Subatomic Physics

6.3. Form Factors 141
The scattering potential V (x) in
Eq. (6.5) at the position of the electron
consists of contributions from the
entire nucleus. Each volume element
d 3 r contains a charge Zeρ(r)d 3 r and
gives a contribution (Eq. 6.7)
dV (x) =− Ze2
z exp
�
so that
V (x) =−Ze 2
�
− z
a
d 3 r ρ(r)
z
�
ρ(r) d 3 r,
�
exp − z
�
a
(6.15)
Figure 6.2: Scattering of a spinless electron by
a spinless nucleus with extended charge distribution.
where z = |z| and the vector z is shown in Fig. 6.2. Introducing V (x) intoEq.(6.5)
and using x = r + z yields
f(q 2 )= mZe2
2π�2 �
d 3 � � �
iq · r
r exp ρ(r) d
�
3 x exp(−z/a)
� �
iq · z
exp .
z
�
For fixed r, d 3 x can be replaced by d 3 z. The integral over d 3 z is then the same as
encountered in the evaluation of Eq. (6.8), and it gives
�
d 3 z exp(−z/a)
z
� �
iq · z
exp =
�
4π�2 q2 4π�2
−→ . (6.16)
+(�/a) 2 q2 The integral over d 3 r is the form factor, defined in Eq. (6.14), and the cross section
dσ/dΩ =|f| 2 becomes
dσ
dΩ =
� �
dσ
|F (q
dΩ R
2 )| 2 . (6.17)
The computation for electrons with spin follows the same lines; Eq. (6.12) is the
correct generalization of Eq. (6.17). One remark is in order concerning the density
ρ(r). By Eq. (6.14), the density ρ(r) has been defined in such a way that
�
ρ(r)d 3 r =1. (6.18)
Equation (6.12) indicates how the form factor |F (q 2 )| can be determined experimentally:
The differential cross section is measured at a number of angles, the
Mott cross section is computed, and the ratio gives |F (q 2 )|. The step from F (q 2 )
to ρ(r) is less easy. In principle, Eq. (6.14) can be inverted and then reads
ρ(r) = 1
(2π) 3
�
d 3 qF(q 2 � �
iq · r
)exp − . (6.19)
�