Chemistry Reference
In-Depth Information
d
σ
d
1
n
0
·
2π
·
b
·
db
·
n
0
b
sin ϑ
·
db
d
ϑ
.
σ
diff
(
ϑ
)
≡
=
d
ϑ
=
(3.103)
2π
·
sin ϑ
·
The factor 2π in (3.103) results from the symmetry of the scattering problem and the
integration over the complete angle ϕ, see Figure 3.17.
The integration over the solid angle provides the total scattering cross section.
σ
diff
·
∂
σ
()
∂
σ
total
=
·
d
=
d
.
(3.104)
The impact parameter for the Coulomb scattering
b
Coul
(
can be found by use of the
invariance of the mechanical energy (3.105) and angular momentum (3.106) of the
charged projectile particle with charge
q
1
and the mass
m
in the electric Coulomb
field of the point charge
q
2
in the center of the laboratory system, similar to the
polarization scattering discussed above.
ϑ
)
dr
dt
2
2
r
2
d
φ
dt
m
2
·
m
2
·
ε
=
v
0
=
+
+
ε
Coul
(
r
)
(3.105)
q
2
4πε
0
·
q
1
·
ε
Coul
(
r
)
=
r
.
d
φ
dt
d
φ
dt
=
v
0
b
r
2
.
r
2
m
·
v
0
·
b
=
m
·
·
→
(3.106)
The rearrangement of Equations 3.105 and 3.106 for the radial and angular velocity,
respectively, and following the ratio of both quantities results in
/
v
0
·
/
d
φ
dt
d
φ
dr
=
b
r
2
dt
=
v
0
−
.
(3.107)
dr
/
v
0
·
b
2
/
r
2
−
2
·
ε
Coul
(
r
)/
mv
0
Replacing the velocity and mass in the laboratory system by the relative velocity and
the reduced mass in the center of mass system, the integral (3.108) has to be solved,
see detailed calculation in [36]
φ
0
∞
b
·
dr
d
φ
=
φ
0
=
m
red
v
r
1
/
2
.
(3.108)
·
1
r
2
−
b
2
/
r
2
−
2ε
Coul
/
0
r
0
Taking into account the scattering angle ϑ
2φ
0
, the impact parameter
b
Coul
(3.109) in dependence on the scattering angle ϑ is found, see [36]
=
π
−
q
1
q
2
4π
1
m
red
·
cot
ϑ
b
Coul
=
ε
0
·
v
r
·
2
.
(3.109)
·
