(Actual question at the very bottom)

In his 1911 paper "The scattering of α and β particles by matter and the structure of the atom" Rutherford starts with this assumption:

Consider an atom which contains a charge +-Ne at its centre surrounded by a sphere of electrification coutaining a charge -+Ne supposed uniformly distributed throughout a sphere of radius R.

He then derives the following:

In order to from some idea of the forces required to deflect an α particle through a large angle, consider an atom containing a positive charge Ne at its centre, and surrounded by a distribution of negative electricity Ne uniformly distributed within a sphere of radius R. The electric force X and the potential V at a distance r from the centre of an atom for a point inside the atom, are given by

$X = N e (\frac{1}{r^2} - \frac{r}{R^3})$



For the time of 1911, this was an invalid assumption. Also, if the assumption is wrong, you can conclude anything.

There wasn't any quantum mechanics around and one can clearly show that a centre charge closely surrounded by the opposite is an impossible structure for it violates Gauss's law.

To picture this, recall that atoms are neutral. Therefore, for any sphere outside the atom we have:

$\oint_S E \,dA = 0$

Therefore, the field lines of the inner charge must surface, which is impossible if the outer charge is continuous.

Since the distance between the outer charge is also a lot closer, the surfacing inner field would be condensed to almost infinite field strength. The positive field lines also restrict any circulation and pose to many constraints for any wave equation to be solvable, since field lines are known to be continuous.

This image of a stylized (non proportional) neon atom created with an EM simulator using the Runge-Kutta algorithm Github: Electric Field Simulator shows the behavior:

enter image description here

Note, that for this image, Gauss's law holds true. A full 2D circle around this structure would lead to a summation of zero. The near field of this structure, however, extends too far out for atoms to be neutral.

The results can also be obtained by separating the charges next to each other, and then using a linear transformation to map one side into the other.

Such a structure cannot exist under classical mechanics.

However, it wouldn't be the first time, valid results have been arrived via wrong assumptions. But it is rather strange that this escaped the peer review.

Now the question:

Since atoms are not classical mechanics, but quantum mechanics governed, what trait of quantum mechanics turns the clearly invalid classical mechanics structure into a valid quantum one? What makes atoms neutral again and solves the problem of the field lines having to surface and still being continuous?

  • 5
    $\begingroup$ He's trying to model the results seen, not prove they can be reproduced in a self-conisistent way by then existing theory. $\endgroup$ Oct 9, 2021 at 15:30
  • 1
    $\begingroup$ I suspect one could assemble a proof that, so long as the probability of finding an electron very near to the sensor (test charge, photon, etc) is low, the space-and-time average of a symmetric probability distribution approximates the space average of a symmetric point charge distribution which approximates a symmetric area charge distribution, and arguments from Gauss's law are valid. $\endgroup$
    – g s
    Oct 9, 2021 at 15:48
  • $\begingroup$ He has assumed spherical symmetry. Do you understand the constant term in the potential? $\endgroup$ Oct 9, 2021 at 18:57
  • $\begingroup$ What StephenG said. Rutherford was just trying to come up with a better model than Thompson's plum pudding that's consistent with the data from the Geiger-Marsden alpha particle scattering experiments. $\endgroup$
    – PM 2Ring
    Oct 9, 2021 at 22:00
  • $\begingroup$ Bear in mind that at the time, it seemed that physics was best modeled using waves & fields, Newton's corpuscle theory of light was as dead as phlogiston. So reviving corpuscles to explain matter was pretty radical, and the peer reviewers would not have had the required skills to evaluate such material properly. FWIW, Mach never accepted the notion of corpuscular atoms. $\endgroup$
    – PM 2Ring
    Oct 9, 2021 at 22:00

1 Answer 1


Lets make it simple and look at the hydrogen atom:

bohr orbits

which is assumed when drawing the electric field lines in the image in your question, fixed locations for the charges.

Quantum mechanics does not have orbits, but orbitals for a given time t, i.e probability of measuring the electron and the nucleus (x,y,z). This is the measured locations of the electron :

orbital H

If you want to image the classical field lines they would be for that instant


The mathematics of quantum mechanics integrates over the space according to the wave-function, at large distances the positive and negative attraction neutralizes. That is why quantum mechanics is successful in fitting the data.

If you assume that the charges in the image in your question are an instantaneous drawing of orbital points then the same argument holds for neutrality.

For large atoms (and higher energy levels of the hydrogen atom) the orbitals can be such that in some directions the positive nucleus field dominates and in others the negative electrons, which allows for bonding of atoms into molecules and lattices, various working models of chemical bonding exist, afaik.

  • $\begingroup$ I agree that at large distances, the positive and negative attractions neutralize. However, only for a nucleus that is not assumed a point charge. For a point charge, the neutrality is never reached. For a non-zero diameter, the near field will for sure not end within a few diameters of the atom. On collision, they should therefore interact and produce magnetic fields (through rotation). This is not observed. On the contrary, ideal gases behave like billiard balls in their behaviour. How is this explained? $\endgroup$ Oct 9, 2021 at 20:15
  • $\begingroup$ On a second thought, Hydrogen makes it even worse, as you can never achieve neutrality. One side will always be positive and the other negative, depending on the location of the electron. But we do not see Hydrogen radicals as dipoles, do we? $\endgroup$ Oct 9, 2021 at 20:27
  • $\begingroup$ The nucleus is not assumed a point charge, it moves in its own orbitals that happen to be of a very tiny radial size with respect with the radial size of the electrons, in the center of mass system of the atom. $\endgroup$
    – anna v
    Oct 10, 2021 at 3:55
  • $\begingroup$ The neutrality comes from the probability distribution . BECAUSE the Schrodinger model fits all known data . The probability postulate (among others) imposed on the model is CHOSEN so that observations and data are fitted with it. $\endgroup$
    – anna v
    Oct 10, 2021 at 4:00
  • $\begingroup$ "One side will always be positive and the other negative, " There is an equal probability for the same instant in time to have exactly the opposite orientation. The probability distribution is given by the $Ψ^*Ψ$ $\endgroup$
    – anna v
    Oct 10, 2021 at 4:12

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