0
$\begingroup$

Struggling to understand the answer to this question.

When alpha particles are directed at a thin gold foil it is found that most of the alpha particles go straight through undeflected. However a very small number are scattered through angles greater than 90°.

Which of the following is not a valid conclusion?

  1. The atom is mainly empty space.
  2. The nucleus must be positively charged.
  3. The nucleus must contain most of the mass of the atom.
  4. There is a large charge concentration in the centre of the atom.

The answer is 2. I assumed that a negative nucleus would not scatter alpha particles in this way. Could it?

Improve this question
$\endgroup$
11
  • $\begingroup$ I am not sure that (3) is a valid conclusion either - unless you make other assumptions about things that you know about the atom. $\endgroup$
    – Floris
    Dec 12, 2017 at 19:10
  • $\begingroup$ @sammygerbil No he's not. He's asking if the differential scattering cross-section (the thing the Geiger-Marsden experiment measured) between like charged particles is different from that between unlike charged particles. $\endgroup$
    – dmckee --- ex-moderator kitten
    Dec 12, 2017 at 19:33
  • $\begingroup$ @dmckee Yes, thanks. I read the final question too literally, and out of context. I have edited the question. $\endgroup$
    – sammy gerbil
    Dec 12, 2017 at 19:49
  • $\begingroup$ (3) is indeed a valid conclusion. To backscatter requires that the mass of the atom is concentrated, and if not in the nucleus where? $\endgroup$
    – Jon Custer
    Dec 12, 2017 at 22:19
  • $\begingroup$ Thank you, I can see now that its a duplicate question!. So how does an attractive force cause scattering of greater than 90 degrees? Surely thats repulsion? $\endgroup$
    – johnsmith4725
    Dec 13, 2017 at 9:34

1 Answer 1

Reset to default
5
$\begingroup$

The Coulomb scattering angular dependence is independent of the sign of the charge. It goes as $(Z_1 Z_2)^2$. See https://en.wikipedia.org/wiki/Rutherford_scattering for instance.

Improve this answer
$\endgroup$
5
  • 1
    $\begingroup$ Indeed, which is why you can use the Rutherford formula for gravitational scattering as well (with appropriate adjustments for the different constants in the potential definitions). $\endgroup$
    – Jon Custer
    Dec 12, 2017 at 16:14
  • $\begingroup$ This is correct! But it doesn't answer the question which of the four statements in the question is not valid. $\endgroup$
    – freecharly
    Dec 12, 2017 at 17:22
  • 1
    $\begingroup$ @freecharly: of course it answers the question: if the scattering is the same for positive and negative nucleae, then the scattering observed cannot prove that the scatter center was positive. Because if it were negative the same result would have been obtained, so the result doesn't help you saying whether it is positive or negative. $\endgroup$
    – entrop-x
    Dec 12, 2017 at 18:30
  • $\begingroup$ @entrop-x - You are, of course right! The explicit conclusion that statement 2. is thus incorrect was only missing in your answer. $\endgroup$
    – freecharly
    Dec 12, 2017 at 18:47
  • 1
    $\begingroup$ In principle the full $\alpha + \mathrm{Au}$ interaction includes the nuclear force (that is it is not just the Coulomb force) in the attractive case and will be slightly different for repulsive and attractive forces for small impact parameters. In practice, I doubt the Geiger-Marsden apparatus had the sensitivity to detect the difference. $\endgroup$
    – dmckee --- ex-moderator kitten
    Dec 12, 2017 at 19:08

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