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# B1.8 Diffraction: x-ray, neutron and electron

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B1.8 Diffraction: x-ray, neutron and electron book

# B1.8 Diffraction: x-ray, neutron and electron

DOI link for B1.8 Diffraction: x-ray, neutron and electron

B1.8 Diffraction: x-ray, neutron and electron book

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## 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.