Diffraction by Organized Layers — Question 7

Section 36-9 · Physics pages 6024–6025

Question 7 A small Bragg-peak shift reveals the sign and magnitude of lattice strain At fixed wavelength, a first-order peak initially occurs at 2θ = 60.0◦. After deformation its scattering angle decreases by 0.100◦. Assume a small uniform change in the relevant plane spacing, with no change in reflection index. d θ d −→d + ∆d Schematic; labels, not drawing lengths, specify the data. (a) Differentiate Bragg’s law to relate fractional spacing change to angular shift. (b) Calculate the strain and determine whether the planes moved farther apart or closer together.
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Question 7 - Solution A small Bragg-peak shift reveals the sign and magnitude of lattice strain d θ d −→d + ∆d (a) Method and calculation 2d sin θ = λ =⇒2 sin θ ∆d + 2d cos θ ∆θ ≃0. ∆d d ≃−cot θ ∆θ . (b) Interpretation and further calculation ∆θ = −0.0500◦= −8.72665 × 10−4 rad. ∆d/d = +0.00151150 = +0.151150% : spacing increases . Physical check The angular shift in the derivative is half the measured 2θ shift and must be expressed in radians.

Question 7

A small Bragg-peak shift reveals the sign and magnitude of lattice strain

At fixed wavelength, a first-order peak initially occurs at 2θ=60.0∘2\theta=60.0^\circ. After deformation its scattering angle decreases by 0.100∘0.100^\circ. Assume a small uniform change in the relevant plane spacing, with no change in reflection index.

Diagram shown in the original SVG worksheet.

  1. Differentiate Bragg’s law to relate fractional spacing change to angular shift.

  2. Calculate the strain and determine whether the planes moved farther apart or closer together.

Question 7 - Solution

A small Bragg-peak shift reveals the sign and magnitude of lattice strain

Diagram shown in the original SVG worksheet.

(a) Method and calculation

2dsin⁡θ=λ⇒2sin⁡θΔd+2dcos⁡θΔθ≃0.2d\sin\theta=\lambda\Longrightarrow2\sin\theta\,\Delta d+2d\cos\theta\,\Delta\theta\simeq0.Δdd≃−cot⁡θΔθ.\boxed{\frac{\Delta d}{d}\simeq-\cot\theta\,\Delta\theta}.

(b) Interpretation and further calculation

Δθ=−0.0500∘=−8.72665×10−4rad.\Delta\theta=-0.0500^\circ=-8.72665\times10^{-4}\,\mathrm{rad}.Δd/d=+0.00151150=+0.151150%:spacing increases.\boxed{\Delta d/d=+0.00151150=+0.151150\%:\ \text{spacing increases}}.

Physical check

The angular shift in the derivative is half the measured 2θ2\theta shift and must be expressed in radians.