Cross Sections for Scattering

Side View of scattering

 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 ± ½Δs, 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) =  2πns ds  = 2πns{-mvo2s2 / [2K cos2(Θ/2)] } dΘ  =  2πns3 {-mvo2/ [2K cos2(Θ/2)] } dΘ

=  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) → 0 but sin(Θ/2) → 1, 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)/sin4o)} / {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)/sin4o)} / {sin(100)/sin4(50)  =  1,050,000