why didn't Rutherford use an aluminium foil, or a silver foil. Why he used gold foil in his gold foil experiment?
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1$\begingroup$ That's really a question you need to ask from Geiger and Marsden: en.wikipedia.org/wiki/Geiger%E2%80%93Marsden_experiment. It might have something to do with the fact that gold can be hammered into extremely thin foils, which is not possible (as far as I know) with either aluminum or silver. That reason is also given in the Wikipedia article. $\endgroup$– CuriousOneJun 5, 2016 at 10:01
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4 Answers
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He actually used also Aluminium, Silver, and Copper. He did so because he wanted to prove that the Rutherford cross section was proportional to $Z^2$.
In any case, he needed to use malleable material (metals) in order to achieve a micrometer-thin foil to prevent the entire $\alpha$ beam to be absorbed by the target.
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$\begingroup$ Hey this looks like a fantastic answer; can you give a citation for it? $\endgroup$ Jun 5, 2016 at 12:55
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$\begingroup$ Professor Longo said it during the Nuclear Physics course at Sapienza university. The material used are cited also on Wikipedia's article: en.wikipedia.org/wiki/Geiger–Marsden_experiment I had forgotten one element:he used tin, too. $\endgroup$ Jun 5, 2016 at 13:12
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Is this true?
In a 1913 paper, The Laws of Deflexion of α Particles through Large Angles...
Geiger and Marsden reused the above apparatus to measure how the scattering pattern varied with the square of the nuclear charge (i.e. if s ∝ Qn2). Geiger and Marsden didn't know what the positive charge of the nucleus of their metals were (they had only just discovered the nucleus existed at all), but they assumed it was proportional to the atomic weight, so they tested whether the scattering was proportional to the atomic weight squared. Geiger and Marsden covered the holes of the disc with foils of gold, tin, silver, copper, and aluminum. They measured each foil's stopping power by equating it to an equivalent thickness of air. They counted the number of scintillations per minute that each foil produced on the screen.
See Wikipedia
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Yes, it is correct that Rutherford used other metallic atoms instead of gold. From using other metallic atoms, he drew the following conclusion that there shall be no change in his prior observations, if and only if the malleability of the metal is sufficive enough for the alpha particles to penetrate through, otherwise there shall be a lack of penetration of the alpha particles, thus different scattering of particles, which would ultimately for-go his previous experiment.
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Geiger and Marsden first used Gold because it is a malleable metal and they could relatively easily produce foils of a thickness of around $1\; \mu$m which still is about 3500 atoms thick.
Even so this was thin enough to observe an incoming alpha particle interacting with only one nucleus and not being absorbed by the foil.
Other malleable metals were then used to see what effect they had on the scattering of alpha particle.
The parameter which they used to categorise a metal was its atomic weight as mentioned by @Mikhail in his question and they did find that the scattering was approximately proportional to the atomic weight squared.
It was Moseley who first systematically associated atomic number $Z$ with the number of positive charges in the nucleus.
