Cross Sections for Scattering
asymptote
Focus for repulsive force
Θ
vo
asymptote
s
Focus for attractive force
vf
Side View of scattering
s
ds
Front view of scattering
Basic Idea: The probability of being scattered through an angle, Θ ± ½ΔΘ, is equal to the
probability of having an impact parameter, s ± ½Δs. The probability of having an impact
parameter, s ± Δs, is equal to the area around the target that has the impact parameter, s ±
½Δs, times the number of target atoms (assuming no overlapping) divided by the total
area of the target (or of the beam).
Mathematically:
P(Θ ± ½ΔΘ) = probability of being scattered by an angle of Θ ± ½ΔΘ
= ΔN / N = Δσ # / A
where ΔN is the number scattered by an angle of Θ ± ½ΔΘ, N is the total number of
particles in the beam, Δσ = 2πs ds is the effective area of each atom that will scatter the
particles into the angle of Θ ± ½ΔΘ, # is the number of scattering atoms, and A is the
total area (of the target or of the beam). Let’s further define n as the number density:
n = #/A.
P(Θ ± ½ΔΘ) = 2πns ds .
Now we need to relate s and ds to Θ and dΘ. From the hyperbolic orbits, we have the
relation:
tan(Θ/2) = [mK2/2EL2]½ ,
where both E and L are constants.
We know L = mv  r = mvo sin(θvr) r z = mvos z , since s/r = sin(θvo-r).
Therefore,
L2 = m2vo2s2 .
We also know that initially, E = ½mvo2 . Therefore, we have
tan(Θ/2) = [mK2/2EL2]½ = [mK2/{2(½mvo2)( m2vo2s2)}]1/2 = K/{mvo2s} .
From this expression, we can solve for s:
s = K cos(Θ/2) / {mvo2 sin(Θ/2)} .
From that same expression, we can solve for ds in terms of dΘ :
d (tan(Θ/2)) = d(K/mvo2s)
{d(tan(Θ/2))/dΘ} dΘ = {d(K/mvo2s)/ds} ds
{1 / [2 cos2(Θ/2)]} dΘ = {-K / (mvo2s2)}ds,
or
ds = {-mvo2s2 / [2K cos2(Θ/2)] } dΘ .
The negative sign merely indicates that as dΘ becomes bigger, ds becomes smaller (due
to the inverse relationship between tan(Θ/2) and s). We will drop the negative sign in the
following expressions.
Therefore, the expression P(Θ ± ½ΔΘ) = 2πns ds becomes:
P(Θ±½ΔΘ) = 2πn [K cos(Θ/2) / {mvo2 sin(Θ/2)}]3 {mvo2 / [2K cos2(Θ/2)] }dΘ
Note that the s2 term in the ds expression must be replaced by the expression for s. This results in the [s ]
term being raised to the third power instead of just the first power.
This can be simplified to become:
P(Θ±½ΔΘ) = dN/N = {nπK2 cos(Θ/2) / [m2vo4 sin3(Θ/2)]} dΘ .
To get a little better form, we use the trig identity:
sin(θ) = sin(θ/2 + θ/2) = sin(θ/2) cos(θ/2) + cos(θ/2) sin(θ/2)
= 2 sin(θ/2) cos(θ/2) , or
cos(θ/2) = sin(θ) / 2sin(θ/2) .
Thus, our expression becomes:
P(Θ±½ΔΘ) = dN/N = {nπK2 sin(Θ) / [2m2vo4 sin4(Θ/2)]} dΘ .
Note that as Θ → 180o, sin(Θ) goes to zero but sin(Θ/2) goes to1, which means P → 0.
Note also that as Θ → 0, both sin(Θ) and sin(Θ/2) both → 0. To see what the
combination will do we can use the Taylor series approximation for sine for small angles:
sin(θ) ≈ θ . This gives (again for small angles):
P(Θ±½ΔΘ) = dN/N = {nπK2 Θ / [2m2vo4 (Θ/2)4]} dΘ
which goes to infinity as Θ → 0. But this doesn’t sound reasonable. However, we must
realize that for Θ → 0, s → ∞ (since they are inversely related). In a real experiment,
either the beam or the target is finite in size, and so there is a limit on how big s can get,
and hence a limit on how small Θ can get.
What we can do is look at the ratio of expected numbers scattered into two angles.
dN/N at 1o / dN/N at 10o = {sin(1o)/sin4(½o)} / {sin(10o)/sin4(5o) = 1,000 ;
dN/N at 10o / dN/N at 100o = {sin(10o)/sin4(5o)} / {sin(100o)/sin4(50o) = 1,052 ;
dN/N at 1o / dN/N at 100 = {sin(1o)/sin4(½o)} / {sin(100)/sin4(50) = 1,050,000