liquid. Ever heard about quicksand? – Ivan Neretin Sep 30 '17 at 7:28
• Any modern chemistry textbook (=less than fifty years old) only uses Bohr as an introduction to orbital theory, except perhaps for the very first year of chemistry in school (8th grade or so, fourteen-year-olds). Exceptions prove to be frauds. ;-) – Karl Sep 30 '17 at 11:55
• @Karl I'm in high school, and still we're being taught the Bohr theory, even worse we are made to solve a huge number of problems with the Bohr theory..and when I ask my teachers about quantum mech, they ward me off by saying I'm too young to know! I can't seem to get the hang of the schrodinger wave equation as it's beyond the level of calc I know.. – YourAverageEuler Sep 30 '17 at 13:45
• @YourAverageEuler Nobody would bother school kids with wave functions, it also works without. Get yourself next years schoolbook if you are really eager. And you can do a lot of instructive things with Bohr, go along with it! In first and second year school chemistry, you should learn things that prepare you for more, not the final "truth". Also Schrödingers model only works quantitativley for single hydrogen atoms. Still it indispensable to be very familliar with it to understand anything further on. – Karl Sep 30 '17 at 14:23
• @Karl Correction: Hydrogen-like atoms. $\ce{He+}$ and $\ce{Li^2+}$ also work ;) – Jan Sep 30 '17 at 15:17

No, the boundary of an atom is not well defined. According Quantum mechanics the electron density around an atoms decreases exponentially, thus getting asymptotically close to zero.

Considering atoms as rigid spheres is only an approximation. The criteria used to decide on the actual value of the atomic radius is arbitrary. You can for example choose the expectation value $$\langle\hat r\rangle$$ of the electronic wavefunction. For the Hydrogren atom, this results in the bohr radius $$a_0$$. But you might as well choose the maximum of the electron density ($$\frac{3}{2}a_0$$ for $$\ce{H}$$), or something based on experiments.

Now the aggregate state is something completely different and not directly related to atomic radius and whether they are rigid sphere or not. Solid means the molecules/atoms of a substance are fixed in their positions. They can vibrate, but do not change their average position with respect to each other. In gases molecules may move freely. And a liquid is something in between, you have some order if you look at a small group of molecules, but not on a larger scale as for solids.

As was already mentioned, the boundary of an atom is everything but well-defined. Quantum mechanics tell us this very well: the probability of finding an electron at the distance $r$ from the nucleus is proportional to $\mathrm e^{-r}$, meaning that it will never hit zero.[1] The lack of a zero-point means that there is no cutoff after which we can clearly say the atom is over. Instead, when performing quantum calculations an arbitrary cutoff value is chosen. The boundary within which the probability of finding the electron is larger than the cutoff value is then plotted as the ‘extent of the orbital’.

However, even if atoms were rigid spheres that does not mean your macroscopic postulate (that rigid spheres cannot form a liquid) holds true because you are misestimating the difference between microscopic and macroscopic properties. I present you the following example:

We will both agree that sand is made up of rigid solid particles even if they are not perfect spheres. Yet, sand shows a few properties that almost make it look liquid if you look from far away. For example, it can ‘flow down’ a slope much in the same way as water could. And also if you try to pile it into a pyramid, it will flow outwards to a certain extent as water (or better: mercury) drops would. You might extend this analogy to marbles which are clearly spherical (usually). You cannot stack them on a blank surface as they will roll away much like you cannot stack water molecules because they will flow to the side.

The main difference between my macroscopic examples and actual molecules is the type and extent of attractive forces between them. While the rough texture of sand can be said to give it some kind of a surface tension (like water and mercury), the attractive forces between the particles are still low and the main force acting on them is gravity (of the Earth, not of other grains of sand). The same thing is true for the marbles example which even lack surface tension and are only attracted to the Earth by gravity. Actual molecules have much stronger intermolecular forces that dominate over the external gravity force source in the microscopic world. That contributes to the final edge of a difference between many grains of sand/many marbles and a sample of water.

All I have discussed above is, however, completely independent of the question whether the spheres that make up atoms are rigid and solid or not.