In particle physics, Rutherford scattering is the elastic scattering of charged particles by the Coulomb interaction. It is a physical phenomenon explained by Ernest Rutherford in 1911^[1] that led to the development of the planetary Rutherford model of the atom and eventually the Bohr model. Rutherford scattering was first referred to as Coulomb scattering because it relies only upon the static electric (Coulomb) potential, and the minimum distance between particles is set entirely by this potential. The classical Rutherford scattering process of alpha particles against gold nuclei is an example of "elastic scattering" because neither the alpha particles nor the gold nuclei are internally excited. The Rutherford formula (see below) further neglects the recoil kinetic energy of the massive target nucleus.

Rutherford scattering is now exploited by the materials science community in an analytical technique called Rutherford backscattering.

Key experiments

Hans Geiger, working in Rutherford's lab, did a series experiments in 1908 showing that alpha particles are "scattered" as they pass through thin layers of mica, and foils of gold and aluminum. In following year, joined by undergraduate Ernest Marsden, they did a series of experiments to untangle confusing results they observed.^[2]: 263

A critical discovery was made by Geiger and Marsden in 1909 when they performed the gold foil experiment in collaboration with Rutherford, in which they fired a beam of alpha particles (helium nuclei) at foils of gold leaf.^[3] At the time of the experiment, the atom was thought to be analogous to a plum pudding (as proposed by J. J. Thomson), with the negatively-charged electrons (the plums) studded throughout a positive spherical matrix (the pudding). The scattering in this model was proposed to occur by many repeated collisions with the electrons and the alpha particles should only be deflected by small angles as they pass through.

However, the intriguing results showed that around 1 in 8,000 ^[2]: 264 alpha particles were deflected by very large angles (over 90°), while the rest passed through with little deflection. From this, Rutherford concluded that the majority of the mass was concentrated in a minute, positively-charged region (the nucleus) surrounded by electrons. When a (positive) alpha particle approached sufficiently close to the nucleus, it was repelled strongly enough to rebound at high angles. The small size of the nucleus explained the small number of alpha particles that were repelled in this way. Rutherford showed, using the method outlined below, that the size of the nucleus was less than about 10⁻¹⁴ m (how much less than this size, Rutherford could not tell from this experiment alone; see more below on this problem of lowest possible size). As a visual example, Figure 1 shows the deflection of an alpha particle by a nucleus in the gas of a cloud chamber.

Maximum nuclear size estimate

Rutherford begins his 1911 paper with 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.

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. Where the particle stops, a distance ${\displaystyle r_{{\textrm {m}}in}}$ the potential energy matches the original kinetic energy:^{[4]: 620 [5]: 320}

$${\displaystyle {\frac {1}{2}}mv^{2}={\frac {1}{4\pi \epsilon _{0}}}\cdot {\frac {q_{1}q_{2}}{r_{\text{min}}}}}$$

Rearranging: $${\displaystyle r_{\text{min}}={\frac {1}{4\pi \epsilon _{0}}}\cdot {\frac {2q_{1}q_{2}}{mv^{2}}}}$$

For an alpha particle:

• m (mass) = 6.64424×10⁻²⁷ kg = 3.7273×10⁹ eV/c²
• q₁ (for helium) = 2 × 1.6×10⁻¹⁹ C = 3.2×10⁻¹⁹ C
• q₂ (for gold) = 79 × 1.6×10⁻¹⁹ C = 1.27×10⁻¹⁷ C
• v (initial velocity) = 2×10⁷ m/s (for this example)

The distance from the alpha particle to the center of the nucleus (r_min) at this point is an upper limit for the nuclear radius. Substituting these in gives the value of about 2.7×10⁻¹⁴ m, or 27 fm. (The true radius is about 7.3 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 fm of the nuclear center, as noted, when the actual radius of gold is 7.3 fm.

Rutherford's 1911 paper^[1] started with a slightly different formula suitable for head-on collision with a sphere of positive charge: $${\displaystyle {\frac {1}{2}}mv^{2}=NeE\cdot ({\frac {1}{b}}-{\frac {3}{2R}}+{\frac {b^{2}}{2R^{3}}})}$$ Rutherford used ${\displaystyle b}$ as the turning point distance called r_min above and ${\displaystyle R}$ 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_min found above.

Single scattering from heavy nuclei

From his results for a head on collision, Rutherford knows that alpha particle scattering occurs close to the center of an atom, at a radius 10,000 times smaller than the atom. Therefore he ignores the effect of "negative electricity". Furthermore 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.^[1]: 672

Rutherford observed that the alpha particle will take a hyperbolic trajectory in the repulsive force near the center of the atom as shown in Fig. 1. He derives his scattering formula by starting with conservation of angular momentum. When the particle of mass ${\displaystyle m_{\alpha }}$ and velocity ${\displaystyle v_{0}}$ is far from the atom, its angular momentum around the center of the atom will be ${\displaystyle m_{\alpha }pv_{0}}$ where ${\displaystyle p}$ is the perpendicular distance between the incoming particle path and the atom, now called the impact parameter. At the point of closest approach, labeled A in the Fig. 1, the angular momentum will be ${\displaystyle m_{\alpha }r_{A}v_{A}}$. Thus Rutherford (in a slightly different notation^[a]) equates $${\displaystyle pv_{0}=r_{A}v_{A}}$$ Next he uses conservation of energy at these two points: $${\displaystyle {\tfrac {1}{2}}m_{\alpha }v_{0}^{2}={\tfrac {1}{2}}m_{\alpha }v_{A}^{2}-{\frac {NeE}{r_{A}}}}$$ 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 particle and atom at the point of closest approach (${\displaystyle A}$)

Next Rutherford rearranges the energy equation and divides by half the mass: $${\displaystyle v_{A}^{2}=v_{0}^{2}(1-{\frac {b}{r_{A}}})}$$ In the same step he implicitly introduces a variable ${\displaystyle b}$ containing the non-geometrical physical constants of the problem: $${\displaystyle b=-{\frac {NeE}{{\tfrac {1}{2}}m_{\alpha }v_{0}^{2}}}.}$$ (This value is the closest approach that Rutherford estimated as ${\displaystyle b\approx 3.4\times 10^{-12}{\textrm {cm}}}$ earlier in the paper.)

Having assumed an inverse square law force between the alpha particle and the compact massive charge, Rutherford can use known results from Newtonian gravity, which also an inverse square law.^[6]: 151 The appropriate orbit will by hyperbolic since the alpha particle has kinetic energy far from the charge. The hyperbola can be written in polar coordinates with the origin between the two foci as $${\displaystyle {\frac {r'}{b'}}={\frac {1}{\sqrt {e^{2}\cos ^{2}\varphi '-1}}}.\,}$$ where ${\displaystyle b'}$ is the semiminor axis and ${\displaystyle e}$ is the eccentricity. As the alpha particle scatters, the angle swings between two extreme values, ${\displaystyle \varphi '_{\text{in}}\rightarrow \varphi '_{\text{out}}.}$ These angles correspond to values where the denominator is zero: $${\displaystyle e^{2}\cos ^{2}\varphi '_{\text{in}}-1=0\implies \cos \varphi '_{\text{in}}=1/e.}$$Zdroj:https://en.wikipedia.org?pojem=Rutherford_scattering
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