Cross Sections

1
Cross Sections

• One of the most important quantities we measure
  in nuclear physics is the cross section. Cross
  sections always have units of area and in
  addition may also be characterized by additional
  observables, such as solid angle, energy,
  momentum, etc.. We start with a simple picture
  first. We want to calculate the scattering rate
  of incident projectiles from targets of area Ds
  distributed randomly, with a density r throughout
  a slab of volume of area S and thickness dx. The
  scattering rate here refers to any process which
  removes an incident projectile by actually
  scattering it or by changing its kinematic
  properties or character in any way. In this case
  we are actually calculating the total cross
  section.
• We assume that if the projectile hits any part of
  the target area Ds it is lost from the incident
  beam. We also assume that the incident beam is
  uniformly distributed over the target area S.

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Suppose there is a target of area Ds in a volume
dx X S. An incident flux, F, of projectiles
strikes the volume.
dx
Ds
F
The density of targets of area Ds Is r.
S
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• The quantity l Dsr, is called the absorption
  coefficient. Usually the total cross section is
  understood to be used in this expression and it
  is simply written as s.
• Target thickness are often quoted in grams/cm2.
  This is because the important quantity for rate
  calculation is rx. What we really use in rate
  calculations is (number of target nuclei)/ cm2.
  We use Avogadros number, A0 to convert mass
  density to particle density.
• A0 6.02x1023 atoms/mole

5
Differential Cross Sections

• We frequently measure particular reaction
  outcomes rather than the total cross section.
  These partial cross sections can be differential
  in the sense that they depend upon rather
  specific final state properties. Consider the
  case of elastic scattering.

N(q,f) incident particles are scattered into the
solid angle dWsin(q)dqdf. In a real experiment
we must use a finite solid angle DW. If N0
incident particles produce N(q,f) scatters for a
target of thickness rx, then the differential
cross section is,
dq
q
F
6
Classical Cross Section and Potential Scattering

• Particles scatter because they change their
  momenta. This necessarily implies a force has
  acted. It can be shown from Newtons equations of
  motion that if two particles interact with each
  other the problem can be separated into two
  pieces. The momentum of the center of mass is
  constant if there are no external forces. The
  coordinate that separates two particles has an
  equation of motion given by

The reduced mass is m and the separation between
particles 1 and 2 is r. Thus, to solve a two body
problem in classical mechanics it is sufficient
to solve two one body problems. We can then focus
on how to solve the scattering problem for a
particle of mass m from a potential V.
7
Potential Scattering

• This section utilizes the discussion in chapter 1
  of the book by Das and Ferbel1. The goal of this
  calculation is to determine the scattering angle
  q for a particle given an initial momentum mv0
  and impact parameter b for a central potential
  V(r).

a0
a0
mv0
r
r0
a
q
b
The particle follows the trajectory ( in blue ).
At a given moment it is a distance r from the
center of the force and at an angle a. The
distance of closest approach to the scattering
center is r0. a0 is the angle between the initial
direction and the point of closest approach. The
asymptotic angle of scattering is q. We assume
the potential goes to zero at infinity so that
the total energy of the particle is its kinetic
energy at infinity. For a central potential the
torque is zero, hence the angular momentum is
constant.
r
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For the position of the particle shown in the
figure the negative square root solution is
appropriate because the radial distance is
decreasing. The positive square root is the
solution for the particle leaving the scattering
center. We could solve this differential equation
by numerical methods for any central potential.
Note that r r(t). If we know r(t) we can then
solve for a(t).
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• We solve the implicit equation for the distance
  of closest approach.

From the diagram we see that the asymptotic
scattering angle is q p - 2a0. Substituting for
L we obtain for a0,
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• From these equations for q we see that different
  values for the impact parameter correspond to
  different angles of scattering. Thus we can make
  a one-to-one correspondence between the
  scattering angle and b. In particular, we see
  that if particles approaching the target pass
  through an annulus of radius b and width db these
  will all emerge at an angle q(b). This annulus
  area is the correct differential cross section
  for the angle of observation. We can invert the
  relationship to get b b(q), then db
  (db/dq)dq. Consider the area of the section of
  the annulus shown.

ds(q) bdfdb
b
df
dWsin(q)dqdf
db
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• Historically this analysis was very important for
  the development of nuclear physics. E. Rutherford
  and his students discovered the very tiny size of
  the nucleus by studying Coulomb scattering.
  Rutherford did the classical analysis, but a
  quantum mechanical analysis of Coulomb scattering
  yields the same result in the first order
  approximation. Since the Coulomb potential goes
  as 1/r, any central potential that has this
  radial behavior will have the same form for its
  differential cross section. The gravitational
  potential is of the form 1/r.

1) Introduction to Nuclear and Particle
Physics, Ashok Das and Thomas Ferbel, John Wiley
and Sons, 1994