Diffraction by Organized Layers — Question 5

Section 36-9 · Physics pages 6020–6021

Question 5 A body-centered translation eliminates odd-sum reflections A conventional cubic cell has identical scatterers at fractional positions (0, 0, 0) and (1/2, 1/2, 1/2). Each has real scattering amplitude f. The structure factor is Fhkl = f[1 + eiπ(h+k+l)]. Ignore all other intensity factors. a center Schematic; labels, not drawing lengths, specify the data. (a) Derive the parity selection rule and the allowed reflection intensity in units of f2. (b) Classify (100), (110), (111), and (200) by this rule.
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Question 5 - Solution A body-centered translation eliminates odd-sum reflections a center (a) Method and calculation eiπ(h+k+l) = (−1)h+k+l. F = 2f for even h + k + l; F = 0 for odd h + k + l . (b) Interpretation and further calculation |F|2 = 4f2 or 0 . (100), (111) : absent; (110), (200) : allowed . Physical check A geometrically possible Bragg angle can still be dark because the basis amplitudes cancel within every unit cell.

Question 5

A body-centered translation eliminates odd-sum reflections

A conventional cubic cell has identical scatterers at fractional positions (0,0,0)(0,0,0) and (1/2,1/2,1/2)(1/2,1/2,1/2). Each has real scattering amplitude f. The structure factor is Fhkl=f[1+eiπ(h+k+l)]F_{hkl}=f[1+e^{i\pi(h+k+l)}]. Ignore all other intensity factors.

Diagram shown in the original SVG worksheet.

  1. Derive the parity selection rule and the allowed reflection intensity in units of f2f^2.

  2. Classify (100), (110), (111), and (200) by this rule.

Question 5 - Solution

A body-centered translation eliminates odd-sum reflections

Diagram shown in the original SVG worksheet.

(a) Method and calculation

eiπ(h+k+l)=(−1)h+k+l.e^{i\pi(h+k+l)}=(-1)^{h+k+l}.F=2ffor even h+k+l;F=0for odd h+k+l.\boxed{F=2f\ \text{for even }h+k+l;\quad F=0\ \text{for odd }h+k+l}.

(b) Interpretation and further calculation

|F|2=4f2or 0.\boxed{|F|^2=4f^2\ \text{or }0}.(100),(111):absent;(110),(200):allowed.\boxed{(100),(111):\ \text{absent};\quad(110),(200):\ \text{allowed}}.

Physical check

A geometrically possible Bragg angle can still be dark because the basis amplitudes cancel within every unit cell.