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© The scientific sentence. 2010

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^{6}m = 1 µm).
Geiger and Marson found about 1 on 10^{4} 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
Z_{1}e : Charge : (Z : atomic number, e : electron charge)
P_{i} : Momentum
N_{i} : Number of incident (alpha) particles
T = (1/2)mv_{i}^{i}2 : kinetic energy of projectile ( a )
P_{i} = mv_{i} : initial momentum
L_{i} = mv_{i}b : initial angular momentum
L_{f} = mv_{f}b : initial angular momentum
E_{α} = E _{i} = (1/2) mv_{i}^{2} : The kinetic energy
of the incident particle
E_{α} = (1/2)mv_{i}^{2}
Target nuclei:
M : Mass of the target
Z_{2}e : Charge
P_{f} : Final momentum
Detection parameters:
ε_{0} : Permittivity of space
θ: Scattering angle
φ : The angle between the vector position and
ω = dφ/dt : The angular velocity of the particle at any moment
the axis of the hyperbola
r : Position vector; the radial unit vector (r = 1).
d : Targettodetector distance
L = mvr = mωr^{2} : Angular momentum at any moment
b : Impact parameter
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 = (1/4πε_{0}) Z_{1}Z_{2}e^{2} r/r^{2} (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 = mωr^{2};
L_{i} = L gives: mv_{i}b = mωr^{2}
Hence:
r^{2}/v_{i}b = 1/ω = dt/dφ (4)
We have then:
ΔP = ∫F_{ΔP} dt ( from &infinit to +&infinit)
= ∫F cos φ dt ( from :  (πθ)/2 to + (πθ)/2)
= ∫dφF cos φ (dt/dφ)
= ∫dφF cos φ(r^{2}/v_{i}b)
= (Z_{1}Z_{2}e^{2}/4πε_{0}v_{i}b) &int:cos φdφ ( from :  (π
θ)/2 to + (πθ)/2)
= (Z_{1}Z_{2}e^{2}/4πε_{0}v_{i}b) . 2 cos(θ/2) (5)
Equating the relationships (1) and (5) we get:
b = (Z_{1}Z_{2}e^{2}/4πε_{0}mv_{i}^{2}) ctg(θ/2)
Since E_{α} = (1/2) mv_{i}^{2}, we have then:
b = (1/2)(Z_{1}Z_{2}e^{2}/4πε_{0} E_{α}) ctg(θ/2) (6)
Let's write :
r_{c} = (Z_{1}Z_{2}e^{2}/4πε_{0}) 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} = Z_{1}Z_{2}e^{2}/4πε_{0}r_{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 10^{5} 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 b^{2} (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 sin^{2}(θ/2) (10)
The equation (9) becomes:
dσ/ dΩ = (r_{c}/2 ctg(θ/2)/sinθ) / r_{c}/2 . (1/2 sin^{2}(θ/2) (11)
With sin θ = 2 sin(θ/2) cos&(θ/2), we find:
dσ/dΩ = (r_{c}/4)^{2} 1/sin^{4}(θ/2)
