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In the Rutherford experiment (where he found the diameter of the nucleus of an Au atom by estimating the probability of an $\alpha$-particle to bounce back on a gold foil), how did Rutherford estimate the number of Au atoms in the foil ?

It seems to be needed in the calculus, since we have:

$$\frac{\text{number of back-bounced $\alpha$-particles/s}}{\text{number of $\alpha$-particles going through/s}} = \frac{N\frac{4}{3}\pi r^3}{V}$$

where:

  • $N$ is the number of Au atoms in the whole volume of the foil
  • $r$ is the radius of the Au atoms
  • $V$ is the total volume of the foil.
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To get an idea of how the experiments were done it is best to refer to the original Geiger and Marsden paper in the Proceedings of the Royal Society A (Volume 82, issue 557, pages 495-500). There one finds various clues, first being:

Since gold can be obtained in very thin and uniform foils, different numbers of these foils were used as reflectors. Each foil was equivalent in stopping power to about 0.4 mm. of air.

Gold leaf, and methods for producing it, far predate Rutherford. The use of distance in air as a measure of the stopping power of materials was quite common in the early 1900's because, well, that was an easy to obtain material and easy to see electrons and ions propagating in air and measure the length of the fluorescence. But, that does not help determine how thick each foil is. However, the next page has:

If the high velocity and mass of the $\alpha$-particle be taken into account, it seems surprising that some of the $\alpha$-paricles, as the experiment shows, can be turned within a layer of $6\times 10^{-5}$ cm. of gold through an angle of $90^{\circ}$, and even more.

So, their 'thin and uniform' foils were 0.6 microns thick. Given that foil thickness and the density of gold it is easy to convert to the areal density (atoms per square centimeter) of gold seen by the beam.

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    $\begingroup$ @xxavier - sigh. Need more coffee this morning. Thanks. $\endgroup$
    – Jon Custer
    Commented May 30 at 13:40

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