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Equation of a plane (crystallography)

A crystal is composed of atoms, molecules or ions that are arranged in a repetitive manner, forming a three-dimensional space lattice. Each point in the space lattice represents either a particle or a collection of particles (known as a basis), and a set of lattice points can be connected by a plane. As there are many possible ways to connect lattice points with planes of different orientations, a system is needed to define these planes. We begin by deriving the vector and scalar equations of a plane.

P(x, y, z) and P0(x0, y0, z0) are two points on a plane with position vectors r and r0 respectively, which makes r – r0 the vector from P0 to P. The plane has a direction defined by the normal vector n(A, B, C), which is perpendicular to the vector r – r0. Therefore,

\textbf{\textit{n}}\cdot (\textbf{\textit{r}}-\textbf{\textit{r}}_0)=0\; \; \; \; \; \; \; (1)

\begin{pmatrix} A &B &C \end{pmatrix}\begin{pmatrix} x-x_0\\y-y_0 \\z-z_0 \end{pmatrix}=0

A(x-x_{0})+B(y-y_{0})+C(z-z_{0})=0

Ax+By+Cz=D

where D = Ax0 + By0 + Cz0.

\frac{x}{D/A}+\frac{y}{D/B}+\frac{z}{D/C}=1\; \; \; \; \; \; \; (2)

Eq1 is the vector equation of a plane, while eq2 is the scalar equation of a plane, where D/A, D/B and D/C are the intercepts of the plane with the x-axis, y-axis and z-axis respectively.

 

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Phase problem (crystallography)

The phase problem in crystallography is the inability to reconstruct an electron density map due to the lack of phase information of the diffracted rays.

Recall that the modulus of a complex function is \left | z \right |=\left | re^{ix} \right |=\left | rcosx+irsinx \right |=\sqrt{r^2cos^2x+r^2sin^2x}=r. Since the structure factor Fhkl is a complex function of the form z = reiΦ, where r = IzI ,

F_{hkl}=\left | F_{hkl} \right |e^{i\phi_{hkl}}\; \; \; \; \; \; \; (49)

Hence, Fhkl is composed of two factors, the modulus IFhklI and the phase e^{i\phi_{hkl}}. If we replace Fhkl in eq48 with just the modulus factor, information with regard to the phase is lost. Without phase information, it is impossible to reconstruct an electron density map. This is known as the phase problem, which fortunately can be overcome by a few methods, one of which is the Patterson method.

 

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Single crystal X-ray diffraction

Single crystal X-ray diffraction is an analytical technique to determine the structure of a single crystal.

Unlike the polycrystalline sample used in powder X-ray diffraction, a monocrystalline sample or single crystal does not have three-dimensional lattices that are randomly oriented in space. The sample has to be rotated in different directions so that scattered X-rays from all lattice planes satisfying Bragg’s law are recorded. This is accomplished by a single crystal X-ray diffractometer called the four circle diffractometer (see diagram below).

Crystal growth usually involves dissolving the chemical species to be analysed in a mixture of solvents, which evaporates over time and therefore allows the concentration of the solute to rise until crystallisation occurs. A single crystal of about 0.1-0.4 mm in all dimensions is then selected and mounted on the head of the spindle.

The single crystal is rotated through one of the four angles at each turn. The four angles are:

  1. 2θ, between the diffracted ray and the incident ray. This angle varies with the rotation of the detector in the equatorial plane that contains the incident and diffracted rays. 2θ = 0 when the detector is in the direction of the incident X-ray.
  2. Ω, defined by the rotation of the cradle around the cradle axis. Ω = 0 when the cradle is perpendicular to the incident X-ray.
  3. χ, between the spindle axis and the cradle axis. This angle varies with the rotation of the cradle in the clockwise or counter-clockwise direction. χ = 0 when the spindle axis is parallel to the cradle axis, at which the Ω and Φ rotations coincide.
  4. Φ, the angle of rotation of the spindle around the spindle axis. The zero position of Φ is defined with reference to the crystal orientation.

We can use the Ewald sphere to visualise various rotations with respect to the reciprocal lattice (see diagram below).

The single crystal at A is irradiated by an incident X-ray IA, scattering a wave vector that intersects the surface of the Ewald sphere at Q. Consider the case where Bragg’s law (or Laue equations) is satisfied, i.e. ∠OAQ = 2θ. Even though reciprocal lattice points spans across and beyond both spheres, we shall focus on those enclosed by the green sphere with the origin of the reciprocal lattice vector at O. As the single crystal at A rotates about the cradle axis, reciprocal lattice points formed by scattered X-rays from the crystal rotate about an axis at O that is parallel to the cradle axis (since O is the origin of the reciprocal lattice vector).

A reciprocal lattice point P therefore satisfies Bragg’s law when it is rotated by a certain value of Ω to Q. As this rotation corresponds to X-rays scattered by a particular lattice plane, the point P is assigned a (hkl) value after the rotation data is analysed by a computer. Similarly, a reciprocal lattice point S when rotated by an angle χ satisfies Bragg’s law when it reaches Q. In short, data is collected by the computer at each turn and analyses of systematic absences, electron densities and so on are carried out to give the structure of the sample.

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Determining the Avogadro constant

The Avogadro constant is accurately determined using X-ray diffraction techniques. Diamond has a face-centred cubic unit cell (figure III below) that is formed on a two-carbon-atom basis, which is shaded yellow in figure I. With reference to figure III, there are 4 x 2 = 8 carbon atoms in each unit cell.

The unit cell of diamond can also be perceived as two interpenetrating face-centred cubic unit cells (figure II) in a way that a line joining the centres of atoms 1, 2, 3 forms a body diagonal of one of the face-centred cubic unit cells (formed by the black atoms). The length of the line joining atoms 1 and 2 is a quarter of that of the body diagonal.

Since the Avogadro constant was previously defined as the number of atoms in 0.012 kilogrammes of carbon-12, a hypothetical way to precisely evaluate it through X-ray diffraction is to synthesise a perfect sphere of pure carbon-12 that weighs exactly 0.012 kilogrammes and calculate the ratio of its molar volume Vmol to the volume of one-eighth of its unit cell (with n = 8 carbon atoms in a unit cell):

N_A=\frac{V_{mol}}{a^3/n}=\frac{nM}{\rho\: a^3}\; \; \; \; \; \; \; (70)

where M is the molar mass, ρ is the density and a is the cubic unit cell dimension.

The shape of a sphere is preferred, as it is easier to measure the size of the crystal and hence its density. In reality, it is impossible to carve a perfect sphere out of diamond with minimal defects, and the Avogadro constant is instead determined using a sphere that is grown from highly enriched silicon-28, which also has the same unit cell structure as diamond. Furthermore, in some experiments, the interplanar distance between the {220} family of planes d220 is usually measured due to the way the crystal is eventually cut and mounted on the diffractometer. From figure IV above, d220 is equal to a/√8. Therefore, eq70 becomes

N_A=\frac{nM}{\rho(d_{220}\sqrt{8})^3}\; \; \; \; \; \; \; (71)

In an internationally coordinated project called the Avogadro Project in 2011, the molar mass of the enriched silicon-28 crystal was determined with high accuracy via mass spectrometry and is independent of the Avogadro constant. The mass of the crystal was calibrated versus the then Pt-Ir kilogram standard in vacuum, while the diameter of the crystal was measured via an interferometer. With the radius of the sphere determined, the volume of the sphere is known, and hence its density. Finally, the interplanar distance d220 is measured using X-ray interferometry, a technique that combines the principles of X-ray diffraction and interferometry. The value of the Avogadro constant was presented by the project team as 6.02214078×1023 mol-1. In 2017, the value was refined to 6.02214076×1023 mol-1 and in Nov 2018, the Avogardo constant was defined as exactly 6.02214076×1023 mol-1.

You can read more about the groundbreaking experiments conducted by the project team in the new eBook, The Mole: Theories Behind the Experiments That Define the Avogadro Constant.

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Patterson method (crystallography)

In 1935, a British scientist named Arthur Patterson developed the Patterson method to address the phase problem by using the value of IFhkl I2 directly from the X-ray diffraction experiments. He modified eq48 by replacing Fhkl with IFhkl I2 to give a density map in the Patterson space of uvw:

P(uvw)=\frac{1}{V}\sum_{h=-\infty }^{\infty }\sum_{k=-\infty }^{\infty }\sum_{l=-\infty }^{\infty }\left | F_{hkl} \right |^2e^{-i2\pi(\frac{hu}{a}+\frac{kv}{b}+\frac{lw}{c})}\; \; \; \; \; \; \; (50)

The density map is related to IFhkl I2, which is dependent on the interatomic vectors of atom pairs (see question below). In other words, the peaks on the Patterson map correspond to (the end points of) interatomic vectors.

Question

Show that IFhkl I2 is dependent on the interatomic vectors of atom pairs.

Answer

Substituting eq29 in eq36 and extending the equation in three dimensions with a = r and letting h = s – s0, we have

F(\textbf{\textit{h}}) =\sum_{j}f_je^{i2\pi \textbf{\textit{r}}_j \cdot\textbf{\textit{h}}}\; \; \; \; \; \; \; (51)

Since I\propto \left | F(\textbf{\textit{h}}) \right |^2=F(\textbf{\textit{h}})^*F(\textbf{\textit{h}})

F(\textbf{\textit{h}}) ^*F(\textbf{\textit{h}}) =(\sum_{k}f_ke^{-i2\pi \textbf{\textit{r}}_k \cdot\textbf{\textit{h}}})(\sum_{j}f_je^{i2\pi \textbf{\textit{r}}_j \cdot\textbf{\textit{h}}})

\left | F(\textbf{\textit{h}}) \right |^2 =\sum_{j}\sum_{k}f_jf_ke^{i2\pi (\textbf{\textit{r}}_j-\textbf{\textit{r}}_k)\cdot\textbf{\textit{h}}}\; \; \; \; \; \; \; (52)

where rj and rk are the position vectors of atoms j and k respectively, which makes rj – rk the interatomic vector of the j–k atom pair.

Whereas the structure factor F(h) depends on the position vector r of the atom, which is reference to some origin, IF(h)I2 depends on the difference between the position vectors of the j and k atoms in the unit cell (interatomic vector), which is independent of the origin. For each rj – rk vector, there is also a rk – rj vector in eq52, i.e. one vector pointing from atom j to atom k and another from atom k to atom j, with both vectors appearing on the Patterson map.

Furthermore, a peak on the Patterson map has a maximum height that is proportional to the product of the scattering factors of the pair of atoms associated with the interatomic vector. Since the maximum value of the scattering factor of an atom is the number of electrons in the atom, the height of a peak on the Patterson map provides important information of the nature of the pair of atoms contributing to the peak. This information is useful for reconstructing the density map of the unit cell.

 

Question

Show that a peak on the Patterson map has a maximum height that is proportional to the product of the scattering factors of the pair of atoms associated with the interatomic vector.

Answer

Comparing eq48 with eq29 and eq51, the vector form of eq48 is

\rho(\textbf{\textit{r}})=\frac{1}{V}\sum_{\textbf{\textit{h}}} F(\textbf{\textit{h}}) e^{-i2\pi\textbf{\textit{r}}\cdot\textbf{\textit{h}}}

The vector form of eq50 is therefore

P(\textbf{\textit{u}})=\frac{1}{V}\sum_{\textbf{\textit{h}}} \left | F(\textbf{\textit{h}}) \right |^2 e^{-i2\pi\textbf{\textit{u}}\cdot\textbf{\textit{h}}} \; \; \; \; \; \; \; (53)

where u is the three-dimensional lattice vector in Patterson space.

Substitute eq52 in eq53

P(\textbf{\textit{u}})=\frac{1}{V}\sum_{\textbf{\textit{h}}}\sum_{j}\sum_{k} f_jf_k e^{-i2\pi[\textbf{\textit{u}}-(\textbf{\textit{r}}_j-\textbf{\textit{r}}_k)]\cdot\textbf{\textit{h}}}

The maximum value of P(u) is when u = (rj – rk),

P(\textbf{\textit{u}})_{max}=\frac{1}{V}\sum_{j}\sum_{k} f_jf_k

 

The diagram below shows three different crystal lattices and their corresponding Patterson maps. Figure Ia has two similar atoms in the unit cell. The peaks in the corresponding Patterson map are at the ends of the interatomic vector (r2 – r1), which is denoted by 1-2, and (r1 – r2), which is denoted by 2-1. There is also a slightly more intense peak at the origin of the Patterson map that is due to the interatomic vector of atom 1 onto itself (r1 – r1) as well as that of atom 2 onto itself (r2 – r2).

By the same logic, a unit cell of three atoms (figure IIa) has nine peaks (N2= 32), three of which are superimposed at the origin, resulting in a relatively intense peak there. The unit cell with five atoms (figure IIIa) has twenty-five peaks (N2= 52) in the Patterson map (figure IIIb) at the following positions:

Position Interatomic vectors Superimposed peaks
1 c → a 1
2 d → a,

c → b

 2
3 d → b 1
4

a → b,

d → c

 2
5 a → c 1
6

b → c,

a → d

 2
7 b → d 1
8 b → a,

c → d

 2
9 e → a 1
10 c → e 1
11 e → b 1
12 d → e 1
13 a → e 1
14 e → c 1
15 b → e 1
16 e → d 1
O a → a,

b → b,

c → c,

d → d,

e → e

 5
  Total 25

Even though some Patterson maps have semblance to the structures of compounds being investigated, they can be very complicated and have to be deconstructed fully to deduce the structures. This used to be done manually, which is a painfully arduous process. In modern days, a computer does the work by considering the symmetry of the Patterson function, analyzing the peak intensities and noting the presence of heavy atoms.

 

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Structure factor (crystallography)

The structure factor, F, is a function that expresses the sum of amplitudes of scattered X-rays from all atoms in a crystal.

We have, in the previous section, discussed the scattering factor of a single atom, which is the sum of amplitudes of waves scattered by the electrons in the atom. We shall now describe the sum of amplitudes of scattered X-rays from all atoms in a unit cell.

Consider a one-dimensional lattice along the a-axis consisting of two types of atoms (see diagram above). If the Laue equation along the a-axis is satisfied for R1 and R3 as well as for R2 and R4, we have:

\delta _{11\: atoms}=\delta _{22\: atoms}=\frac{a}{h}(cos\alpha -cos\alpha _0)=n\lambda

However, if the Laue equation along the -axis is not satisfied for R1 and R2 (or for R2 and R3) we have:

\delta _{12\: atoms}=x(cos\alpha -cos\alpha _0)

Combining the two equations above and letting n = 1,

\delta _{12\: atoms}=\frac{\lambda hx}{a}\; \; \; \; \; \; \; (34)

Substitute eq28 in eq34,

\phi =\frac{2\pi hx}{a}\; \; \; \; \; \; \; (35)

Recall from the previous section that the amplitude of the scattered X-ray received by the detector from an atom is equivalent to the scattering factor f=4\pi \int_{0}^{\infty }\rho \frac{sinkr}{kr}r^2dr. If we have two atoms, the resultant amplitude F at the detector is

F=f_1+f_2e^{i\phi }\; \; \; \; \; \; \; (36)

The second term in eq36 includes the factor eiΦ  as the scattered rays from the second atom may have a phase difference from that from the first atom. If Φ = 0, F = f1+ f2 .

 

Question

Show that the intensity detected from eq36 is consistent with the intensity associated to the resultant amplitude of two general complex waves having a phase difference of Φ.

Answer

Let the waves be y1 = Aei(kx-ωt) and y2 = Bei(kx-ωt+Φ) . The resultant amplitude is y = Aei(kx-ωt) + y2 = Bei(kx-ωt+Φ). As we know, the intensity of the resultant wave is proportional to the square of magnitude of the amplitude of the wave:

I\propto\left | y \right |^2=y^*y=\left [ Ae^{-i(kx-\omega t)}+Be^{-i(kx-\omega t+\phi )} \right ]\left [ Ae^{i(kx-\omega t)}+Be^{i(kx-\omega t+\phi )} \right ]

I\propto A^2+ABe^{i\phi }+ABe^{-i\phi}+B^2

Since e^{\pm i\phi}=cos\phi \pm isin\phi

I\propto A^2+B^2+2ABcos\phi

Similarly, for eq36

I \propto (f_1+f_2e^{-i\phi})(f_1+f_2e^{i\phi})

I\propto {f_{1}}^{2}+{f_{2}}^{2}+2 {f_{1}}{f_{2}}cos\phi

 

Substitute eq35 in eq36,

F=f_1+f_2e^{i\frac{2\pi hx}{a}}\; \; \; \; \; \; \; (37)

In three dimensions, eq37 becomes

F_{hkl}=f_1+f_2e^{i\frac{2\pi hx_2}{a}}+...+f_je^{i\frac{2\pi hx_j}{a}}+

f_1e^{i\frac{2\pi hy_1}{a}}+f_2e^{i\frac{2\pi hy_2}{a}}+ ...+f_je^{i\frac{2\pi hy_j}{a}}+

f_1e^{i\frac{2\pi hz_1}{a}}+f_2e^{i\frac{2\pi hz_2}{a}}+ ...+f_je^{i\frac{2\pi hz_j}{a}}

So,

F_{hkl}=\sum_{j}f_je^{i2\pi (\frac{hx_j}{a}+\frac{ky_j}{b}+\frac{lz_j}{c})}\; \; \; \; \; \; \; (38)

The first term in eq38 is f_1e^{i\frac{2\pi hx_1}{a}} in one-dimension. Since there is no phase difference between the resultant X-ray of the first atom with itself, f_1e^{i\frac{2\pi hx_1}{a}}=f_1 and eq38 is consistent with eq37, i.e. the phases of scattered X-rays of atoms are relative to that of the first atom at the coordinate of x1 = 0. Eq38 can be rewritten in terms of fractional coordinates:

F_{hkl}=\sum_{j}f_je^{i2\pi ({hx_j}'+{ky_j}'+{lz_j}')}\; \; \; \; \; \; \; (39)

where {x_j}'=\frac{x_j}{a} (i.e. xj in units of a), {y_j}'=\frac{y_j}{b}, {z_j}'=\frac{z_j}{c}.

The intensity of a diffraction signal is proportional to the square of the magnitude of the three-dimensional structure factor, i.e. I \propto \left | F_{hkl} \right |^2 and therefore systematic absences appear when F_{hkl}=0.

For example, the body centred cubic unit cell (where a = b = c) has fractional coordinates shown in the diagram below.

As the unit cell is composed of the same atoms, the structure factor for each atom is the same. If an atom is shared by m neighbouring unit cells, we multiply the scattering factor of the atom with 1/m. Therefore, eq39 becomes:

F_{hkl}=\frac{1}{8}f[e^{i2\pi (0+0+0)}+e^{i2\pi (0+k+0)}+e^{i2\pi (h+k+0)}+e^{i2\pi (h+0+0)}+

e^{i2\pi (0+0+l)}+e^{i2\pi (0+k+l)}+e^{i2\pi (h+k+l)}+e^{i2\pi (h+0+l)}]+fe^{i2\pi (\frac{h}{2}+\frac{k}{2}+\frac{l}{2})}

Since ei2π = cos2π + isin2π = 1, eiπ = cosπ + isinπ = -1 and eab = (ea)b

F_{hkl}=\frac{1}{8}f[1+1^k+1^{h+k}+1^h+1^l+1^{k+l}+1^{h+k+l}+1^{h+l}]+f(-1^{h+k+l})

Since 1a = 1

F_{hkl}=f[1+(-1^{h+k+l})]

If h + k + l is even, Fhkl = 2f. If h + k + l is odd, Fhkl = 0. Hence, systematic absences occur when h + k + l is odd. A primitive cubic unit cell is different from a BCC in that it does not have an atom at (\frac{1}{2},\frac{1}{2},\frac{1}{2}) . Its structure factor is Fhkl = f, which means the diffraction intensities of a cubic P cell has no restriction other than the scattering factor of the atom, which is dependent on the Bragg’s angle θ and hence Bragg’s law of sin\theta =\frac{\lambda \sqrt{h^2+k^2+l^2}}{2a} .

For a face-centred cubic unit cell, the coordinates of the atoms are the same as a cubic P unit cell plus six other atoms on the six faces: (\frac{1}{2},\frac{1}{2},0), (\frac{1}{2},0,\frac{1}{2}), (0,\frac{1}{2},\frac{1}{2}), (\frac{1}{2},\frac{1}{2},1), (\frac{1}{2},1,\frac{1}{2}), (1,\frac{1}{2},\frac{1}{2}). These atoms are each shared by two neighbouring cells. The structure factor of a face-centred cubic unit cell is:

F_{hkl}=f+\frac{1}{2}f[e^{i2\pi (\frac{h}{2}+\frac{k}{2}+0)}+e^{i2\pi (\frac{h}{2}+0+\frac{l}{2})}+e^{i2\pi (0+\frac{k}{2}+\frac{l}{2})}+

e^{i2\pi (\frac{h}{2}+\frac{k}{2}+l)}+e^{i2\pi (\frac{h}{2}+k+\frac{l}{2})}+e^{i2\pi (h+\frac{k}{2}+\frac{l}{2})}]

F_{hkl}=f+\frac{1}{2}f[(-1^{h+k})+(-1^{h+l})+(-1^{k+l})+(-1^{h+k+2l})+

(-1^{h+2k+l})+(-1^{2h+k+l})]

F_{hkl}=f+\frac{1}{2}f[(-1^{h+k})+(-1^{h+l})+(-1^{k+l})+(-1^{h+k})(-1^{2l})+

(-1^{h+l})(-1^{2k})+(-1^{k+l})(-1^{2h})]

F_{hkl}=f+f[(-1^{h+k})+(-1^{h+l})+(-1^{k+l})]

If h, k, l are all even or all odd, Fhkl = 4f. If one index is odd and two are even, or vice versa, Fhkl = 0. Therefore, a face-centred cubic unit cell does not have systematic absences when h, k, l are all even or all odd.

 

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Laue equations

The Laue equations are formulated in 1914 by Max Theodor Felix Laue, a German physicist. Like the Bragg equation, they serve to explain X-ray diffraction patterns. 

Consider a one-dimensional row of equally spaced lattice points of a crystal in the diagram above. If \alpha _0\neq \alpha, the path difference δ between R1 and R2 is given by

\delta =AC-BD

Since AC=\frac{a}{h}cos\alpha and BD=\frac{a}{h}cos\alpha _0,

\delta =\frac{a}{h}(cos\alpha -cos\alpha _0)

For constructive interference to occur, the path difference must be an integral multiple of the wavelength of the X-ray radiation. So,

a(cos\alpha -cos\alpha _0)=nh\lambda\; \; \; \; \; \; \; (15)

Similarly, for the other two axes of the crystal, we have:

b(cos\beta -cos\beta _0)=nk\lambda\; \; \; \; \; \; \; (16)

c(cos\gamma -cos\gamma _0)=nl\lambda\; \; \; \; \; \; \; (17)

Eq15, eq16 and eq17 are collectively known as the Laue equations. For constructive interference to occur in three dimensions, the three equations must be simultaneously satisfied.

The Laue equations can also be expressed in vector form (see above diagram), where s and s0 are wave vectors of the scattered and incident X-rays respectively; a is the lattice vector along the a-axis. Again,

\delta =AC-BD

Since AC=\left | \textbf{\textit{a}}\right |cosDAC and \textbf{\textit{a}}\cdot\textbf{\textit{s}}=\left | \textbf{\textit{a}}\right |\left | \textbf{\textit{s}}\right |cosDAC, AC=\textbf{\textit{a}}\cdot \frac{\textbf{\textit{s}}}{\left | \textbf{\textit{s}} \right |}. Similarly, BD=\textbf{\textit{a}}\cdot \frac{\textbf{\textit{s}}_0}{\left | \textbf{\textit{s}} _0\right |}. So,

\delta=\textbf{\textit{a}}\cdot \frac{\textbf{\textit{s}}}{\left | \textbf{\textit{s}} \right |}-\textbf{\textit{a}}\cdot \frac{\textbf{\textit{s}}_0}{\left | \textbf{\textit{s}}_0 \right |}\; \; \; \; \; \; \; (18)

Substituting \left | \textbf{\textit{s}}\right |=\left | \textbf{\textit{s}}_0\right |=\frac{1}{\lambda } (see below for explanation) in eq18,

\delta =\lambda\textbf{\textit{a}}\cdot (\textbf{\textit{s}}-\textbf{\textit{s}}_0)\; \; \; \; \; \; \; (19)

For constructive interference to occur,

\lambda\textbf{\textit{a}}\cdot (\textbf{\textit{s}}-\textbf{\textit{s}}_0)=h\lambda\; \; \; \Rightarrow \; \; \; \textbf{\textit{a}}\cdot (\textbf{\textit{s}}-\textbf{\textit{s}}_0)=h\; \; \; \; \; where\; h\in \mathbb{Z}\; \; \; \; \; \; \; (20)

Similarly, for the other two axes, we have,

\textbf{\textit{b}}\cdot (\textbf{\textit{s}}-\textbf{\textit{s}}_0)=k\; \; \; \; \; where\; k\in \mathbb{Z}\; \; \; \; \; \; \; (21)

\textbf{\textit{c}}\cdot (\textbf{\textit{s}}-\textbf{\textit{s}}_0)=l\; \; \; \; \; where\; l\in \mathbb{Z}\; \; \; \; \; \; \; (22)

Eq20, eq21 and eq22 are the Laue equations in vector form.

 

Question

Why is \left | \textbf{\textit{s}}\right |=\left | \textbf{\textit{s}}_0\right |=\frac{1}{\lambda } ?

Answer

A wave vector k, like any vector, has a direction and magnitude. Its direction is perpendicular to the wavefront, while its magnitude is defined as the number of waves per unit distance, which is 1/λ. Therefore, \left | s \right |=\left | s_0 \right |=\frac{1}{\lambda}.

Wave vectors and have origins in the de Broglie relation, which is p = h/λ = hk where k = 1/λ. Since p is a vector, k must also be a vector, as h is a constant, i.e. p = hk. This implies that k is also a momentum vector with magnitude of 1/λ.

\left | \textbf{\textit{k}}\right |=\sqrt{\textbf{\textit{k}} \cdot\textbf{\textit{k}}}=\frac{1}{\lambda }

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Systematic absences

Systematic absences are ‘missing’ diffraction intensities on a powder X-ray diffraction spectrum of intensity versus 2θ. Consider the possible values of h, k and l that satisfy Bragg’s equation of sin\theta = \frac{\lambda \sqrt{h^2+k^2+l^2}}{2a} for a primitive cubic unit cell:

hkl 100 110 111 200 210 211 220 300 221
h2 + k2 + l2 1 2 3 4 5 6 8 9

9

7 is missing in the h2 + k2 + l2 sequence, as h, k and l are integers. An extension of the sequence reveals that 15, 23, etc. are missing as well. Hence, the spectrum for a primitive cubic unit cell has a specific set of missing intensities. Likewise, different missing intensity sets are expected for other unit cells. For example, a sample with grains that are composed of face-centred cubic unit cells (figure I below) has scattered rays from {200} planes that are out of phase with rays from {100} (figure II), resulting in destructive interference with no {100} intensity peak on the spectrum.

In fact, for all face-centred cubic unit cells (FCC), intensity peaks are observed when the integers h, k and l are either all odd or all even, e.g. {111}, {200}, {220}, …. A mathematical treatment of this requires the knowledge of the structure factor, which is described in a later section. The structure factor for a body-centred cubic unit cell (BCC) also shows that systematic absences do not appear when h + k + l = 2n, i.e. when the sum of the Miller indices is even, e.g. {110}, {200}, {211}, ….

Question

Identify the cubic unit cells of polonium and aluminium and determine their dimensions using the following powder X-ray diffraction spectral data:

2θi 2θ1 2θ2 2θ3 2θ4 2θ5 2θ6 2θ7 2θ8

Po

 12.1o 17.1o 21.0o 24.3o 27.2o 29.9o 34.7o

36.9o

Al

 38.4 o 44.7 o 65.0 o 78.1 o 82.3 o 98.9 o 111.8 o 116.4 o

λAl = 71.0 pm and λPo = 154.1 pm

Answer

Let’s tabulate the systematic absences of the three cubic unit cells for reference, knowing that i) 7 is missing for a primitive cubic cell, ii) the integers h, k and l for an FCC cell are either all odd or all even, and iii) that h + k + l = 2n for a BCC cell.

hkl 100 110 111 200 210 211 220 300

221
h2+ k2 + l2

Primitive

 1 2 3 4 5 6 8 9

9

BCC

 – 2 – 4 – 6 8 –

–

FCC

 – – 3 4 – – 8 – –

Squaring both sides of eq13, we have

sin^2\theta =\frac{\lambda ^2}{4a^2}(h^2+k^2+l^2)\; \; \; \; \; \; \; (14)

If the unit cell is primitive, the successive ratios of \frac{sin^2\theta _i}{sin^2\theta _1} will produce the sequence 1,2,3,4,5,6,8,9… since from eq14, \frac{sin^2\theta _i}{sin^2\theta _1}=\frac{({h}^{2}+{k}^{2}+{l}^{2})_i}{(h^{2}+{k}^{2}+{{l}^{2})_1}} and {(h^{2}+{k}^{2}+{{l}^{2})_1}}=1. If the unit cell is BCC, we need to multiply the successive ratios of \frac{sin^2\theta _i}{sin^2\theta _1} by the factor 2 to obtain the fingerprint sequence of 2,4,6,8… since {(h^{2}+{k}^{2}+{{l}^{2})_1}}=2. By the same logic, the multiplication factor is 3 if the unit cell is FCC. Tabulating the ratios,

 

 \frac{sin^2\theta _1}{sin^2\theta _1} \frac{sin^2\theta _2}{sin^2\theta _1} \frac{sin^2\theta _3}{sin^2\theta _1} \frac{sin^2\theta _4}{sin^2\theta _1} \frac{sin^2\theta _5}{sin^2\theta _1} \frac{sin^2\theta _6}{sin^2\theta _1} \frac{sin^2\theta _7}{sin^2\theta _1} \frac{sin^2\theta _8}{sin^2\theta _1}

Po

 1.00 1.99 2.99 3.99 4.98 5.99 8.01

9.02

Al

 1.00 1.34 2.67 3.67 4.00 5.34 6.34

6.68

it is clear that the unit cell for polonium is primitive cubic while that for aluminium is FCC. To verify the fingerprint sequence of an FCC unit cell, we multiply the aluminium ratios by the factor of 3:

 

 \frac{sin^2\theta _1}{sin^2\theta _1} \frac{sin^2\theta _2}{sin^2\theta _1} \frac{sin^2\theta _3}{sin^2\theta _1} \frac{sin^2\theta _4}{sin^2\theta _1} \frac{sin^2\theta _5}{sin^2\theta _1} \frac{sin^2\theta _6}{sin^2\theta _1} \frac{sin^2\theta _7}{sin^2\theta _1} \frac{sin^2\theta _8}{sin^2\theta _1}

Al x3

 3.00 4.01 8.01 11.01 12.01 16.01 19.02

20.04

Finally, the dimensions of the unit cell of each sample is obtained by substituting any corresponding set of θ and hkl values in eq14:

 

 a

Po

337 pm
Al

406 pm

 

As mentioned above, to mathematically show that the integers h, k and l for an FCC cell are either all odd or all even, and that h + k + l = 2n for a BCC cell, we have to explain what structure factor is. To do that, we need to first understand the Laue equations and the scattering factor.

 

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Powder X-ray diffraction

Powder X-ray diffraction is a scientific method for determining the structure of materials. The diagram below shows the main components of a powder X-ray diffractometer.

The substance to be analysed is placed on the sample stage. It is in its powder (or polycrystalline) form, which contains millions of grains of three-dimensional lattices that are randomly orientated in space (see figure II below). Therefore, every possible orientation of every set of lattice plane is present in the sample.

The position of the X-ray source is fixed, while the sample stage is rotated at an angular rate of ω, where ω = dθ/dt. The detector is rotated in the same direction as the sample stage but at an angular rate of 2ω.

Consider the sample stage being parallel to the line AB and the detector at B when t = 0. At t = t, the sample stage makes an angle θ with the incident X-ray vector, while the detector collects scattered or diffracted X-ray that makes an angle of 2θ with the incident X-ray. By convention, the angles of diffraction are recorded at 2θ instead of θ. 

Question

Why does the detector rotate at a higher angular rate than the sample stage?

Answer

With reference to the above diagram depicting the diffractometer, the detector will always point along a line that is parallel to the sample stage if its angular rate of rotation is the same as that of the sample stage. It must rotate at a higher rate than ω for any meaningful detection. The rate 2ω is conventionally used.

 

If there are sufficient grains with lattice planes parallel to the sample stage at t = t (which is statistically possible due to the presence of millions of randomly orientated grains) and Bragg’s law is satisfied, a diffraction signal is recorded.

Let’s assume that figures III, IV, V show the orientation of three grains of a sample of polonium, which has a primitive cubic unit cell dimension of 337 pm,  at time t = t where 2θ = 12.1o. If Mo X-ray of wavelength 71.0 pm were used, and that figures III, IV and V correspond to the family planes of {100}, {100} and {110} respectively , the only grain that has the required orientation to satisfy Bragg’s law is the one in figure III. This is validated by substituting all the relevant variables in eq13.

Figure IV does not satisfy Bragg’s equation, which requires θ1 = θ2 = θ, i.e. incident ray from the source and the scattered ray received by the detector must make the same angle with a lattice plane. Even though specular angles are observed in figure V, a (110) plane of polonium satisfies Bragg’s equation only if 2θ = 17.1o. This is because the interplanar distance of (110) is different from that of (100) and therefore causes scattered X-rays from adjacent lattice points of (110) planes to interfere destructively at 2θ = 12.1o.

Therefore, only one signal corresponding to the family planes of {100} is observed at 2θ = 12.1o. Using the same logic, signals corresponding to {110}, {111}, {200}, … are recorded at 2θ of 17.1o, 21.0o, 24.3o, …. Finally, a spectrum of intensity versus 2θ is generated by the diffractometer for analysis.

 

Question

Do the planes (010) and (001) for a primitive cubic system give the same diffraction signal as the (100) plane in powder X-ray diffraction?

Answer

Yes, as the grains are randomly distributed. Since the (010) and (001) planes for a primitive cubic unit cell are equivalent to the reference plane (100) by the symmetry of the lattice, the three planes belong to the same family of planes, denoted by {100}, i.e. the signal from {100} includes scattered X-rays from all these planes. For the purpose of unit cell identification, it is not necessary to distinguish the different planes within the same family of planes.

 

How do we identify the unit cell and determine its dimensions for an unknown sample using powder X-ray diffraction? Part of the solution requires the understanding of systematic absences, which is described in the next section.

 

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Interplanar distance (crystallography)

The interplanar distance is the perpendicular distance between two adjacent planes of a family of planes.

Consider a family of parallel planes {hkl} with the reference plane shown in the diagram below.

The interplanar distance, dhkl, is given by

\mathit{d_{hkl}}=OD=\frac{a}{h}cos\alpha \; \; \; \Rightarrow \; \; \; cos\alpha =\frac{h}{a}\mathit{d_{hkl}}

Similarly, we have

cos\beta =\frac{k}{b}d_{hkl}

cos\gamma =\frac{l}{c}d_{hkl}

For orthogonal axes, cos2α + cos2β + cos2γ = 1 (see below for proof). So,

(\frac{h}{a})^2{d_{hkl}}^{2}+(\frac{k}{b})^2{d_{hkl}}^{2}+(\frac{l}{c})^2{d_{hkl}}^{2}=1

\frac{1}{{d_{hkl}}^{2}}=\frac{h^2}{a^2}+\frac{k^2}{b^2}+\frac{l^2}{c^2}\; \; \; \; \; \; \; (3)

Eq3 describes the interplanar distance for an orthorhombic unit cell. For a primitive cubic unit cell, a = b = c and eq3 becomes

\frac{1}{{d_{hkl}}^{2}}=\frac{h^2+k^2+l^2}{a^2}\; \; \; \; \; \; \; (4)

The diagram below shows two parallel planes with Miller indices of (110) and (220).

 

The interplanar distance for (110), i.e. between the orange plane and the origin, is twice of that for (220). If we rewrite (220) as (n1 n1 n0), then d110 = ndn1,n1,n0 , where n = 2. In general, for parallel and equidistant planes,

d_{hkl}=nd_{nh,nk.nl}\; \; \; \; \; \; \; (5)

where n ∈ \mathbb{Z} and n is called the common factor between planes.

Substituting eq5 in eq3 and eq4, we have:

\frac{1}{{d_{nh,nk,nl}}^{2}}=\frac{(nh)^2}{a^2}+\frac{(nk)^2}{b^2}+\frac{(nl)^2}{c^2}\; \; \; \; \; \; \; (6)

\frac{1}{{d_{nh,nk,nl}}^{2}}=\frac{(nh)^2+(nk)^2+(nl)^2}{a^2}\; \; \; \; \; \; \; (7)

 

Question

Show that cos2α + cos2β + cos2γ = 1 for orthogonal axes.

Answer

With reference to the diagram below,

cos\alpha =\frac{a}{\left | N \right |}\; \; \;\; \; \; \; cos\beta =\frac{b}{\left | N \right |}\; \; \; \; \; \; \; cos\gamma =\frac{c}{\left | N \right |}

cos^2\alpha +cos^2\beta +cos^2\gamma =(\frac{a}{\left | N \right |})^2+(\frac{b}{\left | N \right |})^2+(\frac{c}{\left | N \right |})^2

Since a2 + b2 + c2 = INI2

cos^2\alpha +cos^2\beta +cos^2\gamma =1

 

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