9
$\begingroup$

I have a chapter and in it there was a topic on the Rutherford experiment.

It is written that doubly charged helium particles or alpha particles were thrown towards a very thin gold foil and some of the alpha particles where bounced back because there were positive particles (protons) and positive and positive always repel each other.

If this was the reason for the bouncing back of the alpha particles, then why didn't the electrons, which are much smaller than protons, get attracted from alpha particles? Especially when the alpha particles are doubly charged.

$\endgroup$
8
$\begingroup$

The gold foil experiment conducted by Rutherford (or more appropriately by Geiger and Marsden under Rutherford's direction) involves a stream of $\alpha$ particles (which are helium nuclei) bombarding a thin gold foil.

Why didn't some electrons go with the alpha particles because of electrostatic attraction? Some probably did. The experiment was equipped with a fluorescent zinc sulfide screen to detect alpha particles ($\ce{He^{2+}}$ ions). This screen might have also detected $\ce{He+}$ ions (but not necessarily in a way that was different than the detection of alpha particles), but probably not free electrons or helium atoms. The experiment was only set up to detect alpha particles, and so that's all that Rutherford, Geiger, and Marsden observed. We'll never know how many (if any) alpha particles picked up electrons in their experiment. We could, however, repeat the experiment, replacing the fluorescent screen with a mass analyze to detect at least the difference between $\ce{He^{2+}}$ and $\ce{He+}$.

More importantly, we known that many alpha particles passed through (and some were deflected) unchanged. Why didn't these particles pick up electrons? The answer to this question is another question: What is holding those electrons in the gold foil?

Gold atoms have 79 protons holding down those electrons. The first ionization energy of gold is 890.1 kJ/mol and the second is 1980 kJ/mol. However, these values are for isolated gold atoms. Gold (like all metals) is held together by metallic bonds with delocalized electrons occupying a valence band. Each electron is not being held by just one gold nucleus. Once the first few electrons are ejected or snatched by the alpha particles, the gold foil will have a net positive charge. It will be much harder to remove negatively charged particles from the positively charged foil.

$\endgroup$
  • 2
    $\begingroup$ Is a mass difference between $\ce{He+}$ and $\ce{He^{2+}}$ really measurable? $\endgroup$ – Martin - マーチン May 1 '14 at 11:08
  • $\begingroup$ now can i count the weight of each and every helium particle!!! $\endgroup$ – anni May 1 '14 at 15:40
  • 3
    $\begingroup$ @Martin - the mass difference is negligible, but most mass analyzers actually detect mass-to-charge ratio $\endgroup$ – Ben Norris May 1 '14 at 17:42
  • 1
    $\begingroup$ @Ben Thanks! You are right of course, I just did not think about the huge difference in the charge of the particles. $\endgroup$ – Martin - マーチン May 2 '14 at 1:09
  • 2
    $\begingroup$ Sorry, but this explanation is just downright wrong. Positive charge on the foil, if any, will surely try to repel the alpha particles (though that is pretty negligible, as they come with energies of a few MeV), but once they reach the foil, they'll feel as if there were no potential at all. Otherwise we'd have to assume that the potential acts differently on gold atoms within the foil and alpha particle which is already within the foil too. Also, how then alpha particles avoid being neutralized in the source they come from? It is also solid and has no reason to be positively charged. $\endgroup$ – Ivan Neretin Nov 7 '15 at 22:28
3
$\begingroup$

Consider radium decay alpha-particle energies: Ra-226 4.86 MeV, Ra-222 5.59 MeV. The first atomic ionization energy of gold is 9.22 eV. We do not expect charge neutralization of the alpha-particle until it slows by collision. It will leave a short path of dense ionization as it transfers energy, continuously being stripped to a bare nucleus until it slows.

The vacuum work function of bulk gold tops at 5.47 eV, giving the same expectation.

$\endgroup$
0
$\begingroup$

I had a similar, but slightly different question over on the physics exchange.

Relevant to this question: Grabbing electrons would also seem to change the trajectory of the He+ (as would bumping into one of the 79 high-speed electrons per gold atom). While my question was more about the impact this has on the scatter (as this seems important to the measurement of the size of the nucleus of the atoms), seems like deflections should give clues to pick-up of electrons as well.

$\endgroup$
  • $\begingroup$ Wait...are you asking a question? $\endgroup$ – M.A.R. ಠ_ಠ Mar 8 '15 at 17:36
  • $\begingroup$ @MARamezani: wait, are YOU asking a question? $\endgroup$ – Brad Cooper - Purpose Nation Mar 10 '15 at 14:06

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.