Advanced Chemistry - Mono Mole

January 10, 2019July 18, 2024

Stokes’ law: overview

Stokes’ law, which was derived in 1851 by an Irish scientist named George Stokes, describes the drag force exerted on a spherical object in a viscous fluid through the formula:

$F_d=6\pi\mu ru$

where

F_dis the frictional force (drag) acting on the surface of the object

μ is the viscosity of the fluid

r is the radius of the spherical object

u is the flow velocity of the fluid

The derivation of the equation for Stokes’ law involves the following steps:

Step 1: Derive the continuity equation in spherical coordinates for an incompressible fluid.

Step 2: Define the Stokes stream function to satisfy the continuity equation for an incompressible fluid.

Step 3: Derive the Navier-Stokes equations.

Step 4: Derive the differential equation, E²(E²ψ) = 0.

Step 5: Solve the differential equation.

Step 6: Derive the Stokes’ law equation.

next article: Continuity equation

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Reciprocal lattice

A reciprocal lattice is the diffraction image of a real crystal lattice. It is formed by linear combinations of basis vectors of the reciprocal space.

In an earlier article, we mentioned that the three Laue equations must be simultaneously satisfied for constructive interference to occur in three dimensions. In other words, the solution to the Laue equations is a mathematical description of the diffraction pattern of an X-ray diffraction experiment.

Let h = (s – s₀) and we can rewrite eq20, eq21 and eq22 as:

$\textbf{\textit{a}}\cdot\textbf{\textit{h}}=h\; \; \; \; \; \; \; (24a)$

$\textbf{\textit{b}}\cdot\textbf{\textit{h}}=k\; \; \; \; \; \; \; (24b)$

$\textbf{\textit{c}}\cdot\textbf{\textit{h}}=l\; \; \; \; \; \; \; (24c)$

A general solution to the simultaneous equations of eq24a, eq24b and eq24c is:

$\textbf{\textit{h}}=h\textbf{\textit{a}}^*+k\textbf{\textit{b}}^*+l\textbf{\textit{c}}^*\; \; \; \; \; \; \; (24d)$

where $\textbf{\textit{a}}^*=\frac{\textbf{\textit{b}}\times\textbf{\textit{c}}}{V}$ , $\textbf{\textit{b}}^*=\frac{\textbf{\textit{c}}\times\textbf{\textit{a}}}{V}$ and $\textbf{\textit{c}}^*=\frac{\textbf{\textit{a}}\times\textbf{\textit{b}}}{V}$ . V is the volume of the unit cell formed by the basis vectors a, b and c.

Question

Show that eq24d is a general solution to the Laue equations.

Answer

Substitute eq24d in eq24a

$\textbf{\textit{a}} \cdot (h\textbf{\textit{a}}^*+k\textbf{\textit{b}}^*+l\textbf{\textit{c}}^*)=h$

$h\frac{\textbf{\textit{a}}\cdot (\textbf{\textit{b}}\times \textbf{\textit{c}})}{V} +k\frac{\textbf{\textit{a}}\cdot (\textbf{\textit{c}}\times \textbf{\textit{a}})}{V}+l\frac{\textbf{\textit{a}}\cdot (\textbf{\textit{a}}\times \textbf{\textit{b}})}{V}=h$

Since (c × a) is a vector that is perpendicular to the plane formed by c and a, a · (c × a) = IaI Ic × aI cos90º = 0. By the same logic, a · (a × b) = 0. Therefore,

$h\frac{\textbf{\textit{a}}\cdot (\textbf{\textit{b}}\times \textbf{\textit{c}})}{V} =h$

Since a · (b × c) = V (see below for proof), LHS = RHS. We can substitute eq24d in eq24a and eq24b with similar results.

The vector h in eq24d is also a lattice vector but with basis vectors of a*, b* and c* instead of a, b and c. The dimension of a* is length^-1because $\textbf{\textit{a}}^*=\frac{\textbf{\textit{b}}\times\textbf{\textit{c}}}{V}=\frac{ \left |\textbf{\textit{b}} \right | \left | \textbf{\textit{c}}\right |sin\theta\textbf{\textit{n}}}{V}$ , where n is a unit vector in the direction of a* (the unit vector n is necessary because a* is a vector but $\frac{\left | \textbf{\textit{b}}\right |\left | \textbf{\textit{c}}\right |sin\theta }{V}$ is a scalar). Similarly, b* and c* have dimensions of the reciprocal of length. Hence, h is known as a reciprocal lattice vector.

Just as a lattice vector describes the positions of lattice points in a space lattice, the reciprocal lattice vector mathematically defines the positions of lattice points in a reciprocal space lattice, which means that the diffraction pattern is constructed in a reciprocal space (see diagram below).

Even though the reciprocal lattice and the Ewald sphere allow crystallographers to visualise diffraction patterns in relation to various X-ray diffraction techniques, Bragg’s law is most often used to explain X-ray diffraction concepts. This is because Bragg’s law is easier to apply than the Laue equations, and in some cases involves the knowledge of just an angle θ to deduce the unit cell types and unit cell dimensions of a sample.

Question

Show that a · (b × c) = V.

Answer

With reference to the above diagram, a, b, c are the basis vectors of the unit cell. The area of the base of the unit cell, which is a parallelogram, is:

$base\: area=\vert\boldsymbol{\mathit{b}}\vert\vert\boldsymbol{\mathit{c}}\vert sin\theta=\vert\boldsymbol{\mathit{b}}\times\boldsymbol{\mathit{c}}\vert$

The volume of the unit cell, V, is:

$V=(base\: area)(height)=\left | \textbf{\textit{b}}\times\textbf{\textit{c}}\right | \left |\textbf{\textit{a}} \right |cos\phi=\textbf{\textit{a}}\cdot (\textbf{\textit{b}}\times\textbf{\textit{c}})$

a · (b × c) is known as the scalar triple product.

next article: ewald sphere

previous article: equivalence of the laue equations and the bragg equation

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December 13, 2018September 26, 2024

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 F_hkl = 0.

Similarly, F_hkl = 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 $-\infty$ to $\infty$ without affecting the values of the integrals because $\rho(xyz)$ is zero outside the crystal, we have

$F_{hkl}=\frac{V}{abc}\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}\rho (xyz)e^{i2\pi(\frac{hx}{a}+\frac{ky}{b}+\frac{lz}{c})}dxdydz\; \; \; \; \; \; \; (48a)$

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 F_hkl (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.

next article: phase problem

previous article: structure factor

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December 13, 2018October 15, 2024

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’ = Ae^i(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.

next article: structure factor

previous article: single crystal X-ray diffraction

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December 13, 2018May 12, 2025

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 $d$ -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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December 13, 2018September 26, 2024

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 d_nh,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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December 13, 2018September 26, 2024

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 r₀respectively, which makes r – r₀the vector from P₀to 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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December 13, 2018September 26, 2024

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 F_hkl is a complex function of the form z = re^iΦ, where r = IzI ,

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

Hence, F_hkl is composed of two factors, the modulus IF_hklI and the phase $e^{i\phi_{hkl}}$ . If we replace F_hkl 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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December 13, 2018October 29, 2024

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:

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.
Ω, defined by the rotation of the cradle around the cradle axis. Ω = 0 when the cradle is perpendicular to the incident X-ray.
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.
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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December 13, 2018October 26, 2024

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 V_molto 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 d₂₂₀is 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 6.02214078×10²³mol^-1. In 2017, the value was refined to 6.02214076×10²³mol^-1 and in Nov 2018, the Avogardo constant was defined as exactly 6.02214076×10²³mol^-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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