Structure Factor and X-ray Diffraction Peaks

August 19, 2020

Laue condition, Miller Index, and Structure Factor

Laue condition needs to be satisfied to have constructive X-ray diffraction interference (i.e., the peaks on X-ray spectrum). Using the reciprocal lattice, the condition states that

If the difference between the incident $(\mathit{k})$ and scattered $\left({\mathit{k}}^{\prime }\right)$ wave vectors is equal to a reciprocal lattice vector, the diffracted intensity may be nonzero. This is the Laue condition. With $\mathit{K}={\mathit{k}}^{\prime }-\mathit{k},$ this leads to the simple equation $\mathit{K}=\mathit{G}$, where $\mathit{G}$ is the vectors on reciprocal lattice.

Let the base vectors of the reciprocal space be ${\mathit{g}}_i$, then the Miller index is the coordinates on the reciprocal space, i.e.,

where $h,k,l$ are integers. For any family of lattice planes separated by a distance $d$, there is a reciprocal vector with length $\Vert \mathit{G}\Vert =2\pi /d$, and this vector is perpendicular to the lattice planes.

The Laue condition is based solely on the Bravais lattice, so the positions of the diffraction peaks are independent of the atomic basis. However, the intensities of the peaks are strongly influenced by the basis. The structure factor, $S(\mathit{G}),$ and the form factor, $f_{\alpha },$ tell us how the intensities of the peaks depend on the atoms making up the crystals. These quantities are calculated as a sum (or integral) within the unit cell. In the simplest approximation the scattering depends on the atomic charge distribution ${\rho }_{\alpha }(r),$ and the intensity is proportional to the absolute value squared of

and

where $e$ is the electron charge, the sum is over the atoms in the unit cell, and the integration is over the volume of an atom. Similar formulae work for electron and neutron scattering, except the form factor integral is different depending on the microscopic interaction at play. Even for $\mathrm{X}$ -rays, the calculation of the form factor as an integral over the charge density works only for the simplest cases. For a realistic calculation of scattered radiation intensities one has to perform a Rietveld analysis.

Problem: Structure of $R{b}_x{C}_{60}$

Assume you perform a powder X-ray diffraction measurement on a Rb doped ${\mathrm{C}}_{60}$ material with $\lambda =0.9$ Angstrom $\mathrm{X}$ -rays. What are the positions $(2\theta ,$ in degrees $)$ of the first five diffraction peaks for the three observed structures (doping $x=3,4,$ and 6)?

$R{b}_x{C}_{60}$ has an $fcc$ structure for $x=3,bct$ for $x=4,$ and $bcc$ for $x=6.$ To calculate the X-ray peak positions, $2\theta ,$ we must use

where $b_1=b_2=\frac{2\pi }{a},$ and $b_3=\frac{2\pi }{a}$ for $fcc$ and $bcc$ or $b_3=\frac{2\pi }{c}$ for bct. We can then use the condition $2\frac{2\pi }{\lambda }\sin \theta =G$ to calculate values of $2\theta$. Since we are using the conventional cubic unit cells, we must calculate the structure factor, Eq. (1), to determine which $(h,k,l)$ values are allowed.

for $n$ scatterers at positions ${\mathbf{d}}_{\mathbf{j}}$ in a unit cell. For the $fcc$ unit cell, there is a $C_{60}$ at position ${\mathit{d}}_{\mathbf{1}}=0,$ and on the three faces ${\mathit{d}}_{\mathbf{2}}=\frac{a}{2}(\hat{x}+\hat{y}),{\mathit{d}}_{\mathbf{3}}=\frac{a}{2}(\hat{x}+\hat{z}),{\mathit{d}}_{\mathbf{4}}=\frac{a}{2}(\hat{y}+\hat{z}).$ The general vector is $\mathit{K}=\frac{2\pi }{a}(h\hat{x}+k\hat{y}+l\hat{z})$. Therefore

When $S_K\ne 0$, the Bragg peak associated with the reciprocal lattice vector $\mathit{K}$ at scattering angle $2\theta$ is allowed. Similarly we can look at the two-molecule per unit cell bet and bee structures to determine that their structure factors are given by

The resultant calculated values for the first five allowed diffraction peaks for the fcc structures is shown below

Problem: Powder Diffraction of hcp and fcc Crystals

Cobalt has two forms: $\alpha$-Co, with hcp structure (lattice spacing of $a=$ $2.51)$Å and $\beta -$ Co, with $fcc$ structure (lattice spacing of $a_{cubic}=3.55$Å). Assume that the $hcp$ structure has an ideal $c/a$ ratio. Calculate and compare the position of the first five X-ray powder diffraction peaks. The quantity $K=4\pi /\lambda \sin \theta$ can be used to characterize the peak positions (here $\lambda$ is the wavelength of the X-ray radiation and $2\theta$ is the scattering angle).

To satisfy the powder diffraction condition, the length of the reciprocal lattice vector must be equal to $G=K=4\pi /\lambda \sin \theta$. To calculate the position of the peaks for the $fcc$ structure we use the simple cubic unit cell with four atoms per cell. The reciprocal lattice is cubic, with lattice spacing of $G_0=2\pi /a_{cubic}=1.77{\AA }^{-1},$ and the structure factor is nonzero only if the (hkl) indices are all odd or all even. The length of the reciprocal lattice vector is $G=G_0\sqrt{h^2+k^2+l^2}$.

The hcp lattice has two atoms per unit cell. The reciprocal lattice is constructed from a simple hexagonal lattice by assigning a zero structure factor to some of the points, resulting in alternating hexagonal and honeycomb arrays. We will index the reciprocal lattice points in terms of $\mathit{G}=h{\mathit{a}}^{\mathit{\prime }}+k{\mathit{b}}^{\mathit{\prime }}+l{\mathit{c}}^{\mathit{\prime }}$, where the angle between ${\mathit{a}}^{\mathit{\prime }}$ and ${\mathit{b}}^{\mathit{\prime }}$ is ${120}^{\circ }$ and ${\mathit{c}}^{\mathit{\prime }}$ is perpendicular to ${\mathit{a}}^{\mathit{\prime }}$ and ${\mathit{b}}^{\mathit{\prime }}$. The lengths of the primitive vectors are calculated as $a^{\prime }=b^{\prime }=4\pi /\sqrt{3}a=2.89{\AA }^{-1}$, $c^{\prime }=2\pi /c=\sqrt{3/8}2\pi /a=1.53{\AA }^{-1}$.

For $\beta -Co$, we have

For $\alpha -Co$,

The nearest-neighbor distance is the same for the two structures. This is why some of the X-ray diffraction peak positions coincide.