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Electron density as a Fourier transform of the structure factor

The electron densities of atoms in a crystal are Fourier transforms of the structure factor.

 

In the article on scattering factor, we have restricted the electron density ρ to a lattice point, which is too simplistic. We have also described the scattering factor (eq27) as

df=\rho e^{i\phi}dV

or in its integrated form

f=\int \rho e^{i\phi}dV\; \; \; \; \; \; \; (40)

If we now extend the distribution of ρ through the unit cell, eq40 becomes the structure factor:

F_{hkl}=\int \rho 'e^{i\phi}dV

where ρ’ = ρ(xyz) is the electron density at coordinates xyz in the unit cell and \phi=2\pi(\frac{hx}{a}+\frac{ky}{b}+\frac{lz}{c} ), i.e.

F_{hkl}=\int \rho(xyz)e^{i2\pi(\frac{hx}{a}+\frac{ky}{b}+\frac{lz}{c} )}dV\; \; \; \; \; \; \; (41)

Let dV be an infinitesimal volume of the unit cell with edges dx, dy, dz that are parallel to the unit cell axes of a, b, c (volume of unit cell is V). The ratio of dV/V must be equal to (dxdydz)/abc and we can rewrite eq41 as

F_{hkl}=\frac{V}{abc}\int_{0}^{a}\int_{0}^{b}\int_{0}^{c}\rho (xyz)e^{i2\pi(\frac{hx}{a}+\frac{ky}{b}+\frac{lz}{c})}dxdydz\; \; \; \; \; \; \; (42)

A Fourier series is an expansion series used to represent a periodic function and is given by:

f(x)=\sum_{n=0}^{\infty }[A_ncos(nx)+B_nsin(nx)]

or by its complex form

f(x)=\sum_{n=-\infty }^{\infty }c_ne^{i\frac{2\pi nx}{a}}

Due to the repetitive arrangement of atoms in a crystal, the electron density in a crystal is also periodic and can be expressed as a Fourier series:

\rho (x)=\sum_{n=-\infty }^{\infty }c_ne^{i\frac{2\pi nx}{a}}\; \; \; \; \; \; \; (43)

In three dimensions, eq43 becomes

\rho (xyz)=\sum_{n=-\infty }^{\infty }\sum_{m=-\infty }^{\infty }\sum_{o=-\infty }^{\infty }c_{nmo}e^{i2\pi(\frac{nx}{a}+\frac{ my}{b}+\frac{ oz}{c})}\; \; \; \; \; \; \; (44)

Substitute eq44 in eq42

F_{hkl}=\frac{V}{abc}\int_{0}^{a}\int_{0}^{b}\int_{0}^{c}\sum_{n=-\infty }^{\infty }\sum_{m=-\infty }^{\infty }\sum_{o=-\infty }^{\infty }c_{nmo}e^{i2\pi(\frac{nx}{a}+\frac{my}{b}+\frac{oz}{c})} e^{i2\pi(\frac{hx}{a}+\frac{ky}{b}+\frac{lz}{c})}dxdydz

F_{hkl}=\frac{V}{abc}\sum_{n=-\infty }^{\infty }\sum_{m=-\infty }^{\infty }\sum_{o=-\infty }^{\infty }c_{nmo}\int_{0}^{a}e^{i2\pi\frac{x}{a}(n+h)}dx\int_{0}^{b}e^{i2\pi\frac{y}{b}(m+k)}dy\int_{0}^{c}e^{i2\pi\frac{z}{c}(o+l)}dz\; \; \; \; (45)

If n ≠ –h,

\int_{0}^{a}e^{i2\pi\frac{x}{a}(n+h)}dx=\int_{0}^{a}[cos2\pi\frac{x}{a}(n+h)+isin2\pi\frac{x}{a}(n+h)]dx

=\int_{0}^{a}cos2\pi\frac{x}{a}(n+h)dx=0

which makes Fhkl = 0.

Similarly, Fhkl = 0 if m ≠ -k or o ≠ -l. Therefore, the surviving term in the triple summation in eq45 corresponds to the case of n = –h, m = –k, o = –l, giving

F_{hkl}=\frac{V}{abc}c_{-h,-k,-l}\int_{0}^{a}dx\int_{0}^{b}dy\int_{0}^{c}dz=Vc_{-h,-k,-l}

c_{-h,-k,-l}=\frac{F_{hkl}}{V}\; \; \; \; \; \; \; (46)

When n = –h, m = –k, o = –l, eq44 becomes

\rho (xyz)=\sum_{-h=-\infty }^{\infty }\sum_{-k=-\infty }^{\infty }\sum_{-l=-\infty }^{\infty } c_{-h,-k,-l}e^{-i2\pi(\frac{hx}{a}+\frac{ky}{b}+\frac{lz}{c})}

Since \sum_{-h=-\infty }^{\infty }\sum_{-k=-\infty }^{\infty }\sum_{-l=-\infty }^{\infty }=\sum_{h=\infty }^{-\infty }\sum_{k=\infty }^{-\infty }\sum_{l=\infty }^{-\infty }=\sum_{h=-\infty }^{\infty }\sum_{k=-\infty }^{\infty }\sum_{l=-\infty }^{\infty }

\rho (xyz)=\sum_{h=-\infty }^{\infty }\sum_{k=-\infty }^{\infty }\sum_{l=-\infty }^{\infty }c_{-h,-k,-l}e^{-i2\pi(\frac{hx}{a}+\frac{ky}{b}+\frac{lz}{c})}\; \; \; \; \; \; \; (47)

Substitute eq46 in eq47

\rho (xyz)=\frac{1}{V}\sum_{h=-\infty }^{\infty }\sum_{k=-\infty }^{\infty }\sum_{l=-\infty }^{\infty }F_{hkl}e^{-i2\pi(\frac{hx}{a}+\frac{ky}{b}+\frac{lz}{c})}\; \; \; \; \; \; \; (48)

Since the limits of the integrals in eq42 can be changed to run from to without affecting the values of the integrals because is zero outside the crystal, we have

Eq48 and eq48a are a Fourier transform pair. As mentioned in the previous article, 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. If we know the value of Fhkl (which in principle is the square root of the intensity of a peak from an X-ray diffraction experiment \sqrt{\left | F_{hkl} \right |^2}=\left | F_{hkl} \right | ) and having indexed the plane contributing to this intensity peak (i.e. knowing the h, k, l values), we can determine ρ(xyz) using a mathematical software. The solution to ρ(xyz) is an electron density map that elucidates bond lengths and bond angles of the compound. However, a problem called the phase problem arises.

 

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Scattering factor (crystallography)

The scattering factor describes the amplitude of scattered rays from a single atom.

The presence or absence of an intensity peak depends on the interference of scattered X-rays at the detector of the powder X-ray diffractometer. Since the interference of scattered X-rays is the sum of the amplitudes of X-ray waves from many atoms in a sample, we must begin with the analysis of the resultant amplitude of scattered rays from a single atom (scattering factor), which is, in turn, the sum of amplitudes of waves scattered by the electrons in the atom.

The amplitude of a travelling wave is expressed by the equation:

y=Acos(kx-\omega t+\phi )

We can also expressed the general wave equation in its complex form:

y=Ae^{i(kx-\omega t+\phi )}\; \; \; \; \; \; \; (25)

The complex form of the wave equation is equivalent to the travelling wave equation if we consider only the real part of it: Re(y), which is the value that is physically observed in an X-ray diffraction experiment. The reason eq25 is preferred over the travelling wave equation is that it is easier to manipulate mathematically to derive the scattering factor. Recall that ω = 2π/T (angular frequency), k = 2π/λ (wave number) and Φ is the phase difference. If we assume that the scattering of X-rays by an atom is elastic, i.e. there is no change in frequency of the X-ray between the incident and scattered rays, the term kx–ωt becomes a constant and we can simplify eq25 to:

y=A'e^{i\phi }\; \; \; \; \; \; \; (26)

where A’ = Aei(kx-ωt).

The amplitudes of the scattered X-rays not only differ in terms of their phase as suggested by eq26 but are also proportional to the number of electrons in the atom, ρdV, where ρ and V are electron density and volume of the atom respectively. If we define A’ as the number of electrons in the atom, eq26 becomes

df=\rho e^{i\phi }dV\; \; \; \; \; \; \; (27)

where the amplitude of the scattered X-rays from an atom is denoted by f, the scattering factor of an atom.

To determine an expression for Φ, we note that the ratio of path difference δ and wavelength is equal to the ratio of phase difference and 2π, i.e.

\frac{\delta }{\lambda }=\frac{\phi }{2\pi }\; \; \; \; \; \; \; (28)

Substitute eq19 in eq28,

\phi =2\pi\textbf{\textit{a}}\cdot (\textbf{\textit{s}}-\textbf{\textit{s}}_0)\; \; \; \; \; \; \; (29)

Since \textbf{\textit{a}}\cdot (\textbf{\textit{s}}-\textbf{\textit{s}}_0)=\left |\textbf{\textit{a}} \right |\left | \textbf{\textit{s}}-\textbf{\textit{s}}_0\right |cos\mu and from eq23, \left | \textbf{\textit{s}}-\textbf{\textit{s}}_0\right |=\frac{2sin\theta }{\lambda }, eq29 becomes:

\phi =\frac{4\pi }{\lambda }\left | \textbf{\textit{a}}\right |sin\theta cos\mu

In three dimensions, we let a = r and IaI = IrI = r,

\phi =\frac{4\pi }{\lambda }rsin\theta cos\mu\; \; \; \; \; \; \; (30)

Substitute eq30 in eq27

df=\rho e^{ikrcos\mu }dV\; \; \; \; \; \; \; (31)

where k=\frac{4\pi }{\lambda }sin\theta

Since ρ is spherically symmetrical, we can represent the volume element dV in eq31 in spherical coordinates where dV = r2sinμdrdμdΦ (see above diagram). Eq31 becomes:

df=\rho e^{ikrcos\mu }r^2sin\mu drd\mu d\phi

Integrating both sides,

f=\int_{0}^{\infty }\int_{0}^{\pi }\int_{0}^{2\pi }\rho e^{ikrcos\mu }r^2sin\mu drd\mu d\phi

f=2\pi \int_{0}^{\infty }\rho r^2dr\int_{0}^{\pi }e^{ikrcos\mu }sin\mu d\mu \; \; \; \; \; \; \; (32)

Note that d(cosμ) = -sinμdμ; when μ = π, cosμ = -1; when μ = 0, cosμ = 1. So the second integral in eq32 becomes

\int_{0}^{\pi }e^{ikrcos\mu }sin\mu d\mu =-\int_{1}^{-1 }e^{ikrcos\mu } d(cos\mu)=\int_{-1}^{1 }e^{ikrcos\mu }d(cos\mu)

=\left [ \frac{1}{ikr}e^{ikrcos\mu } \right ]_{-1}^{1}=\frac{e^{ikr}-e^{-ikr}}{ikr}

Eq32 now becomes

f=2\pi \int_{0}^{\infty }\rho r^2 \frac{e^{ikr}-e^{-ikr}}{ikr}dr

Since e^{ikr}-e^{-ikr}=coskr+isinkr-(coskr-isinkr)=2isinkr

f=4\pi \int_{0}^{\infty }\rho \frac{sinkr}{kr}r^2dr\; \; \; \; \; \; \; (33)

where k=\frac{4\pi }{\lambda }sin\theta. ρ is a function of r and remains within the integral.

So far, we have developed an expression (eq33) that describes the resultant amplitude of scattered rays from a single atom. To further analyse the interference of scattered X-rays from multiple atoms in a sample, we have to derive another expression called the structure factor.

 

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Ewald sphere (crystallography)

The Ewald sphere is a mathematical construct that relates the geometry of the incident and scattered wave vectors to the reciprocal lattice.

Consider a crystal at A being irradiated by an incident wave vector s0, scattering a wave vector s that makes an angle 2θ with s0, thereby satisfying Bragg’s law (see diagram below).

Since X-ray scattering is elastic, IsI = Is0I and the two wave vectors become radii of a sphere called the Ewald sphere. The vector OP is the reciprocal lattice vector (s – s0) that is denoted by h with the origin at O.

\frac{OP}{IO}=\frac{\left | \textbf{\textit{h}}\right |}{IO}=sin\theta\; \; \; \; \; \; \; (24e)

Since h = (s – s0), eq24 becomes

d_{nh,nk,nl}=\frac{1}{\left | \textbf{\textit{h}}\right |}\; \; \; \; \; \; \; (24f)

From eq11, sin\theta =\frac{\lambda }{2d_{nh,nk,nl}}. Substitute sin\theta =\frac{\lambda }{2d_{nh,nk,nl}} and eq24f in eq24e, we have:

IO=\frac{2}{\lambda }

This means that IA = AO = AP = 1/λ. Therefore, to satisfy Bragg’s law that results in constructive interference of scattered X-rays, the head of the reciprocal lattice vector must lie on any point on the surface of the Ewald sphere. From eq24f and eq11,

\left | \textbf{\textit{h}}\right |=\frac{2sin\theta }{\lambda }

Since -1 ≤ sinθ ≤ 1,

\left | \textbf{\textit{h}}\right |_{max}=\frac{2 }{\lambda }\; \; \; \; \; \; \; (24g)

As the maximum magnitude of the reciprocal lattice vector is 2/λ, all reciprocal lattice vectors that potentially satisfy the Laue equations or Bragg’s law are enclosed in a sphere of radius known as the limiting sphere. The Ewald sphere is used to visualise different X-ray diffraction techniques including single crystal X-ray diffraction.

 

Question

Is the Ewald sphere really necessary? Can all diffraction patterns be discerned without the construct?

Answer

While it’s true that the fundamental condition for diffraction is Bragg’s Law or the Laue conditions, the Ewald sphere is an incredibly valuable and practically necessary construct for several key reasons:

Visualisation of Multiple Reflections: Bragg’s Law and the Laue conditions describe the diffraction from a single set of crystal planes. However, a crystal contains a vast number of different sets of planes, each with its own -spacing and orientation. The Ewald sphere provides a single, elegant way to simultaneously visualise which of these many reciprocal lattice points (representing all possible sets of planes) are in the correct orientation to satisfy the diffraction condition for a given incident X-ray wavelength and crystal orientation. Without it, you would have to consider each set of planes individually, making it much harder to grasp the overall diffraction pattern. 

Understanding the Influence of Wavelength and Orientation: The Ewald sphere clearly illustrates how changing the incident X-ray wavelength (which changes the radius of the sphere) or rotating the crystal (which rotates the reciprocal lattice) affects which reflections will be observed. This is crucial for designing and interpreting diffraction experiments. When you rotate the crystal, the reciprocal lattice (which is mathematically linked to the crystal lattice) also rotates by the same amount around the origin of the reciprocal space. As the reciprocal lattice rotates, different reciprocal lattice points will, at certain crystal orientations (and thus reciprocal lattice orientations), pass through the surface of the Ewald sphere. Each time a reciprocal lattice point touches the sphere, the conditions for diffraction from the corresponding set of crystal planes are met, and a diffraction spot at a specific angle is observed. In practice, software used in X-ray diffraction experiments utilises the Ewald sphere concept to simulate diffraction patterns.

Interpretation and Indexing: The diffraction pattern observed on a detector is a projection of the reciprocal lattice points that intersect the Ewald sphere. Understanding the Ewald sphere helps in indexing these diffraction spots and relating them back to the crystal’s reciprocal lattice and ultimately its real-space structure. The software uses the Ewald sphere concept and the known or refined unit cell parameters to index the reflections, assigning the correct Miller indices (hkl) to each spot by finding the reciprocal lattice point that was in diffracting condition.

Although it is theoretically possible to discern some diffraction patterns without explicitly drawing an Ewald sphere, especially for very simple cases or with extensive calculations based directly on Bragg’s Law and crystal geometry, the Ewald sphere is the essential tool that connects these laws to the observable diffraction pattern in a comprehensive and practical way.

 

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Equivalence of the Laue equations and the Bragg equation

The Laue equations are equal to the Bragg equation if we denote the angle between the wave vectors s and s0  by 2θ (see diagram below), which makes the vector s – s0  normal to the lattice plane.

As the scattering of X-rays by an atom is assumed to be elastic (no loss in momentum), the magnitudes of the wave vectors are the same. Thus, s – s0 becomes the base of an isosceles triangle with equal sides of Is0I and IsI. We have:

\left | \textbf{\textit{s}}-\textbf{\textit{s}}_0\right |=\left | \textbf{\textit{s}}\right |sin\theta +\left | \textbf{\textit{s}}_0\right |sin\theta

Since \left | \textbf{\textit{s}}\right |=\left | \textbf{\textit{s}}_0\right |=\frac{1}{\lambda } (see previous article)

\left | \textbf{\textit{s}}-\textbf{\textit{s}}_0\right |=\frac{2sin\theta }{\lambda }\; \; \; \; \; \; \; (23)

To find an expression for dnh,nk,nl (see above diagram), we determine the scalar projection of a/h (the vector linking two plane-intercept points on the a-axis) on the vector s – s0 to give:

d_{nh,nk,nl}=\frac{\textbf{\textit{a}}}{h}\cdot \frac{\textbf{\textit{s}}-\textbf{\textit{s}}_0}{\left | \textbf{\textit{s}}-\textbf{\textit{s}}_0\right |}

Dividing both sides of eq20 by h, we have \frac{\textbf{\textit{a}}}{h}\cdot ( \textbf{\textit{s}}-\textbf{\textit{s}}_0) =1. So,

d_{nh,nk,nl}=\frac{1}{\left | \textbf{\textit{s}}-\textbf{\textit{s}}_0\right |}\; \; \; \; \; \; \; (24)

Substitute eq23 in eq24, we have the Bragg equation:

2d_{nh,nk,nl}\: sin\theta =\lambda

Since the Laue equations are equal to the Bragg equation if we denote the angle between the wave vectors s and s0  by 2θ, the Bragg equation is a specific form of the Laue equations with the condition of 2θ imposed.

 

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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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