Introduction to Solid State Physics · Problem 4

WAVE DIFFRACTION AND THE RECIPROCAL LATTICE — Charles Kittel

Physics — Solid State · diffraction-maximum-width

Problem

Width of diffraction maximum. We suppose that in a linear crystal there are identical point scattering centers at every lattice point ρm=ma\boldsymbol{\rho}_m = m\mathbf{a}, where mm is an integer. By analogy with (20), the total scattered radiation amplitude will be proportional to F=∑exp⁡[−ima⋅Δk]F = \sum \exp[-im\mathbf{a}\cdot\Delta\mathbf{k}]. The sum over MM lattice points is F=1−exp⁡[−iM(a⋅Δk)]1−exp⁡[−i(a⋅Δk)]  ,F = \frac{1 - \exp[-iM(\mathbf{a}\cdot\Delta\mathbf{k})]}{1 - \exp[-i(\mathbf{a}\cdot\Delta\mathbf{k})]} \;, by the use of the series ∑m=0M−1xm=1−xM1−x\displaystyle\sum_{m=0}^{M-1} x^m = \frac{1-x^M}{1-x}. (a) The scattered intensity is proportional to ∣F∣2|F|^2. Show that ∣F∣2≡F∗F=sin⁡212M(a⋅Δk)sin⁡212(a⋅Δk)  .|F|^2 \equiv F^*F = \frac{\sin^2 \frac{1}{2}M(\mathbf{a}\cdot\Delta\mathbf{k})}{\sin^2 \frac{1}{2}(\mathbf{a}\cdot\Delta\mathbf{k})} \;. (b) We know that a diffraction maximum appears when a⋅Δk=2πh\mathbf{a}\cdot\Delta\mathbf{k} = 2\pi h, where hh is an integer. We change Δk\Delta\mathbf{k} slightly and define ϵ\boldsymbol{\epsilon} in a⋅Δk=2πh+ϵ\mathbf{a}\cdot\Delta\mathbf{k} = 2\pi h + \epsilon such that ϵ\boldsymbol{\epsilon} gives the position of the first zero in sin⁡12M(a⋅Δk)\sin\frac{1}{2}M(\mathbf{a}\cdot\Delta\mathbf{k}). Show that ϵ=2π/M\epsilon = 2\pi/M, so that the width of the diffraction maximum is proportional to 1/M1/M and can be extremely narrow for macroscopic values of MM. The same result holds true for a three-dimensional crystal.

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