ABSTRACT
To evaluate this integral, first let Q = 2π(sf − si ), let Q = |Q|, let r = |r| and let α be the angle between r and Q. Now |sf − si | = 2|si | sin θ = 2 sin θ/λ, so that
r · (sf − si ) = 2r|si | sin θ cosα = 2r sin θ cosα/λ. The area of a ring around Q with width dα at radius r is 2πr2 sin α dα, so
dτ = 2πr2 sin α dα dr. Because ρ(r) is spherically symmetric, the number of electrons in this volume element is ρ(r) dτ . Letting x = Qr cosα, dα = −dx/(Qr sin α). Then, making all substitutions,
f (Q) = C ∫ ∞
2πr2
Qr ρ(r) dr
exp(ix) dx (B1.8.2a)
= 4πC ∫ ∞
0 r2ρ(r)
sin(Qr) Qr
dr. (B1.8.2b)
If θ = 0, so that Q = 0, this reduces to
f (0) = 4πC ∫ ∞
0 r2ρ(r) dr (B1.8.3)
so that the integral is the total charge in the electron cloud. The constant C has the units of a length, and is conventionally chosen so that f (Q) is a multiple of the ‘classical electron radius’, 2.818 × 10−15 m.
