On the basis of the observations, Rutherford drew the following conclusions regarding the structure of an atom:

  1. Most of the space in the atom is empty as most of the alpha particles passed through the foil undeflected.

  2. A few positively charged alpha particles were deflected. The deflection must be due to enormous repulsive force showing that the positive charge of the atom is not spread throughout the atom as Thomson had presumed. The positive charge had to be concentrated in a very small volume that repelled and deflected the positively charged alpha particles.

  3. Calculations by Rutherford showed that the volume occupied by the nucleus was negligibly small compared to the total volume of the atom. The radius of the atom is about $\mathbf{10^{–10}\,\mathrm{m}}$, while that of the nucleus is $\mathbf{10^{–15}\,\mathrm{m}}$

I know this model is unsatisfactory, but how did Rutherford calculate the radius of the atom to be $10^{-10}\,\mathrm{m}$?

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    $\begingroup$ For that I expect he used existing estimates such as reasoning from kinetic theory and properties of gases, including Brownian motion, to obtain Avogadro's number, and then using e.g. the density of a solid containing a known number of atoms (obtained from molar mass). $\endgroup$ Commented Oct 30, 2021 at 9:44
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    $\begingroup$ I think perhaps you mean to ask about estimating the size of the nucleus, not the entire atom. As answers point out, that was already known from calculations. The estimate of the nucleus' size was presumably made from the fraction of alpha particles that were deflected. Simplistically, if 99% go straight through, the nucleus occupies 1% of the cross section. (Which also defines the size of the nucleus as the radius where the electric field is strong enough to deflect the particles.) $\endgroup$
    – jamesqf
    Commented Oct 31, 2021 at 1:30
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    $\begingroup$ @jamesqf, the question is perfectly clear: 5 dots wants to know where Rutherford got the figure of $\mathbf{10^{–10}\,\mathrm{m}}$ for the radius of the atom. $\endgroup$
    – TonyK
    Commented Oct 31, 2021 at 19:34
  • $\begingroup$ Rutherford was not aware that there is no specific boundary of an atom. These things became evident somewhat later, including as a result of his works. His estimate implied spherical atoms with defined boundary. $\endgroup$
    – fraxinus
    Commented Nov 1, 2021 at 13:19
  • $\begingroup$ @TonyK: Then the answer seems to be that he looked it up in prior work :-) $\endgroup$
    – jamesqf
    Commented Nov 2, 2021 at 2:42

3 Answers 3


Rutherford probably estimated the size of gold atoms as already sketched by @AndrewSteane in his comment.

The density of gold is $\rho=19.3\text{ g/cm}^3$.
The molar mass of gold was known from chemistry: $m_\text{mol}= 197 \text{ g/mol}$.
From this you get the molar volume $$V_\text{mol}=\frac{m_\text{mol}}{\rho}$$

Early estimations of Avogadro's constant (i.e. the number of atoms per mol) were already known from physical experiments before Rutherford's time. Later experiments refined this value: $$N_A=6.02\cdot 10^{23}\text{/mol}$$

Using this you get the volume per atom $$V_\text{atom}=\frac{V_\text{mol}}{N_A}$$

Let us assume the gold atoms form a cubic lattice (this is wrong, but good enough for an estimation). Then each atom occupies a cube of edge length $$d=\sqrt[3]{V_\text{atom}}$$

Doing the calculation we get $$\begin{align} d&=\sqrt[3]{V_\text{atom}} =\sqrt[3]{\frac{V_\text{mol}}{N_A}} =\sqrt[3]{{\frac{m_\text{mol}}{\rho\ N_A}}} \\ &=\sqrt[3]{{\frac{197 \text{ g/mol}}{19.3\text{ g/cm}^3 \cdot 6.02\cdot 10^{23}\text{/mol}}}} \\ &=\sqrt[3]{1.70\cdot 10^{-23}\text{cm}^3} =\sqrt[3]{1.70\cdot 10^{-29}\text{m}^3} =2.6\cdot 10^{-10}\text{ m} \end{align}$$

And the radius of an atom is half of this cube edge length $$r=\frac{d}{2}=1.3\cdot 10^{-10}\text{ m}$$

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    $\begingroup$ Rather than hypothesize a cubic lattice, you can assign them to specific volumes of approximate diameter $\sqrt[3]{V_\text{atom}}$... $\endgroup$ Commented Oct 31, 2021 at 15:59
  • $\begingroup$ Why is it reasonable to assume that the radius of an atom is half of the distance between atoms in the cubic lattice? Isn't that a bit like assuming that the radius of a billiard ball is half of the average distance between balls on a billiard table? $\endgroup$ Commented Nov 1, 2021 at 12:25
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    $\begingroup$ @MichaelKay we assume that atoms are packed densely enough to make the material resist compression. Such behavior must mean that the atoms interact with each other due to some kind of effective size. If we talk about billiard balls, we should visualize arrangement of the balls in a rack. This arrangement does resist compression, even if one ball or two are missing. $\endgroup$
    – Ruslan
    Commented Nov 1, 2021 at 12:59
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    $\begingroup$ @MichaelKay - adding to Ruslan's explanation: it's not assuming the radius is half the distance between atoms, it's assuming the radius is half the distance between nuclei of atoms of the same element. The billiard ball can be assumed to have uniform density. The atom's density is largely concentrated in the nucleus. Imagine if you can only see the pinpoint centers of those racked billiard balls but not the surface boundaries. Halving the distance between two centers is pretty reasonable. $\endgroup$
    – Forbin
    Commented Nov 1, 2021 at 14:38
  • $\begingroup$ @ Thomas Fritsch could you or anyone give a brief description of how the molar mass of gold was known, and how Avagadro's number was known, it would make the answer more convincing. $\endgroup$ Commented Nov 1, 2021 at 22:17

The size of the atom was estimated before Rutherford did his alpha particle experiment.

One way is to take a drop of oil of known radius and put in on water. It spreads out, given some time, to a large circle, of a small thickness.

From formulae for volume of a sphere and cylinder, the thickness can be calculated, if it's assumed to be one atom thick. That method calculates the size of an atom, originally done by Lord Rayleigh in about 1890, before Rutherford's experiment in 1908.

See also the grey box half way down this.

The value obtained was $1.6\times 10^{-9}$m. Other scientists continued this work and refined the value.

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    $\begingroup$ Why the downvote? $\endgroup$ Commented Oct 30, 2021 at 10:58
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    $\begingroup$ Oil molecules are much larger than a single atom. In particular, the thickness should be comparable to the length of an oil molecule. I didn't downvote though. $\endgroup$
    – Ruslan
    Commented Oct 30, 2021 at 11:43
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    $\begingroup$ sorry sir, but you aren't address the question . $\endgroup$
    – 5 Dots
    Commented Oct 30, 2021 at 12:24
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    $\begingroup$ @ 5 Dots Ok, maybe you could take this as an upper limit for the size of an atom, good look with finding an answer for the atom rather than a molecule, perhaps someone will post one $\endgroup$ Commented Oct 30, 2021 at 12:26
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    $\begingroup$ It does answer the question, Rutherford was probably aware of prior estimates of atomic sizes (noted in the answer to be of order $10^{-10} m$) and had no need to do any calculations. $\endgroup$
    – jim
    Commented Oct 30, 2021 at 17:51

Even if the electron cloud around an atom is diffuse, when packed together atoms take up a well-defined volume

The previous answers explain how the average volume taken up by an atom is calculated. And this is, indeed, what was done pre-Rutherford. This leaves the question of why the fuzzy region of space occupied by the electrons in an atom can be said to occupy a specific volume.

This needs an understanding of the forces involved when atoms come close together. Just because the isolated atom has a cloud of electrons which is "fuzzy" (at least in the sense that there is a small probability of finding an electron a long way away from the nucleus) this doesn't mean that two atoms interacting don't settle a definite (or, at least, fairly precise) distance apart.

That distance depends on the balance of attractive and repulsive forces among the atoms that are interacting. Some isolated atoms see strong forces when they come close (two isolated hydrogen atoms actually form a bond when they come close as energy can be released by sharing the electrons. This results in a bond with a very specific length. Crudely, the attractive forces of the bond counteract the repulsive forces driving the nuclei apart. But a quantum-mechanical calculation is needed to give a fuller picture taking into account things like the pauli exclusion principle).

A simpler situation arises when molecules or noble gas atoms not keen on making further bonds come into contact. Despite the "fuzzy" electron clouds they still see a mix of repulsive and attractive forces. The forces can be thought of as arising from quantum fluctuations in the electron clouds leading to very short lived dipoles that create short term forces pulling molecules or atoms together until the repulsive forces balance them out. The form of this overall potential is well understood (and can be derived from some fairly complex quantum calculations) but the details are not important. What matters is that atoms settle a fixed distance apart when the forces balance. Chemists tend to call this distance the atomic radius (or the Van der Walls radius after the name of the forces involved) and this is often considered the "size" of an atom. Many molecular solids are held together with these forces.

Other compounds have further types of bonding. Some solids, like diamond, are held together with an infinite array of strong covalent bonds. In these the atoms sit a specific distance apart caused by the length of the bond which, in turn is caused by the equilibrium of quantum forces pulling the atoms closer and others pushing them apart. Metals have many metal atoms sitting in a sea of free electrons holding them together against atomic repulsion.

The point, in all of these cases, is that what determines the definite and specific size of atoms in solids or molecules is a balance between repulsive and attractive forces. those forces reach an equilibrium at a fairly specific and definite point which can be used to define a fairly precise size for an atom despite the apparent "fuzziness" of the single atom's electron cloud.

IF you look at the forces involved in interacting atoms you get a much less fuzzy view of atomic size than you do by trying to draw an arbitrary boundary on electron density of the atom's electron cloud. That is how Rutherford could define the size of a gold atom.

  • $\begingroup$ Nitpick, but... "Some solids, like diamond, are held together with an infinite array of strong covalent bonds." is a bit misleading. Clearly no diamond actually has an infinite number of bonds... It's four bonds per atom, with atoms arranged in an infinite array (of finite size, obviously). +1, great info/read - just that one line really didn't read well for me. I assume it's probably a more naturally understood sentence for physicists - as a curious layman, it's too easy to interpret literally on first pass. $\endgroup$
    – TCooper
    Commented Nov 1, 2021 at 18:49
  • $\begingroup$ @TCooper When you are dealing with something as small as an atom the numbers in a finite macroscopic sample are so large that the difference between infinite and nearly infinite are insignificant. A 1cm cube of diamond has about 10^23 carbon atoms (about 40m along each side of a cube). So chemists and physicists, for example, tend to say "infinite" when they mean "nearly-infinite" as the difference is not worth mentioning unless you are a surface scientist. But, strictly, you are right. $\endgroup$
    – matt_black
    Commented Nov 2, 2021 at 10:29
  • $\begingroup$ Huh, if I considered graphite, wouldn't it have 30^23 covalent bonds compared to 40^23 covalent bonds in diamond (based on your estimates/provided example, 3 bonds per atom in graphite, 4 bonds per atom, right?). By the idea of "it's such a large number, we can assume infinite for simplicity", would you then say that both graphite and diamond have infinite covalent bonds? which would imply equal, and yet as I understand, the lacking of a fourth covalent bond is what makes the two materials so drastically different (conductive vs not, 1-2 mohs vs 10mohs, etc) What am I missing? or diff convo? $\endgroup$
    – TCooper
    Commented Nov 2, 2021 at 16:32
  • $\begingroup$ @TCooper I didn't mean to imply that there were an equal number of bonds or that they were the same sort of bonds. In chemistry the structure of the bonds matters for the properties not the total number in the solid. The properties of a flake of graphite with 10^20 atoms or 30^ 20 bonds will be the same as one with 10^23 atoms or 10^18 atoms was what I meant. Same with diamond: once the structure is set the total number of atoms and bonds is irrelevant (unless we are talking only a few thousand atoms when surface effects might dominate). $\endgroup$
    – matt_black
    Commented Nov 2, 2021 at 17:10
  • $\begingroup$ Okay, the word choice of infinite will still bother me, as I'm infinitely pedantic ;) but thanks for taking the time to explain. That makes a lot of sense re: properties. $\endgroup$
    – TCooper
    Commented Nov 2, 2021 at 19:03

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