
323Collisions and Scatterings
© 2010 Taylor & Francis Group, LLC
Differentiating with respect to θ
s
, we obtain
d
d
bk
Eθ
θ
=−
4
1
2
2
sin(
(10.28)
so that σ(θ) becomes, by Equation 10.22,
σθ
θ
θ
θ
()
cot( /)
sin
sin(/)
=
1
22
21
2
2
2
k
E
or
σθ
θ
()
sin(
=
1
42
1
2
2
4
k
E
(10.29)
where
e
=
′
2
2
.
This is the famous Rutherford scattering cross section, originally derived by Rutherford to
explain the experimental results of Geiger and Marsden on the scattering of alpha particles by
heavy nuclei (gold, Z′ = 79). In his derivation, this cross section is strongly dependent on both the
velocities of the incident particle and the scattering angle. It also increases rapidly with ...