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532 15 SOLIDS
P15B.6 �e intensity of a re�ection is proportional to the square modulus of the struc-
ture factor given by [15B.3–650], Fhk l = ∑ j f jeiϕhk l ( j), where the phase is ϕhk l( j) =
2π(hx j + ky j + lz j).
For atoms at positions (0,0,0), (0, 12 ,
1
2 ), (
1
2 ,
1
2 ,
1
2 ), and (
1
2 , 0,
3
4 ) with the same
scattering factor f
Fhk l = f (e0i + eiπ(k+l) + eiπ(h+k+l) + eiπ(h+3 l/2))
For the (114) re�ection, F114 = f (1 + e5πi + e6πi + e7πi). Using the result eiπn =
(−1)n it follows that F114 = f (1−1+1−1) = 0 .�e I atoms therefore contribute
no net intensity to the (114) re�ection.
P15B.8 �e scattering intensity is given by the Wierl equation [15B.8–654],
I(θ) =∑
i , j
f i f j sin (sR i j)/(sR i j)
where s = 4π sin (θ/2)/λ and the sum is over all pairs of atoms in the species.
(a) For theBr2molecule, f1 = f2 = f withR12 = R.�us I(θ) = f 2 sin (sR)/(sR).
To �nd the positions of the �rst maximum and minimum, di�erentiate I
with respect to θ and set equal to zero.
dI
dθ
= dI
ds
ds
dθ
= 2π f
2
s2Rλ
[sR cos(sR) − sin(sR)] cos (θ/2) = 0
Rearranging sR cos(sR)− sin(sR) = 0 gives tan(sR) = sR which is solved
numerically to give solutions at sR = 0, 4.493, 7.725, .... Inspection of
the form of the function I(θ) shows that the solution sR = 0 corresponds
to the �rst maximum, therefore sR = 4.493 corresponds to the �rst min-
imum. �e angle θ is computed from sR using θ = 2 sin−1(sλ/4π) =
2 sin−1(sRλ/4πR). Taking the Br2 bond length as 228.3 pm gives for
neutron scattering
θmax = 0 θmin = 2 × sin−1
(4.493) × (78 pm)
4π(228.3 pm)
= 14.0○
and for electron scattering
θmax = 0 θmin = 2 × sin−1
(4.493) × (4.0 pm)
4π(228.3 pm)
= 0.72○
(b) For CCl4 there are four carbon-chlorine pairs and six chlorine-chlorine
pairs
I = 4 fC fCl
sin sRCCl
sRCCl
+ 6 f 2Cl
sin sRClCl
sRClCl
= 4 × 6 × 17 × f 2 sin sRCCl
sRCCl
+ 6 × (17)2 × f 2
sin [(8/3)1/2 sRCCl]
(8/3)1/2 sRCCl
I
f 2
= 408 sin sRCCl
sRCCl
+ 1061.8... ×
sin [(8/3)1/2 sRCCl]
sRCCl