Rutherford scattering

1. The description of the collision

The main idea, here is the discovery of the proton. Geiger and Marson experimented the collision of α particles (of energy 5.5 Mev) from the Radon R_n(222,86); that is the nuclei of the H_e(4,2) incident on the gold A_u(197,79) target (of width 10^-6m = 1 µm).
Geiger and Marson found about 1 on 10⁴ α are returned back (scattered in the direction θ>π/2).

J.J. Thomson had thought; that is the the atom was like a pulm ( or raisin) pudding, where the pudding was positive charges and raisin electrons. This is the first atomic model that it doesn't work at all.

With the rutherford experiment, It came out that the incident particles had a small amount of deflections and even some particles returned from the target.
The conclusion was that the atom is made up of the positive nucleus and electrons where the biggest part is simply a vacuum. This second model gave a good understanding about the atom.

Features of the collision:
Projectile:
m : Mass of the projectile
v_i : Initial speed of the projectile
v_f : Final speed of the projectile
Z₁e : Charge : (Z : atomic number, e : electron charge)
P_i : Momentum
N_i : Number of incident (alpha) particles
T = (1/2)mv_iⁱ2 : kinetic energy of projectile ( a )
P_i = mv_i : initial momentum
L_i = mv_ib : initial angular momentum
L_f = mv_fb : initial angular momentum
E_α = E _i = (1/2) mv_i² : The kinetic energy of the incident particle
E_α = (1/2)mv_i²
Target nuclei:
M : Mass of the target
Z₂e : Charge
P_f : Final momentum

The scattering under the influence of the Coulomb force involve the dependence of the scattering θ angle and the impact parameter b which is the crucial parameter for nuclear scattering.

We Consider the M>>m. So,the magnitudes of the initial and final momenta are the same considering their conservation and the assumption that the target recoil is negligible:

Let:
ΔP = ||ΔP|| = ||P_f - P_i|| = 2 mv_i sin(θ/2) (1)
The Coulomb vector force is expressed by:
F = (1/4πε₀) Z₁Z₂e² r/r² (2)
r : is the position of the projectile from the target at any moment.

In view of the symmetry of the scattering geometry, the net change of this force is located in the symmetry axis of the hyperbola. Therefore:
F_ΔP = F cos φ (3)

Recall that F = dP/dt; hence : P = ∫ F dt.
And L:the angular momentum is conserved; that is L_i = L_f = L = mvr = mωr²;
L_i = L gives: mv_ib = mωr²
Hence:
r²/v_ib = 1/ω = dt/dφ (4)

We have then:
ΔP = ∫F_ΔP dt ( from -&infinit to +&infinit)
= ∫F cos φ dt ( from : - (π-θ)/2 to + (π-θ)/2)
= ∫dφF cos φ (dt/dφ)
= ∫dφF cos φ(r²/v_ib)
= (Z₁Z₂e²/4πε₀v_ib) &int:cos φdφ ( from : - (π -θ)/2 to + (π-θ)/2)
= (Z₁Z₂e²/4πε₀v_ib) . 2 cos(θ/2) (5)

Equating the relationships (1) and (5) we get:
b = (Z₁Z₂e²/4πε₀mv_i²) ctg(θ/2)
Since E_α = (1/2) mv_i², we have then:
b = (1/2)(Z₁Z₂e²/4πε₀ E_α) ctg(θ/2) (6)

Let's write :
r_c = (Z₁Z₂e²/4πε₀) x 1/ E_α
That is obtained with φ = 0 and b = 0
In this case, the kinetic enrgy of the incident particle is null; wheras the potential energy takes the expression:
E_p = Z₁Z₂e²/4πε₀r_c

Recall that the size of the nucleus is given by the formula :
r_nucleus = 1.2 A ^1/3. Where A is the mass of the nucleus
For the gold element A = 197, thus r_nucleus = 8 fm
For the current Rutherford collision : r_c ≈ 30 fm
and the Bohr radius ≈0.5 10⁵ fm

The expression of the impact parameter takes then the form:

b = (r_c/2) ctg (θ/2) (7)

wich is the Rutherford formula.

2. The related cross section

The differential cross section gives information about the probability for the insident α to be scattered in a certain direction.
The total cross section is expressed by:
σ = π b²    (8)
Then:
dσ = 2π b db
dΩ = 2πsinθdθ
dσ = 2π b db sinθdθ / sinθdθ = dΩ b db / sinθdθ
Where Ω is the related solid angle.
Hence;
dσ/ dΩ = (b / sinθ)(db /dθ)    (9)
From the equation (7), we get:
|db / dθ| = r_c/2 . (1/2 sin²(θ/2)    (10)

The equation (9) becomes:
dσ/ dΩ = (r_c/2 ctg(θ/2)/sinθ) / r_c/2 . (1/2 sin²(θ/2)
   (11) With sin θ = 2 sin(θ/2) cos&(θ/2), we find:

dσ/dΩ = (r_c/4)² 1/sin⁴(θ/2)