Editing Rutherford scattering experiments (section)

== Rutherford's scattering model{{anchor|Rutherford scattering}}<!-- Rutherford scattering redirects here --> ==
{{Split section|Coulomb scattering |discuss={{TALKPAGENAME}}#Splitting proposal |date=February 2025}}


Rutherford begins his 1911 paper<ref name="Rutherford 1911"/> with a discussion of Thomson's results on scattering of [[beta decay|beta particles]], a form of radioactivity that results in high velocity electrons. Thomson's model had electrons circulating inside of a sphere of positive charge. Rutherford highlights the need for compound or multiple scattering events: the deflections predicted for each collision are much less than one degree. He then proposes a model which will produce large deflections on a single encounter: place all of the positive charge at the centre of the sphere and ignore the electron scattering as insignificant. The concentrated charge will explain why most alpha particles do not scatter to any measurable degree – they fly past too far from the charge – and yet particles that do pass very close to the centre scatter through large angles.<ref name=Heilbron1968/>{{rp|285}}

===Maximum nuclear size estimate===
Rutherford begins his analysis by considering a head-on collision between the alpha particle and atom. This will establish the minimum distance between them, a value which will be used throughout his calculations.<ref name="Rutherford 1911"/>{{rp|670}}

Assuming there are no external forces and that initially the alpha particles are far from the nucleus, the [[inverse-square law]] between the charges on the alpha particle and nucleus gives the potential energy gained by the particle as it approaches the nucleus. For head-on collisions between alpha particles and the nucleus, all the [[kinetic energy]] of the alpha particle is turned into [[potential energy]] and the particle stops and turns back.<ref name="BelyaevRoss2021"/>{{rp|5}} 
[[File:Rutherford model rest distance.svg|thumb|center|upright=2|Schematic view of a head-on collision between an alpha particle and an atom. The radius of the atom is on the order of 10<sup>−10</sup>&nbsp;m and the minimum stopping distance is on the order of 10<sup>−14</sup>&nbsp;m.]]

Where the particle stops at a distance <math>r_{\text{min}}</math> from the centre, the potential energy matches the original kinetic energy:<ref>[https://archive.org/details/in.ernet.dli.2015.60140/page/n647/mode/2up?q=appendix "Electrons (+ and -), Protons, Photons, Neutrons, Mesotrons and Cosmic Rays"] 
By Robert Andrews Millikan. Revised edition. Pp. x+642. (Chicago: University of Chicago Press; London: Cambridge University Press, 1947.)</ref>{{rp|620}}<ref name=Cooper1970>Cooper,&nbsp;L.&nbsp;N.&nbsp;(1970).&nbsp;[https://archive.org/details/introductiontom00coop/page/320/mode/2up "An Introduction to the Meaning and Structure of Physics"].&nbsp;Japan:&nbsp;Harper & Row.</ref>{{rp|320}}

<math display="block">\frac{1}{2} mv^2 = k \frac{q_\text{a} q_\text{g}}{r_\text{min}}</math>

where

<math display="block">k = \frac{1}{4\pi \epsilon_0}</math>

Rearranging:<ref name="Rutherford 1911"/>{{rp|671|q=It will be seen that b is an important quantity in later calculations}} 
<math display="block">r_\text{min} = k \frac{2 q_\text{a} q_\text{g}}{mv^2}</math>

For an alpha particle:
* {{math|''m''}} (mass) = {{val|6.64424e-27|u=kg}} = {{val|3.7273e9|u=eV/''c''<sup>2</sup>}}
* {{math|''q''<sub>a</sub>}} (for the alpha particle) = 2 × {{val|1.6e-19|u=C}} = {{val|3.2e-19|u=C}}
* {{math|''q''<sub>g</sub>}} (for gold) = 79 × {{val|1.6e-19|u=C}} = {{val|1.27e-17|u=C}}
* {{math|''v''}} (initial velocity) = {{val|2e7|u=m/s}} (for this example)
The distance from the alpha particle to the centre of the nucleus ({{math|''r''<sub>min</sub>}}) at this point is an upper limit for the nuclear radius.
Substituting these in gives the value of about {{val|2.7e-14|u=m}}, or 27&nbsp;[[femtometre|fm]]. (The true radius is about 7.3&nbsp;fm.)  The true radius of the nucleus is not recovered in these experiments because the alphas do not have enough energy to penetrate to more than 27&nbsp;fm of the nuclear centre, as noted, when the  radius of the nucleus of a gold atom is 7.3&nbsp;fm.
[[File:RutherfordConcentrated.png|thumb|Figure 1. Potential energy diagram for Rutherford's atom model illustrating concentration in the nucleus.]]
Rutherford's 1911 paper<ref name="Rutherford 1911"/> started with a slightly different formula suitable for head-on collision with a sphere of positive charge:

<math display="block">\frac{1}{2}mv^2 = NeE \cdot \left (\frac{1}{b} - \frac{3}{2R} + \frac{b^2}{2R^3} \right )</math>

In Rutherford's notation, ''e'' is the [[elementary charge]], ''N'' is the charge number of the nucleus (now also known as the [[atomic number]]), and ''E'' is the charge of an alpha particle. The convention in Rutherford's time was to measure charge in [[electrostatic units]], distance in centimeters, force in [[dyne]]s, and energy in [[erg]]s. The modern convention is to measure charge in [[coulomb]]s, distance in meters, force in newtons, and energy in [[joule]]s. Using coulombs requires using the [[Coulomb constant]] (''k'') in the equation. Rutherford used ''b'' as the turning point distance (called ''r''<sub>min</sub> above) and ''R'' is the radius of the atom. The first term is the Coulomb repulsion used above. This form assumes the alpha particle could penetrate the positive charge. At the time of Rutherford's paper, Thomson's [[plum pudding model]] proposed a positive charge with the radius of an atom, thousands of times larger than the ''r''<sub>min</sub> found above. Figure 1 shows how concentrated this potential is compared to the size of the atom.
Many of Rutherford's results are expressed in terms of this turning point distance ''r''<sub>min</sub>, simplifying the results and limiting the need for units to this calculation of turning point.
{{Clear}}

===Single scattering by a heavy nucleus {{anchor|Single scattering from a heavy nucleus}} ===
From his results for a head on collision, Rutherford knows that alpha particle scattering occurs close to the centre of an atom, at a radius 10,000 times smaller than the atom. The electrons have negligible effect. He begins by assuming no energy loss in the collision, that is he ignores the recoil of the target atom. He will revisit each of these issues later in his paper.<ref name="Rutherford 1911"/>{{rp|672}}
Under these conditions, the alpha particle and atom interact through a [[Classical central-force problem|central force]], a physical problem studied first by [[Isaac Newton]].<ref>{{Cite journal |last=Speiser |first=David |date=1996 |title=The Kepler Problem from Newton to Johann Bernoulli |url=http://link.springer.com/10.1007/BF02327155 |journal=Archive for History of Exact Sciences |language=en |volume=50 |issue=2 |pages=103–116 |doi=10.1007/BF02327155 |bibcode=1996AHES...50..103S |issn=0003-9519|url-access=subscription }}</ref> 
A central force only acts along a line between the particles and when the force varies with the inverse square, like [[Coulomb's law|Coulomb force]] in this case, a detailed theory was developed under the name of the [[Kepler problem]].<ref name=Goldstein1st/>{{rp|76}} The well-known solutions to the Kepler problem are called [[orbits]] and unbound orbits are [[hyperbolas]].
Thus Rutherford proposed that the alpha particle will take a [[hyperbolic trajectory]] in the repulsive force near the centre of the atom as shown in Figure 2.

[[File:Rutherford scattering geometry 2.svg|thumb|center|upright=2|'''Figure 2.''' The geometry of Rutherford's scattering formula, based on a diagram in his 1911 paper. The alpha particle is the green dot and moves along the green path, which is a hyperbola with O as its centre and S as its external focus. The atomic nucleus is located at S. A is the [[apsis]], the point of closest approach. ''b'' is the impact parameter, the lateral distance between the alpha particle's initial trajectory and the nucleus.]]

To apply the hyperbolic trajectory solutions to the alpha particle problem, Rutherford expresses the parameters of the hyperbola in terms of the scattering geometry and energies. He starts with [[conservation of angular momentum]]. When the particle of mass <math>m</math> and initial velocity <math>v_0</math> is far from the atom, its angular momentum around the centre of the atom will be <math>m b v_0</math> where <math>b</math> is the [[impact parameter]], which is the lateral distance between the alpha particle's path and the atom. At the point of closest approach, labeled A in Figure 2, the angular momentum will be <math>m r_\text{A} v_\text{A}</math>. Therefore<ref name=Heilbron1968/>{{rp|270}}

<math display="block">m b v_0 = m r_\text{A} v_\text{A}</math>

<math display="block">v_\text{A} = \frac{b v_0}{r_\text{A}}</math>

Rutherford also applies the law of [[conservation of energy]] between the same two points:

<math display="block">\tfrac{1}{2}m v_0^2 = \tfrac{1}{2} m v_\text{A}^2 + \frac{k q_\text{a} q_\text{g}}{r_\text{A}}</math>

The left hand side and the first term on the right hand side are the kinetic energies of the particle at the two points; the last term is the potential energy due to the Coulomb force between the alpha particle and atom at the point of closest approach (A). ''q''<sub>a</sub> is the charge of the alpha particle, ''q''<sub>g</sub> is the charge of the nucleus, and ''k'' is the [[Coulomb constant]].{{refn|These equations are in [[SI]] units Rutherford<ref name=Rutherford1911/>{{rp|673}} uses [[cgs units]]}}

The energy equation can then be rearranged thus:

<math display="block"> v_\text{A}^2 =  v_0^2 \left (1 - \frac{k q_\text{a} q_\text{g}}{\tfrac{1}{2} m v_0^2 r_\text{A}} \right)</math>

For convenience, the non-geometric physical variables in this equation<ref name="Rutherford 1911"/>{{rp|674}} can be contained in a variable <math>r_\text{min}</math>, which is the point of closest approach in a head-on collision scenario<ref name="Rutherford 1911"/>{{rp|671|q=It will be seen that b is an important quantity in later calculations}} which was explored in a previous section of this article:

<math display="block">r_\text{min} = \frac{k q_\text{a} q_\text{g}}{\tfrac{1}{2} m v_0^2}</math>

This allows Rutherford simplify the energy equation to:

<math display="block"> v_\text{A}^2 =  v_0^2 \left (1 - \frac{r_\text{min}}{r_\text{A}} \right)</math>

This leaves two simultaneous equations for <math>v_\text{A}^2</math>, the first derived from the conservation of momentum equation and the second from the conservation of energy equation. Eliminating <math>v_\text{A}</math> and <math>v_0</math> gives at a new formula for <math>r_\text{min}</math>:

<math display="block">v_\text{A}^2 =  \frac{b^2v_0^2}{r_\text{A}^2} = v_0^2 \left (1 - \frac{r_\text{min}}{r_\text{A}} \right)</math>

<math display="block">r_\text{min} = r_\text{A} - \frac{b^2}{r_\text{A}}</math>

The next step is to find a formula for <math>r_\text{A}</math>.  From Figure 2, <math>r_\text{A} </math> is the sum of two distances related to the hyperbola, SO and OA.  Using the following logic, these distances can be expressed in terms of angle <math>\Phi</math> and impact parameter <math>b</math>.

[[File:Hyperbel-def-ass-e 2.svg|right|thumb|upright=2|'''Figure 3''' The basic components of the geometry of [[hyperbola]]s.]]
The [[Eccentricity (mathematics)|eccentricity]] of a hyperbola is a value that describes the hyperbola's shape. It can be calculated by dividing the focal distance by the length of the semi-major axis, which per Figure 2 is {{sfrac|SO|OA}}. As can be seen in Figure 3, the [[Hyperbola#Polar coordinate eccentricity|eccentricity is also equal to <math>\sec\Phi</math>]], where <math>\Phi</math> is the angle between the major axis and the asymptote.<ref name=Casey1885>Casey, John, (1885) [https://archive.org/details/cu31924001520455/page/n219/mode/2up "A treatise on the analytical geometry of the point, line, circle, and conic sections, containing an account of its most recent extensions, with numerous examples"]</ref>{{rp|219}} Therefore:

<math display="block">\frac{\text{SO}}{\text{OA}} = \sec\Phi</math>

As can be deduced from Figure 2, the focal distance SO is

<math display="block"> \text{SO} = b \cdot \csc\Phi</math>

and therefore

<math display="block"> \text{OA} = \frac{\text{SO}}{\sec\Phi} = b \cdot \cot\Phi</math>

With these formulas for SO and OA, the distance <math>r_\text{A}</math> can be written in terms of <math>\Phi</math> and simplified using a trigonometric identity known as a [[List of trigonometric identities#Half-angle formulae|half-angle formula]]:<ref name=Rutherford1911/>{{rp|673}}

<math display="block">r_\text{A} = \text{SO} + \text{OA}</math>

<math display="block">= b \cdot \csc\Phi + b \cdot \cot\Phi</math>

<math display="block">= b \cdot \cot\frac{\Phi}{2}</math>

Applying a trigonometric identity known as the [[List of trigonometric identities#Double-angle formulae|cotangent double angle formula]] and the previous equation for <math>r_\text{A}</math> gives a  simpler relationship between the physical and geometric variables:

<math display="block">r_\text{min} = r_\text{A} - \frac{b^2}{r_\text{A}}</math>

<math display="block">= b\cdot\cot\frac{\Phi}{2} - \frac{b^2}{b\cdot\cot\frac{\Phi}{2}}</math>

<math display="block">= b \frac{\cot^2\frac{\Phi}{2} - 1}{\cot\frac{\Phi}{2}}</math>

<math display="block">= 2 b \cdot \cot \Phi</math>

The scattering angle of the particle is <math>\theta = \pi - 2 \Phi</math> and therefore <math>\Phi = \tfrac{\pi - \theta}{2}</math>. With the help of a trigonometric identity known as a [[List of trigonometric identities#Reflections|reflection formula]], the relationship between ''θ'' and ''b'' can be resolved to:<ref name="Rutherford 1911"/>{{rp|673}}

<math display="block">r_\text{min} = 2b \cdot \cot \frac{\pi - \theta}{2}</math>

<math display="block">= 2b\cdot\tan \frac{\theta}{2}</math>

<math display="block"> \cot\frac{\theta}{2} = \frac{2b}{r_\text{min}}</math>

which can be rearranged to give

<math display="block">\theta = 2 \arctan \frac{r_\text{min}}{2b} = 2 \arctan \frac{k q_\text{a} q_\text{g}}{m v_0^2 b}</math>
[[File:RutherfordHyperbolas.png|thumb|right|Hyperbolic trajectories of alpha particles scattering from a [[Gold]] nucleus (modern radius shown as gray circle) as described in Rutherford's 1911 paper]]

Rutherford gives some illustrative values as shown in this table:<ref name="Rutherford 1911"/>{{rp|673}}
{| class="wikitable"
|+ Rutherford's angle of deviation table
|-
| <math>\tfrac{b}{r_\text{min}}</math> || 10 || 5 || 2 || 1 || 0.5 || 0.25 || 0.125
|-
|  <math>\theta</math>  || 5.7° || 11.4° || 28° || 53° || 90° || 127° || 152°
|}

Rutherford's approach to this scattering problem remains a standard treatment in textbooks<ref name=HandFinch3rd>{{Cite book |last1=Hand |first1=Louis N. |url=https://www.cambridge.org/highereducation/books/analytical-mechanics/7145C0BF476388E6841B44064515909E |title=Analytical Mechanics |last2=Finch |first2=Janet D. |date=1998-11-13 |language=en |doi=10.1017/cbo9780511801662|isbn=978-0-521-57572-0 }}</ref>{{rp|151}}<ref>{{Cite book |last1=Fowles |first1=Grant R. |title=Analytical mechanics |last2=Cassiday |first2=George L. |date=1993 |publisher=Saunders College Pub |isbn=978-0-03-096022-2 |edition=5th |series=Saunders golden sunburst series |location=Fort Worth}}</ref>{{rp|240}}<ref>{{Cite journal |last1=Webber |first1=B.R. |last2=Davis |first2=E.A. |date=February 2012 |title=Commentary on 'The scattering of α and β particles by matter and the structure of the atom' by E. Rutherford (''Philosophical Magazine'' 21 (1911) 669–688) |url=http://www.tandfonline.com/doi/abs/10.1080/14786435.2011.614643 |journal=Philosophical Magazine |language=en |volume=92 |issue=4 |pages=399–405 |doi=10.1080/14786435.2011.614643 |bibcode=2012PMag...92..399W |issn=1478-6435|url-access=subscription }}</ref>{{rp|400|q=Rutherford's theory of scattering, assuming an inverse square law of electrostatic repulsion from a central fixed charge, is now textbook physics. }} on [[classical mechanics]].
{{clear}}

===Intensity vs angle===
[[File:ScatteringDiagram.svg|thumb|Geometry of differential scattering cross-section]]
To compare to experiments the relationship between impact parameter and scattering angle needs to be converted to probability versus angle.  The scattering cross section gives the relative intensity by  angles:<ref name=Goldstein1st/>{{rp|81}}
<math display="block">\frac{\mathrm{d} \sigma}{\mathrm{d} \Omega}(\Omega) \mathrm{d} \Omega = \frac{\text{number of particles scattered into solid angle } \mathrm{d} \Omega \text{ per unit time}}{\text{incident intensity}}</math>

In classical mechanics, the scattering angle <math>\theta</math> is uniquely determined the initial kinetic energy of the incoming particles and the impact parameter {{math|''b''}}.<ref name=Goldstein1st/>{{rp|82}} Therefore, the number of particles scattered into an angle between <math>\theta</math> and <math>\theta + \mathrm{d}\theta</math> must be the same as the number of particles with associated impact parameters between {{math|''b''}} and {{math|''b'' + ''db''}}. For an incident intensity {{math|''I''}}, this implies:

<math display="block">2\pi I b \cdot \left|\mathrm{d}b\right| =-2 \pi \cdot \sigma (\theta) \cdot I \cdot \sin(\theta) \cdot \mathrm{d}\theta </math> 

Thus the cross section depends on scattering angle as:

<math display="block">\sigma (\theta) = - \frac{b}{\sin\theta} \cdot \frac{\mathrm{d}b}{\mathrm{d}\theta} </math> 

Using the impact parameter as a function of angle, {{math|''b''(''θ'')}}, from the single scattering result above produces the Rutherford scattering cross section:<ref name=Goldstein1st/>{{rp|84}}

[[File:Rutherford's scattering equation illustrated.svg|thumb]]
<math display="block">
s = \frac {Xnt\cdot \csc^4{\frac{\phi}{2}}}{16r^2} \cdot {\left(\frac {2 k q_\text{n} q_\text{a}}{mv^2}\right)}^2
</math>

*''s'' = the number of alpha particles falling on unit area at an angle of deflection ''Φ''
*''r'' = distance from point of incidence of α rays on scattering material
*''X'' = total number of particles falling on the scattering material
*''n'' = number of atoms in a unit volume of the material
*''t'' = thickness of the foil
*''q''<sub>n</sub> = positive charge of the atomic nucleus
*''q''<sub>a</sub> = positive charge of the alpha particles
*''m'' = mass of an alpha particle
*''v'' = velocity of the alpha particle

[[File:RutherfordCrosssection2Scales.png|thumb|Rutherford scattering cross-section is strongly peaked around zero degrees, and yet has nonzero values out to 180 degrees.]]
This formula predicted the results that Geiger measured in the coming year. The scattering probability into small angles greatly exceeds the probability in to larger angles, reflecting the tiny nucleus surrounded by empty space. However, for rare close encounters, large angle scattering occurs with just a single target.<ref>Karplus, Martin, and Richard Needham Porter. "Atoms and molecules; an introduction for students of physical chemistry." Atoms and molecules; an introduction for students of physical chemistry (1970).</ref>{{rp|19}}

At the end of his development of the cross section formula, Rutherford emphasises that the results apply to single scattering and thus require measurements with thin foils. For thin foils the degree of scattering is proportional to the foil thickness in agreement with Geiger's measurements.<ref name="Rutherford 1911"/>

=== Comparison to JJ Thomson's results ===
{{See also| Plum_pudding_model#Thomson's_1910_scattering_model}}
In 1910, Thomson presented a model for [[beta particle]] scattering<ref name="ThomsonScattering1910">{{cite journal |author=J. J. Thomson |year=1910 |title=On the Scattering of rapidly moving Electrified Particles |journal=Proceedings of the Cambridge Philosophical Society |volume=15 |pages=465–471 |url=https://archive.org/details/proceedingsofcam15190810camb/page/464/mode/2up}}</ref> which predicted that in the plum pudding model, a beta particle could be scattered by a significant angle after a series of atomic collisions. Rutherford's model produced stronger scattering by concentrating the positive charge of the atom at a central point rather than spread it over the volume of the atom. Then a collision with just one atom could produce a larger effect on a beta particle than Thomson's model. Rutherford completed his analysis including the effects of density and foil thickness, then concluded that thin foils are governed by single collision scattering, not multiple collision scattering.<ref name=Heilbron1968/>{{rp|298}}

But Thomson's scattering model could not account for large scattering when it came to [[alpha particle]]s, which have much more momentum than [[beta particle]]s.{{dubious|date=March 2025}}  Even with multiple collisions, the possibility of an alpha particle being deflected by any measurable amount is so low as to be meaningless.

=== Target recoil ===
Rutherford's analysis assumed that alpha particle trajectories turned at the centre of the atom but the exit velocity was not reduced.<ref name=GilibertiLovisetti/>{{rp|253}} This is equivalent to assuming that the concentrated charge at the centre had infinite mass or was anchored in place.  Rutherford discusses the limitations of this assumption by comparing scattering from lighter atoms like aluminium with heavier atoms like gold. If the concentrated charge is lighter it will recoil from the interaction, gaining momentum while the alpha particle loses momentum and consequently slows down.<ref name=Rutherford1911/>{{rp|676|q=It is seen that the reduction of velocity of the alpha particle becomes marked on this theory for encounters with the lighter atoms.}}

Modern treatments analyze this type of Coulomb scattering in the [[centre of mass]] reference frame.  The six coordinates of the two particles (also called "bodies") are converted into three relative coordinates between the two particles and three centre-of-mass coordinates moving in space (called the lab frame). The interaction only occurs in the relative coordinates, giving an equivalent one-body problem<ref name=Goldstein1st/>{{rp|58}} just as Rutherford solved, but with different interpretations for the mass and scattering angle.

Rather than the mass of the alpha particle, the more accurate formula including recoil uses [[reduced mass]]:<ref name=Goldstein1st/>{{rp|80}}

<math display="block">\mu = \cfrac{m_1 m_2}{m_1 + m_2}.</math> 

For Rutherford's alpha particle scattering from gold, with mass of 197, the reduced mass is very close to the mass of the alpha particle:

<math display="block">\mu_\text{Au} = \cfrac{4\times 197}{4 + 197} = 3.92 \approx 4</math>

For lighter aluminium, with mass 27, the effect is greater:

<math display="block">\mu_\text{Al} = \cfrac{4\times 27}{4 + 27} = 3.48 </math>

a 13% difference in mass.  Rutherford notes this difference and suggests experiments be performed with lighter atoms.<ref name="Rutherford 1911"/>{{rp|677}}

The second effect is a change in scattering angle. The angle in the relative coordinate system or centre of mass frame needs to be converted to an angle in the lab frame.<ref name=Goldstein1st>Goldstein,&nbsp;Herbert.&nbsp;Classical Mechanics.&nbsp;United States,&nbsp;Addison-Wesley,&nbsp;1950.</ref>{{rp|85}}  In the lab frame, denoted by a subscript L, the scattering angle for a general central potential is

<math display="block">\tan \Theta_\text{L} = \frac{\sin\Theta}{\cos\Theta + \frac{m_1}{m_2}}</math>  

For a heavy particle like gold used by Rutherford, the factor <math>\tfrac{m_1}{m_2} = \tfrac{4}{197} \approx 0.02</math> can be neglected at almost all angles. Then the lab and relative angles are the same, <math>\Theta_\text{L} \approx \Theta</math>.

The change in scattering angle alters the formula for differential cross-section needed for comparison to experiment. In general the calculation is complex. For the case of alpha-particle scattering from gold atoms, this effect on the cross section is quite small.<ref name=Goldstein1st/>{{rp|88}}