Fig. 4.1 The geometry involved in calculating the integrated intensity from a small single crystal, which is rotated at constant angular velocity ω about an axis normal to the paper.

32

22

122

max)( NNNFII Tep =

∫∫ ∫∫∫ γβ2 dddtRIdtdAIE pp ==

ω/)α(ddt=

∫∫∫ γβαω

2

dddIR

E p=

11211

12

112 Δ

λ

πsin)

λ

Δ(πsin)

λ

Δ(πsin aNS

aNShNaN

SH hkl =+=+

)(λ

Δ332211 bpbpbp

S++=

112

113322112

112 πsin)(πsin

λ

Δπsin pNaNbpbpbpaNS

=++=

∫∫∫ γβαπsin

πsin

πsin

πsin

πsin

πsin

ω 32

332

22

222

12

112

22

dddp

pN

p

pN

p

pNF

RIE Te=

α)Δ( α dSd = β)Δ( β dSd = γ)Δ( γ dSd =

,

,

γβαθ2sin)Δ()Δ()Δ( αγβ dddSdSdSddV =×=

∫∫∫3

233

2

22

222

12

11222

πsin

πsin

πsin

πsin

πsin

πsin

θ2sinωdV

p

pN

p

pN

p

pNFRIE

T

e=

3213

3213

332211 )/λ(λλλλ dpdpdpvdpdpdpvdpbdpbdpbdV ab ==×=

∫ ∫ ∫∞

∞

∞

∞

∞

∞ 323

332

222

222

121

112232

)π(

πsin

)π(

πsin

)π(

πsin

θ2sinω

λdp

p

pNdp

p

pNdp

p

pN

v

FRIE

a

T

e ×=

321

232

θ2sinω

||λNNNN

v

FNRIE

a

Te ==

+=

2

θ2cos1 2

242

4

Rcm

eII oe

+

=

θθδλ

ω 2sin22cos1|| 2

2

23

42

40

a

T

v

FV

cm

eIE

……Total energy collected from a small crystal of volume δV

Fig. 4.2 The geometry involved in calculating the integratedintensity for an extended face mosaic crystal

∫∞

0

θsin/μ22

2

23

42

40

θsinδθ2sin2

θ2cos1δλ

ω =

+=

z

oz

a

T

V

dzAe

v

VF

cm

eIE

+=

θ2sin2

θ2cos1

μ2

λ

ω

2

2

23

42

4

a

To

v

F

cm

epE

+

=

θθ

µλ

ω 2sin2

2cos1

2

2

2

23

42

4

a

To

v

F

cm

epE

……Total energy collected from a mosaic crystal witha linear absorption coefficient of µ

In a powder sample , only a fraction (φ) of the irradiated crystals is effective in intensity contribution.

Δαθcos2

Δαθcosπ2π4

Φ 22

pH

H

p==

p = multiplicity factor

Fig. Representation of the number of crystal in powder sample whose hkl planes make angles between θ+α and θ+α+dα with the primary beam.

α∆

Total intensity collected from a powder sample by a pinhole camera:

+=

+=

θsin2

θ2cos1

4

λ

μ2

Δα

ω

θ2sin2

θ2cos1

μ2

λ

ωΦ

2

2

23

42

4

2

2

23

42

4

a

Too

a

To

v

pF

cm

eAI

v

F

cm

epE

μ2θsin→ ∫

∞

0

θsin/)μ2( o

z

oz AdzAeV =

=

With effective volume V:

+

=

θθλ

sin22cos1

4

|| 2

2

23

42

4

a

To

v

FpV

cm

eIP

θ2sinπ2 R

PP =′

+

=′

θθθλ

π 2sinsin2cos1||

16

2

2

23

42

4

a

To

v

FpV

cm

eR

IP

Lorentz-polarization factor LP=(1+cos²2θ)/(sinθsin2θ)

Integrated intensity collected from a powder sample by a pinhole camera:

Integrated intensity collected from a powder sample by a diffractometer:

2θR

Length of Debye ring = 2πRsin2θ

Lorentz-Polarization factor

Absorption factor

e2µt/sinθ

Perfect crystal

Extinction• Primary extinction: after diffracted at A an B,

K2 has a phase shift of π and thus interfere destructively with Ko (happens in perfect crystal).

• Secondary extinction: In mosaic crystal, if mis-orientations among domains are not large enough, incident beam on an interior domain has been weakened by previous correctly oriented domains due to diffraction

Integrated intensity from a thick perfect crystal with negligible absorption, N = number of unit cells per unit volume

Reflectivity = 100% over a range of 2s in the rocking curve

For CuKα, λ=1.542 Å, F211=26.2 per CaCO3, N=1.6x1022,for an unpolarized beam

Reflectivity = 1 over 2s in the rocking curve