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4.4. Green’s function solution for elastic scattering
all solid angles dΩ, to obtain the total cross section σ e for elastic
scattering as follows:
σ e =
σ(q) dΩ,
(4.103)
4π
where dΩ = 2π sin ϑ dϑ. We notice that (4.100)
dΩ
dΩ dϑ
1
2πq
=
= 2π sin ϑ
=
,
(4.104)
dq
dϑ dq
k cos(ϑ/2)
k 2
where 0 ≤ q ≤ 2k. It follows that (4.99, 4.103, 4.104)
2k
2π
σ e =
σ(q) q dq
k 2 0
mZze
2 2
4π
=
.
(4.105)
2πf 0 h ¯
2
α 2 (4k 2 + α 2 )
In the limit α → 0, where the nuclear charge is unscreened, the
total cross section becomes infinite. The Coulomb force is therefore
said to have infinite range. Physically, α represents the reciprocal
of the radius of the atomic electron cloud. It is approximately
Z
1/3
α ≈
(4.106)
a 0
where a 0 is the Bohr radius of the hydrogen atom given by
h
2
4πf 0 ¯
137 λ C
a 0 =
=
= 0.0531 nm.
(4.107)
2
e m
2π
For incident energy H > 1 KeV, we have 4k
2 � α
2 , in which case
(4.77, 4.105)
mZze
2 2
π
σ e ≈
h
2
k 2 α 2
2πf 0 ¯
2πm mZze
2 2
=
.
(4.108)
h
2
H
2πf 0 ¯
Taking
H =
1
2
mβ
2 c
2 ,
β = v/c,
z = 1,
(4.109)