Editing Rutherford scattering experiments (section)

===Partial deflection by the positive sphere===
Consider an alpha particle passing by a sphere of pure positive charge (no electrons) with a radius ''R''. The sphere is so much heavier than the alpha particle that we do not account for recoil. Its position is fixed. The alpha particle passes just close enough to graze the edge of the sphere, which is where the electric field of the sphere is strongest.

[[File:Thomson model alpha particle scattering 4.svg|center|thumb|upright=2|'''Figure 4''']]

[[#Single scattering by a heavy nucleus|An earlier section of this article]] presented an equation which models how an incoming charged particle is deflected by another charged particle at a fixed position (ie infinite mass).

<math display="block">\theta = 2 \arctan {\frac{k q_1 q_2}{m v^2 b}}</math>

This equation can be used to calculate the deflection angle in the special case in Figure 4 by setting the impact parameter ''b'' to the same value as the radius of the sphere ''R''. So long as the alpha particle does not penetrate the sphere, there is no difference between a sphere of charge and a point charge, a mathematical result known as the [[Shell theorem]].

* ''q''<sub>g</sub> = positive charge of the gold atom = {{val|79|u=''q''<sub>e</sub>}} = {{val|1.26|e=-17|u=C}}
* ''q''<sub>a</sub> = charge of the alpha particle = {{val|2|u=''q''<sub>e</sub>}} = {{val|3.20|e=-19|u=C}}
* ''R'' = radius of the gold atom = {{val|1.44|e=-10|u=m}}
* ''v'' = speed of the alpha particle = {{val|1.53|e=7|u=m/s}}
* ''m'' = mass of the alpha particle = {{val|6.64|e=-27|u=kg}}
* ''k'' = [[Coulomb constant]] = {{val|8.987|e=9|u=N·m<sup>2</sup>/C<sup>2</sup>}}

<math display="block">\theta_2 = 2 \arctan {\frac{k q_\text{a} q_\text{g}}{m v^2 R}} \approx 0.02  \text{ degrees}</math>

This shows that the largest possible deflection will be very small, to the point that the path of the alpha particle passing through the positive sphere of a gold atom is almost a straight line.  Therefore in computing the average deflection, which will be smaller still, we will treat the particle's path through the sphere as a [[chord (geometry)|chord]] of length ''L''.

[[File:Thomson model alpha particle scattering 3.svg|center|thumb|upright=2|'''Figure 5''']]

Inside a sphere of uniformly distributed positive charge, the force exerted on the alpha particle at any point along its path through the sphere is<ref>{{cite web | url=http://hyperphysics.phy-astr.gsu.edu/hbase/electric/elesph.html | title=Electric Field, Spherical Geometry }}</ref><ref name=BeiserPerspectives/>{{rp|106}}

<math display="block">F = \frac{k q_\text{a} q_\text{g}}{r^2} \cdot \frac{r^3}{R^3}</math>

The lateral component of this force is

<math display="block">F_\text{y} = \frac{k q_\text{a} q_\text{g}}{r^2} \cdot \frac{r^3}{R^3} \cdot \cos\varphi = \frac{b k q_\text{a} q_\text{g}}{R^3}</math>

The lateral change in momentum ''p''<sub>y</sub> is therefore

<math display="block">\Delta p_\text{y} = F_\text{y} t =\frac{b k q_\text{a} q_\text{g}}{R^3} \cdot \frac{L}{v}</math>

The deflection angle <math>\theta_2</math> is given by

<math display="block">\tan\theta_2 = \frac{\Delta p_\text{y}}{p_\text{x}} = \frac{b k q_\text{a} q_\text{g}}{R^3} \cdot \frac{L}{v} \cdot \frac{1}{mv}</math>

where ''p''<sub>x</sub> is the average horizontal momentum, which is first reduced then restored as horizontal force changes direction as the alpha particle goes across the sphere. Since the deflection is very small, <math>\tan\theta_2</math> can be treated as equal to <math>\theta_2</math>. The chord length <math>L = 2 \sqrt{R^2 - b^2}</math>, per [[Pythagorean theorem]].

The average deflection angle <math>\bar\theta_2</math> sums the angle for values of ''b'' and ''L'' across the entire sphere and divides by the cross-section of the sphere:

<math display="block">\bar\theta_2 = \frac{1}{\pi R^2} \int_0^R \frac{b k q_\text{a} q_\text{g}}{R^3} \cdot \frac{2\sqrt{R^2 - b^2}}{v} \cdot \frac{1}{mv} \cdot 2\pi b \cdot \mathrm{d}b</math>

<math display="block">= \frac{\pi}{4} \cdot \frac{k q_\text{a} q_\text{g}}{mv^2R}</math>

This matches Thomson's formula in his 1910 paper.