30
$\begingroup$

Geiger and Marsden's experiment led Rutherford to believe that the positive charge and most of the mass of the atom was concentrated in a small region. I understand what led him to conclude the way the positive charge is positioned in the atom. But how did he conclude that most of the mass was in a small region (the nucleus)?

How did the distribution of the mass matter after all? Given that the electric force is greater than the gravitational force by many magnitudes, the force between the positice charge and the electrons was predominantly electric.

So how did Rutherford conclude that most of the mass is in the nucleus?

$\endgroup$
  • $\begingroup$ Rutherford assumed that the target nucleus did not recoil when the alpha particle interacted with it. $\endgroup$ – Farcher Apr 13 '17 at 6:18
  • $\begingroup$ @Farcher Merely an assumption? $\endgroup$ – Kunal Pawar Apr 13 '17 at 7:26
  • $\begingroup$ There is an effect. sjsu.edu/faculty/watkins/scattering2.htm $\endgroup$ – Farcher Apr 13 '17 at 7:49
  • 2
    $\begingroup$ @Farcher Using a fixed center assumption to build the comparison cross-section doesn't mean he assumed the mass in a vacuum. If you assume that the alpha interacts elastically with only one atom., then scattering angle much larger than 90 degrees force the mass ratio on you. This is one of many powerful results from the classical theory of scattering that relies only on the conservation of energy and momentum (a subject that tend to get short shrift in the pressure of a modern curriculum, but they were well known). $\endgroup$ – dmckee Apr 13 '17 at 15:12
  • 2
    $\begingroup$ Since the alpha particles rebounded in many of the cases it is certain that most of the positive charge was concentrated in a small volume. What puzzles me is how was it ascertained that that same region which contained the positive charge was also the one which carried most of the mass of the atom. Couldn't the mass have been thought to be distributed uniformly through the atom? Surely the mass isn't as significant is it? There is no actual collision taking place? The alpha particles comes as close as it can without actual touching the 'nucleus'. $\endgroup$ – Kunal Pawar Apr 13 '17 at 16:28
39
$\begingroup$

This is a good example of how Science works.

Geiger and Marsden observed that some of the alpha particles were being backscattered. This is inconceivable if the alpha particle is scattered by a lighter particle.

If one considers a particle of mass $m$ and initial velocity $v_1$ striking a target of mass $m'$ at rest, without changing its direction, then its final velocity $v_2$ can assume two possible values, $$v_2=v_1,\quad\mathrm{or}\quad v_2=-v_1\left(\frac{m'-m}{m'+m}\right).$$ The second solution gives that backscattering is only possible if the target has greater mass than the incident particle. By the time, the mass of the electron was known to be much smaller than the mass of the alpha particle so a backscattering event would imply that the scattering centers were in fact heavy positive nuclei. And indeed those scatterings were observed.

To formally check this, Rutherford obtained a formula for the number of scattered particles as a function of the scattering angle using the hypothesis of heavy nuclei (which is justified by the above paragraph). Geiger and Marsden did the experiment and the data agreed with Rutherford formula.

$\endgroup$
  • 1
    $\begingroup$ This. Classical scattering theory forces you to understand the nucleus as considerably heavier than the alpha. $\endgroup$ – dmckee Apr 13 '17 at 15:12
  • $\begingroup$ There is no actual collision taking place is there? The alpha particle never actually touched the nucleus. It just came as close as its energy permitted and turned around (if it were destined for a head-on collision) or just scattered. $\endgroup$ – Kunal Pawar Apr 13 '17 at 16:34
  • 1
    $\begingroup$ @KunalPawar A collision or crash is an event in which two or more bodies exert forces on each other for a relatively short time. Although the most common colloquial use of the word "collision" refers to incidents in which two or more objects collide, the scientific use of the word "collision" implies nothing about the magnitude of the force. $\endgroup$ – Shashaank Apr 13 '17 at 18:01
  • 10
    $\begingroup$ @KunalPawar: No electrons or nuclei make contact with each other when two billiard balls collide, either. The forces involved in both cases are the same. $\endgroup$ – Michael Seifert Apr 13 '17 at 19:46
36
$\begingroup$

Wikipedia explains this rather well but I'll pick out the relevant stuff for you.

Before the Geiger–Marsden experiment, the general idea was that atoms were built of some permeable positive substrate in which some negative particles were floating around; the so called plum-pudding model.

If we shoot $\alpha$ particles on this setup they should all pass through the atoms since the positive substrate is thought to be permeable! (left side of the figure)

But when people did the experiment they saw that most particles went through while some scattered $180^\circ$ backwards, some even bent a small angle! (right side of the figure)

Experiment

The plum-pudding model had no problem with the particles that went through undisturbed but what about the ones that were backscattered? People theorized that there must be some solid core in the atom against which the $\alpha$ particles could scatter. The core couldn't be too big since only a small fraction of the $\alpha$ particles backscattered.

This leads to a model in which most of the mass (to which $\alpha$ particles can scatter) are in the center of the atom with the negative charges around it!

$\endgroup$
8
$\begingroup$

Rutherford performed a series of experiments where he measured back scattering of $\alpha$ particles a function of thickness of metal foil. In the third part of the experiment he observed that around 1 in 8000 $\alpha$ particles were able to back scatter. This was possible only if it encountered large Electric field. Such large electric field can't be created by distributed charge, but has to be created by some concentrated charge. So this explains the concept of nucleus.

And since nucleus has most of the charge (protons are heavier than electron) it has most of the mass of the atom.

$\endgroup$
  • $\begingroup$ This doesn't answer the question as posed. At the time when Rutherford postulated his model, the proton was unknown, let alone its mass. $\endgroup$ – Emilio Pisanty Apr 13 '17 at 12:02
  • 1
    $\begingroup$ @ Emilio Pisanty I agree that proton was not discovered until 1920 and Rutherford model actually supports the fact that nucleus could have carried +ve as well as -ve charge. But since $\alpha$ particles were deflected, nucleus had to be +ve charge. It was also observed that scattering angle where maximum $\alpha$ particles were detected was directly proportional to the atomic mass of metal foil. This supports the fact that indeed nucleus had most of the mass. If mass was not a factor then, scattering angle would not be a function of atomic mass where we detect maximum $\alpha$ particles. $\endgroup$ – Draco_1125 Apr 13 '17 at 12:23
  • 1
    $\begingroup$ That $\alpha$ particles being deflected isn't the reason for concluding the nucleus is positive. IIRC, the scattering pattern is not that sensitive to the charge of the targets. Positivity of the nucleus follows from matter being electrically neutral and electrons being negative. $\endgroup$ – Sean E. Lake Apr 13 '17 at 12:29

protected by Qmechanic Apr 14 '17 at 13:50

Thank you for your interest in this question. Because it has attracted low-quality or spam answers that had to be removed, posting an answer now requires 10 reputation on this site (the association bonus does not count).

Would you like to answer one of these unanswered questions instead?

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