1. N number of scattering scattered particles
Incident mono-energetic beam
v D t A
d W
N = number density in beam
(particles per unit volume)
Solid angle d W represents
detector counting the dN
particles per unit time that
scatter through q into d W
N number of scattering
centers in target
intercepted by beamspot
FLUX = # of particles crossing through unit cross section per sec
= Nv Dt A / Dt A = Nv
Notice: qNv we call current, I, measured in Coulombs.
dN N F d W dN = s(q)N F d W dN = N F d s -

2. dN = FN s(q)d W  N F d s(q)
the “differential” cross section R  R R R R

3. the differential solid angle d for integration is sin d d
Rsind
Rd
Rsind
Rd
Rsin

4. Symmetry arguments allow us to immediately integrate  out
and consider
rings defined
by  alone R
Rsind  R R R
Nscattered = N F sTOTAL
Integrated over all solid angles

5. Nscattered = N F sTOTAL The scattering rate per unit time
Particles IN (per unit time) = FArea(of beam spot)
Particles scattered OUT (per unit time) = F N sTOTAL

6. Earth
Moon

7. Earth
Moon

8. for some sense of spacing consider the ratio
orbital diameters
central body diameter
~ 10s for moons/planets
~100s for planets orbiting sun
In a solid
interatomic spacing: 1-5 Å (1-5  m)
nuclear radii: f (1.5-5  m)
the ratio
orbital diameters
central body diameter
~ 66,666 for atomic electron
orbitals to their own nucleus
Carbon 6C
Oxygen 8O
Aluminum 13Al
Iron 26Fe
Copper 29Cu
Lead 82Pb

9. n= rNA / A where NA = Avogadro’s Number
A solid sheet of lead offers how much of a (cross
sectional) physical target (and how much empty
space) to a subatomic projectile?
82Pb207 w
Number density, n: number of individual
atoms (or scattering centers!) per unit volume
n= rNA / A where NA = Avogadro’s Number
A = atomic weight (g)
r = density (g/cc)
n= (11.3 g/cc)(6.021023/mole)/(207.2 g/mole)
= 3.28  1022/cm3

10. n(Volume)  (atomic cross section) = n(surface areaw)(pr2)
82Pb207 w
For a thin enough layer
n(Volume)  (atomic cross section)
= n(surface areaw)(pr2)
as a fraction of the target’s area:
= n(w)p(5  10-13cm)2
For 1 mm sheet of lead:
1 cm sheet of lead:

11. Actually a projectile “sees” but Znw electrons per unit area!
nw nuclei per unit area
but Znw electrons per unit area!

13. let’s augmented with the specific example of
that general description of cross section
let’s augmented with the specific example of
Coulomb scattering

14.  q1 q2 Recoil of target q1   BOTH target and projectile
will move in response to
the forces between them. q1   q2 
Recoil of
target
But here we are
interested only
in the scattered
projectile q1

16. impact parameter, b

17. A beam of N incident particles strike a (thin foil) target. b q2 d
A beam of N incident particles strike a (thin foil) target.
The beam spot (cross section of the beam) illuminates n scattering centers.
If dN counts the average number of particles scattered between and d
dN/N = n d
using
d = 2 b db
becomes:

18.  b q2 d
and so

19.  b q2 d

20. 1/ (3672 Z) What about the ENERGY LOST in the collision?
the recoiling target carries energy
some of the projectile’s energy was surrendered
if the target is heavy
the recoil is small
the energy loss is insignificant
Reminder:
1/ (3672 Z)