PENELOPE 2003 - OECD Nuclear Energy Agency

212 Appendix A. Collision kinematics
cos θ 3 = 1 κ
(
cos θ 4 = (κ + 1)
(
κ + 1 − 1 )
, (A.19)
τ
1 − τ
κ [2 + κ(1 − τ)]
• Annihilation of positrons with free electrons at rest.
) 1/2
. (A.20)
Projectile: Positron m 1 = m e , W 1 = E + m e c 2 ≡ γ m e c 2 .
Target: Electron m 2 = m e , W 2 = m e c 2 .
Annihilation photons: m 3 = 0, W 3 ≡ ζ(E + 2m e c 2 ).
m 4 = 0, W 4 = (1 − ζ)(E + 2m e c 2 ).
cos θ 3 = ( γ 2 − 1 ) −1/2
(γ + 1 − 1/ζ) ,
(A.21)
cos θ 4 = ( γ 2 − 1 ) (
−1/2
γ + 1 − 1 )
. (A.22)
1 − ζ
A.1.1
Elastic scattering
By definition, elastic collisions keep the internal structure (i.e. the mass) of the projectile
and target particles unaltered. Let us consider the kinematics of elastic collisions of a
projectile of mass m (= m 1 = m 3 ) and kinetic energy E with a target particle of mass M
(= m 2 = m 4 ) at rest (see fig. A.2). After the interaction, the target recoils with a certain
kinetic energy W and the kinetic energy of the projectile is reduced to E ′ = E −W . The
angular deflection of the projectile cos θ and the energy transfer W are related through
eq. (A.15), which now reads
cos θ =
E(E + 2mc 2 ) − W (E + mc 2 + Mc 2 )
√E(E + 2mc 2 ) (E − W )(E − W + 2mc 2 ) . (A.23)
The target recoil direction is given by eq. (A.16),
cos θ r =
(E + mc 2 + Mc 2 )W
√E(E + 2mc 2 ) W (W + 2mc 2 ) . (A.24)
Solving eq. (A.23), we obtain the following expression for the energy transfer W
corresponding to a given scattering angle θ,
W =
[
√
]
(E + mc 2 ) sin 2 θ + Mc 2 − cos θ M 2 c 4 − m 2 c 4 sin 2 θ
×
E(E + 2mc 2 )
(E + mc 2 + Mc 2 ) 2 − E(E + 2mc 2 ) cos 2 θ . (A.25)