Editing Rutherford scattering experiments (section)

===Partial deflection by the electrons===
Consider an alpha particle passing through an atom of radius ''R'' along a path of length ''L''. The effect of the positive sphere is ignored so as to isolate the effect of the atomic electrons. As with the positive sphere, deflection by the electrons is expected to be very small, to the point that the path is practically a straight line.

[[File:Thomson model electron scattering.svg|center|thumb|upright=2|'''Figure 6''']]

[[#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.

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

''m'' and ''v'' are the mass and velocity of the incoming particle and ''b'' is the impact parameter.

For the electrons within an arbitrary distance ''s'' of the alpha particle's path, their mean distance will be {{sfrac|s|2}}. Therefore, the average deflection per electron will be

<math display="block">2 \arctan \frac{k q_\text{a} q_\text{e}}{mv^2 \tfrac{s}{2}} \approx \frac{4k q_\text{a} q_\text{e}}{mv^2 s}</math>

where ''q''<sub>e</sub> is the [[elementary charge]]. The average net deflection by all the electrons within this arbitrary cylinder of effect around the alpha particle's path is

<math display="block">\theta_1 = \frac{4k q_\text{a} q_\text{e}}{mv^2 s} \sqrt{N_0 \pi s^2 L}</math>

where ''N''<sub>0</sub> is the number of electrons per unit volume and <math>\pi s^2 L</math> is the volume of this cylinder.

[[File:LbR relationship.svg|border|right|thumb|'''Figure 7''']]

Treating ''L'' as a straight line,  <math>L = 2\sqrt{R^2 - b^2}</math> where ''b'' is the distance of this line from the centre. The mean of <math>\sqrt{L}</math> is therefore

<math display="block">\frac{1}{\pi R^2} \int_0^R \sqrt{2 \sqrt{R^2 - b^2}} \cdot 2\pi b \cdot \mathrm{d}b = \frac{4}{5} \sqrt{2R}</math>

To obtain the mean deflection <math>\bar{\theta}_1</math>, replace <math>\sqrt{L}</math> in the equation for <math>\theta_1</math>:

<math display="block">\bar{\theta}_1 = \frac{4k q_\text{a} q_\text{e}}{mv^2 s} \sqrt{N_0 \pi s^2} \cdot \frac{4}{5} \sqrt{2R}</math>

<math display="block">= \frac{16}{5} \cdot \frac{k q_\text{a} q_\text{e}}{m v^2 R} \sqrt{\frac{3N}{2}}</math>

where ''N'' is the number of electrons in the atom, equal to <math>N_0 \tfrac{4}{3} \pi R^3</math>.