Hello there! Welcome. You've got a great set of questions on X-ray Diffraction, which is one of the most powerful tools we have for looking at the arrangement of atoms in solid materials. Think of it as our 'atomic-scale vision'.
The whole idea is based on a simple principle:
crystals are periodic, 3D arrangements of atoms
. This periodic structure acts like a
diffraction grating
for X-rays. Why X-rays? Because their wavelength (λ ~ 0.1 nm or 1 Å) is on the same order of magnitude as the spacing between atoms in a crystal.
Let's build up all the theory you'll need, starting from the ground up.
💡
Tutor's Note:
The two most important things to get from XRD are:
Peak Positions:
These tell you about the
geometry
of the crystal (its size, shape, and symmetry - the unit cell).
Peak Intensities:
These tell you about the
contents
of the unit cell (
what
atoms are present and
where
they are).
1. Bragg's Law: The Foundation of Diffraction
This is the single most important equation in XRD. It describes the condition for
constructive interference
from X-rays scattering off a set of parallel crystal planes.
Derivation of the Bragg Condition
Imagine two parallel X-ray beams (Ray 1 and Ray 2) hitting two parallel planes of atoms (Plane 1 and Plane 2) separated by a distance
$d$
. They strike the surface at an angle
$\theta$
(the 'Bragg angle').
Ray 1 strikes an atom on Plane 1 and is scattered.
Ray 2 travels
further
, strikes an atom on Plane 2, and is also scattered.
For the scattered rays to be
in-phase
(to interfere constructively and give us a signal), the extra distance traveled by Ray 2 must be an
integer multiple ($n$)
of the X-ray wavelength ($\lambda$).
Let's look at the geometry. The extra path consists of two parts: the path 'in' (segment AB) and the path 'out' (segment BC).
In the right-angled triangle OAB, $\sin\theta = AB / OB = AB / d$. So,
$AB = d \sin\theta$
.
Similarly, in triangle OBC,
$BC = d \sin\theta$
.
The total path difference is $AB + BC = d \sin\theta + d \sin\theta = 2d \sin\theta$.
The condition for constructive interference is:
Path Difference = $n\lambda$
.
This gives us the famous
Bragg's Law
:
$$ 2d \sin\theta = n\lambda $$
$d$
: The interplanar spacing (distance between the planes).
$\theta$
: The Bragg angle of incidence.
$n$
: The 'order' of reflection (an integer: 1, 2, 3...). A second-order reflection ($n=2$) from (100) planes is physically the same as a first-order reflection ($n=1$) from a set of planes with half the spacing, which we call the (200) planes. Because of this, we often set $n=1$ and just talk about the (200), (300), etc. planes.
Use Bragg's Law to find $d$:
$2d_{100} \sin\theta = (1)\lambda$
$d_{100} = \frac{\lambda}{2 \sin\theta} = \frac{1.54 \text{ Å}}{2 \sin(20.0^\circ)} = \frac{1.54 \text{ Å}}{2 \times 0.342} = 2.25$
Å
Relate $d$ to 'a':
For a cubic system, the spacing for (hkl) planes is $d_{hkl} = \frac{a}{\sqrt{h^2+k^2+l^2}}$.
For (100) planes, $d_{100} = \frac{a}{\sqrt{1^2+0^2+0^2}} = \frac{a}{1} = a$.
Conclusion:
The lattice parameter
$a = d_{100} = 2.25$ Å
.
2. Describing Crystal Planes and Spacing
Miller Indices (hkl)
Bragg's Law talks about 'planes', but how do we describe them? We use
Miller Indices (hkl)
, which are a set of three integers that uniquely define a set of parallel planes in the crystal.
How to find them:
Find Intercepts:
Find where the plane intercepts the crystallographic axes (a, b, c). Express these intercepts as fractions of the unit cell lengths. (e.g., 1a, ½b, ∞c).
Take Reciprocals:
Take the reciprocal of these numbers. (e.g., 1/1, 1/(½), 1/∞ which gives 1, 2, 0).
Clear Fractions:
Reduce these numbers to the smallest set of integers (not necessary here).
So, the plane is the
(120)
plane.
A plane parallel to an axis has an intercept at ∞, so its Miller index is 0. (e.g., (100) is parallel to b and c).
A plane that cuts a negative axis is shown with a bar: (1$\bar{1}$0).
(hkl)
: A specific plane.
{hkl}
: A family of symmetrically equivalent planes. In a cubic crystal, {100} includes (100), (010), (001), ($\bar{1}$00), etc.
Interplanar Spacing ($d_{hkl}$)
The distance 'd' in Bragg's law depends on which set of (hkl) planes you're talking about. For an
orthorhombic
system (where $a \neq b \neq c$ but all angles are 90°), the formula is:
This is a more advanced, but incredibly powerful, way to think about diffraction.
Real Lattice:
The 3D grid of atoms/points in real space.
Reciprocal Lattice:
A 3D grid of points in 'reciprocal space' (which has units of 1/distance). Each
point (hkl)
in the reciprocal lattice corresponds to a
set of planes (hkl)
in the real lattice. The vector from the origin (000) to the point (hkl) is perpendicular to the (hkl) planes and has a length $1/d_{hkl}$.
Ewald Sphere Construction
This is a beautiful geometric picture of Bragg's Law.
Draw your reciprocal lattice, with the origin (000) at one point.
The incident X-ray beam (a vector
k
in
with length $1/\lambda$) comes in and points to the origin (000) of the reciprocal lattice.
Draw a sphere (the
Ewald Sphere
) with a radius of
$1/\lambda$
, such that the tail of the
k
in
vector is at the center of the sphere.
Diffraction occurs
if and only if another reciprocal lattice point (hkl) lies
exactly
on the surface of this sphere.
If this happens, the vector from the center of the sphere to that point (hkl) is the vector of the diffracted X-ray beam,
k
out
.
This shows why diffraction is a 'special' event. For a fixed crystal and a fixed wavelength, it's very unlikely any points will be on the sphere. That's why, in experiments, we have to
change
something:
Rotating Crystal / Powder Method:
We
rotate the crystal
, which
rotates the reciprocal lattice
, sweeping points
through
the sphere.
Laue Method:
We use
polychromatic
X-rays, which means we are using a 'nested set' of Ewald spheres with different radii ($1/\lambda$).
4. Experimental Methods: How We Measure XRD
There are three main 'flavors' of XRD experiments.
The Laue Method
Beam:
Polychromatic (white) X-rays.
Sample:
Fixed
single crystal
.
How it works:
The crystal sits still. The range of wavelengths ensures that
some
$\lambda$ will satisfy the Bragg condition for
some
plane (hkl).
Use:
Primarily for determining crystal
orientation
and
symmetry
. The pattern of 'Laue spots' reveals the internal symmetry (e.g., a 4-fold symmetry in the pattern means the crystal was aligned on a 4-fold axis).
Types:
Back-reflection (for thick samples) and transmission.
Rotating Crystal Method
Beam:
Monochromatic ($\lambda$ is fixed).
Sample:
Single crystal
, which is rotated.
How it works:
As the crystal rotates, it sweeps reciprocal lattice points through the Ewald sphere, causing diffraction. The diffracted beams form spots on a cylindrical film, which appear as layers ('layer lines').
Use:
Was historically crucial for determining
lattice parameters
. The spacing of the layer lines tells you the unit cell dimension along the axis of rotation.
Debye-Scherrer (Powder) Method
This is the most common method you'll encounter in a chemistry lab.
Beam:
Monochromatic ($\lambda$ is fixed).
Sample:
Polycrystalline (powder)
. This is the key! A powder is a huge collection of tiny, randomly oriented crystallites.
How it works:
Because of the random orientation,
some
crystallites will
always
be at the correct Bragg angle ($\theta$) for
every
possible (hkl) plane.
For a given (hkl) plane, the diffracted beams from all the correctly-oriented crystals don't form a single spot, but a
cone
(a 'Debye cone') with a semi-angle of
$2\theta$
.
A detector (either a film strip or a modern electronic detector) intercepts these cones, recording them as lines or peaks at specific $2\theta$ values.
Use:
The
#1 tool for phase identification
(fingerprinting) and for precise
lattice parameter determination
of known structures.
Comparison Summary
Technique
X-ray Beam
Sample
Primary Use
Laue Method
Polychromatic
Fixed Single Crystal
Orientation & Symmetry
Rotating Crystal
Monochromatic
Rotating Single Crystal
Lattice Parameters
Debye-Scherrer (Powder)
Monochromatic
Polycrystalline (Powder)
Phase ID & Lattice Parameters
5. Intensity of Diffraction: What's in the Unit Cell
Okay, so Bragg's law tells us
where
the peaks can be. But why are some strong and some weak? And why are some
missing entirely
? This is all about what's
inside
the unit cell.
Atomic Scattering Factor ($f$)
This is the scattering power of a
single atom
.
It's roughly proportional to the
number of electrons (Z)
. A Pb atom (Z=82) scatters much more strongly than a C atom (Z=6).
It's
angle-dependent
. Atoms are 'fluffy' electron clouds, not points. At high angles ($\theta$), interference
within
the atom's own electron cloud causes the scattering to drop off. It's highest at $\theta = 0^\circ$.
Structure Factor ($F_{hkl}$)
This is the
key
. The Structure Factor is the
total scattering
from
all atoms within one unit cell
, considering their positions. It's a sum of all the atomic scattering factors ($f_j$), but with
phase shifts
($\phi_j$) that depend on each atom's position ($u_j, v_j, w_j$).
The
Intensity
we measure is proportional to the
square
of the
magnitude
of the structure factor:
$I_{hkl} \propto |F_{hkl}|^2$
.
Systematic Absences (Extinction Rules)
This is what happens when the atoms are arranged in such a high-symmetry way that the scattered waves from different atoms
perfectly cancel out
for certain (hkl) planes. This means
$F_{hkl} = 0$
, and the peak is
absent
. This is
extremely
useful for identifying the lattice type!
Derivation for Body-Centered Cubic (BCC)
Atoms:
Two identical atoms ($f$) per cell.
Positions:
Atom 1 at (0, 0, 0) and Atom 2 at (½, ½, ½).
Derivation:
$F_{hkl} = f_1 e^{2\pi i (h\cdot 0 + k\cdot 0 + l\cdot 0)} + f_2 e^{2\pi i (h\cdot \frac{1}{2} + k\cdot \frac{1}{2} + l\cdot \frac{1}{2})}$
$F_{hkl} = f \cdot e^0 + f \cdot e^{\pi i (h+k+l)}$
$F_{hkl} = f (1 + e^{\pi i (h+k+l)})$
Now, use $e^{\pi i N} = (-1)^N$.
Case 1: ($h+k+l$) is EVEN
(e.g., (110), (200), (222))
$F_{hkl} = f (1 + e^{\pi i (\text{even})}) = f (1 + 1) =$
$2f$
(Reflection is
PRESENT
)
Case 2: ($h+k+l$) is ODD
(e.g., (100), (111), (210))
$F_{hkl} = f (1 + e^{\pi i (\text{odd})}) = f (1 - 1) =$
$0$
(Reflection is
ABSENT
)
BCC Rule:
Reflections are only seen when
$h+k+l = \text{even}$
.
Derivation:
$F_{hkl} = f [e^0 + e^{\pi i (h+k)} + e^{\pi i (h+l)} + e^{\pi i (k+l)}]$
$F_{hkl} = f [1 + e^{\pi i (h+k)} + e^{\pi i (h+l)} + e^{\pi i (k+l)}]$
This is a bit trickier, but the result is:
Case 1: $h, k, l$ are all EVEN or all ODD
(e.g., (111), (200), (220))
All exponents are even multiples of $\pi i$, so all terms are +1.
$F_{hkl} = f (1 + 1 + 1 + 1) =$
$4f$
(Reflection is
PRESENT
)
Case 2: $h, k, l$ are MIXED
(e.g., (100), (110), (210))
Two of the $e$ terms will be -1, and two will be +1.
$F_{hkl} = f (1 + ... ) =$
$0$
(Reflection is
ABSENT
)
FCC Rule:
Reflections are only seen when
$h, k, l$ are all even or all odd (unmixed)
.
Example: NaCl Structure
Structure:
FCC lattice. Basis is Na$^+$ at (0,0,0) and Cl$^-$ at (½, ½, ½).
Derivation:
We can group this:
$F_{hkl} = F_{hkl}(\text{Na}^+) + F_{hkl}(\text{Cl}^-)$
$F_{hkl} = \{ f_{Na^+} [1 + e^{\pi i (h+k)} + e^{\pi i (h+l)} + e^{\pi i (k+l)}] \} + \{ f_{Cl^-} [e^{\pi i (h+k+l)} + e^{\pi i (2h+k+l)} + e^{\pi i (h+2k+l)} + e^{\pi i (h+k+2l)}] \}$
This simplifies to:
$F_{hkl} = [1 + e^{\pi i (h+k)} + e^{\pi i (h+l)} + e^{\pi i (k+l)}] \times [f_{Na^+} + f_{Cl^-} e^{\pi i (h+k+l)}]$
Conditions:
1. $h,k,l$ mixed:
The first bracket is 0. So
$F_{hkl} = 0$
. (Absent)
2. $h,k,l$ all odd:
First bracket is 4. ($h+k+l$) is odd.
$F_{hkl} = 4 \times [f_{Na^+} + f_{Cl^-} (-1)] =$
$4(f_{Na^+} - f_{Cl^-})$
(
Weak
reflections)
3. $h,k,l$ all even:
First bracket is 4. ($h+k+l$) is even.
$F_{hkl} = 4 \times [f_{Na^+} + f_{Cl^-} (+1)] =$
$4(f_{Na^+} + f_{Cl^-})$
(
Strong
reflections)
Other Factors Affecting Measured Intensity
The final measured intensity $I_{hkl}$ is not just $\propto |F_{hkl}|^2$. We have to correct for several other factors:
Lorentz-Polarization (Lp) Factor:
This corrects for two effects. The
Polarization
part accounts for the fact that the unpolarized incident X-rays become polarized upon scattering. The
Lorentz
part accounts for a geometric effect: planes at different $2\theta$ values spend different amounts of time in the correct diffracting orientation. The combined factor is $\frac{1+\cos^2(2\theta)}{\sin^2\theta \cos\theta}$.
Multiplicity Factor ($p$):
How many
equivalent
planes contribute to the same peak? In a powder sample, (100) diffraction is the same as (010), (001), ($\bar{1}$00), etc. For a cubic crystal, the {100} family has $p=6$. The {111} family has $p=8$. The {110} family has $p=12$. More equivalent planes = higher intensity.
Temperature Factor (Debye-Waller Factor):
Atoms are not static; they vibrate. This thermal vibration 'smears' the electron density, which
reduces
the scattering intensity. This effect is
worse at high angles
(high $2\theta$).
Absorption Factor:
The sample itself absorbs X-rays.
So, the full intensity formula is more like:
$I_{hkl} = (\text{Scale}) \times p \times (Lp) \times |F_{hkl}|^2 \times (\text{Temp. Factor}) \times (\text{Absorption})$
.
6. Applications: What We Do With XRD Data
A. Phase Identification (JCPDS/ICDD)
💡
Tutor's Note:
This is the #1 use of powder XRD. The powder pattern is a
unique fingerprint
for a crystalline solid. If your pattern matches a database pattern, you've identified your compound.
JCPDS/ICDD:
The Joint Committee on Powder Diffraction Standards (now the International Centre for Diffraction Data) maintains a
massive database
of known diffraction patterns.
The 'Card File':
Each entry (historically on a card, now digital) contains:
The chemical formula and name.
A list of
$d$-spacings
(calculated from $2\theta$ and $\lambda$).
A list of
relative intensities
($I/I_1$), where the strongest peak is set to 100%.
Lattice parameters, crystal system, etc.
Hanawalt Method:
The classic search-match procedure. You take your unknown pattern, find the $d$-spacings for your
three strongest peaks
, and use those to search the database. This narrows down the possibilities very quickly. You then check if the rest of your peaks match the candidate pattern.
B. Powder Pattern Indexing
This is how you figure out the unit cell (lattice type and 'a') from an
unknown
cubic pattern.
From your list of $2\theta$ values, calculate $\sin^2\theta$ for each peak.
From Bragg's Law and the cubic $d$-spacing formula:
$\sin^2\theta = \frac{n^2 \lambda^2}{4 d^2}$
(Let's assume $n=1$)
$d^2 = \frac{a^2}{h^2+k^2+l^2}$
Combining these:
$\sin^2\theta = \frac{\lambda^2}{4a^2} (h^2+k^2+l^2)$
Let $N = h^2+k^2+l^2$ (an integer) and $C = \frac{\lambda^2}{4a^2}$ (a constant). So,
$\sin^2\theta = C \times N$
.
This means the
ratios
of your $\sin^2\theta$ values must be
ratios of integers (N)
.
You divide all your $\sin^2\theta$ values by the
smallest
$\sin^2\theta$ value. The resulting numbers should look like a clear integer sequence.
Once you identify the lattice type and assign (hkl) values (i.e., $N$ values) to each peak, you can calculate the constant $C$ for each peak. Average them, and then solve for 'a':
$a = \sqrt{\frac{\lambda^2}{4C}}$
.
C. Lattice Parameter & Density Calculation
If you
know
the structure (like Mo is BCC) and can
index
a peak (like (211) at $2\theta = 73.6^\circ$), you can find the density.
Given: Mo (BCC), (211) peak at $2\theta = 73.6^\circ$, $\lambda = 1.54$ Å, M = 95.96 g/mol.
Find Density ($\rho$):
$\rho = \frac{n \times M}{N_A \times V_{\text{cell}}}$
$n$: atoms per cell. For BCC,
$n=2$
.
$M$: Molar Mass =
95.96 g/mol
.
$N_A$: Avogadro's Number =
$6.022 \times 10^{23}$ mol$^{-1}$
.
$V_{\text{cell}}$: Volume of unit cell = $a^3$.
Crucial:
Convert 'a' from Å to cm!
$a = 3.147 \text{ Å} = 3.147 \times 10^{-8}$ cm
$V_{\text{cell}} = (3.147 \times 10^{-8} \text{ cm})^3 = 3.118 \times 10^{-23}$ cm$^3$
$\rho = \frac{2 \times 95.96 \text{ g/mol}}{(6.022 \times 10^{23} \text{ mol}^{-1}) \times (3.118 \times 10^{-23} \text{ cm}^3)}$
$\rho = \frac{191.92}{18.778} \text{ g/cm}^3 =$
$10.22 \text{ g/cm}^3$
(The real value is 10.28 g/cm$^3$, so we're very close!)
D. Crystallite Size & The Scherrer Equation
One last thing. Perfect, infinite crystals give infinitely sharp diffraction peaks. But real crystals are small.
When the crystal domains (crystallites) are very small (typically
< 100-200 nm
), the diffraction peaks become
broad
.
This is an uncertainty principle effect: a small crystal (small $\Delta x$) leads to uncertainty in momentum, which 'smears' the diffraction angle.
The
Scherrer Equation
relates this broadening to the average crystallite size ($L$).
$$ L = \frac{K\lambda}{\beta \cos\theta} $$
$L$
: Average crystallite size (diameter).
$K$
: The Scherrer constant, a shape factor, usually just approximated as
$K \approx 0.9$
.
$\lambda$
: X-ray wavelength.
$\beta$
: The
peak broadening
, measured as the Full-Width at Half-Maximum (FWHM).
CRITICAL:
$\beta$ must be in
RADIANS
! And it's the
instrument-corrected
broadening ($\beta_{\text{sample}}^2 = \beta_{\text{measured}}^2 - \beta_{\text{instrument}}^2$).
$\theta$
: The Bragg angle (
not
$2\theta$).
This is a cornerstone of
nanoscience
, as it's a simple lab-based way to get a quick estimate of nanoparticle size.
And that's the grand tour! You can see how we go from a simple observation of spots, to deriving Bragg's Law, to understanding
why
certain spots are missing (Structure Factor), and finally to using this to identify materials, determine their structures, and even measure their size.
🔬 Answers to Your X-ray Diffraction Questions
Excellent! Let's go through those questions one by one. I'll break down the answer for each in a clear, step-by-step manner, just like we'd do in a tutorial session.
1. Bragg's Law Derivation and Application
Part A: Derivation of Bragg's Law
The Bragg condition describes when X-rays scattering from parallel crystal planes interfere
constructively
, creating a diffracted beam.
The Setup:
Imagine two parallel X-ray beams (Ray 1, Ray 2) hitting two parallel crystal planes separated by a distance
$d$
. The beams strike at an angle
$\theta$
.
The Path Difference:
Ray 1 hits the top plane and scatters. Ray 2 travels an
extra distance
(segments AB + BC) to hit the second plane before scattering.
Geometry:
Using the right-angled triangle OAB, the extra path 'in' is segment AB. We have $\sin\theta = AB / d$, so
$AB = d \sin\theta$
. The extra path 'out' (segment BC) is identical.
Total Path Difference:
The total extra distance traveled by Ray 2 is $AB + BC = d \sin\theta + d \sin\theta = 2d \sin\theta$.
Constructive Interference:
For the two scattered rays to be in-phase, this extra path difference must be an
integer multiple ($n$)
of the X-ray wavelength ($\lambda$).
The Condition:
This gives us Bragg's Law:
$$ 2d \sin\theta = n\lambda $$
Part B: Application
Given:
$n=1$
(first-order),
$(hkl) = (100)$
,
$\theta = 20.0^\circ$
,
$\lambda = 1.54$
Å. We need to find the lattice parameter 'a'.
Find $d$-spacing:
Start with Bragg's Law and solve for $d$:
$2d_{100} \sin\theta = n\lambda$
$d_{100} = \frac{n\lambda}{2 \sin\theta} = \frac{(1) \times (1.54 \text{ Å})}{2 \sin(20.0^\circ)}$
Relate $d$ to 'a':
For a cubic system, the formula for interplanar spacing is:
$d_{hkl} = \frac{a}{\sqrt{h^2+k^2+l^2}}$
Solve for (100):
$d_{100} = \frac{a}{\sqrt{1^2+0^2+0^2}} = \frac{a}{\sqrt{1}} = a$
Conclusion:
Therefore, the lattice parameter
$a = d_{100} = 2.25$ Å
.
2. The Debye-Scherrer Method
This is the workhorse powder diffraction technique.
Principle:
This method uses a
monochromatic
X-ray beam and a
polycrystalline (powder)
sample. A powder contains millions of tiny crystallites in random orientations. This randomness is key: it guarantees that for
any
possible set of crystal planes
$(hkl)$
,
some
crystallites will be perfectly oriented to satisfy the Bragg condition ($2d_{hkl} \sin\theta = \lambda$).
Experimental Setup:
A monochromatic X-ray beam is produced and
collimated
(made narrow and parallel).
The beam strikes the powder sample, which is often held in a thin glass capillary or mounted on a flat plate at the
center
of a cylindrical camera.
A
detector
(historically, a strip of photographic film; now, an electronic detector) is placed in a circle around the sample.
Resulting Pattern:
For a single set of planes
$(hkl)$
, the diffracted rays don't just go in one direction. Because the crystallites are randomly oriented, the diffracted beams form a
cone
. The axis of this cone is the incident beam, and the semi-angle of the cone is
$2\theta$
.
The detector intercepts these concentric cones, recording them as curved
lines
(on film) or
peaks
(with an electronic detector) at specific
$2\theta$
positions. Each cone corresponds to a different
$d$
-spacing in the crystal.
3. Systematic Absences and Lattice Type
Systematic Absences
(or extinction rules) are reflections that are "missing" (have zero intensity) in a diffraction pattern. This isn't an accident; it's a direct consequence of destructive interference from atoms located at high-symmetry positions within the unit cell (like body-centers or face-centers). They are the key to telling different lattice types apart.
Derivation for Body-Centered Cubic (BCC)
Atoms:
2 identical atoms per cell, at coordinates (0, 0, 0) and (½, ½, ½).
Structure Factor ($F_{hkl}$):
$F_{hkl} = \sum_{j} f_j e^{2\pi i (hu_j + kv_j + lw_j)}$
$F_{hkl} = f e^{2\pi i (0)} + f e^{2\pi i (h/2 + k/2 + l/2)}$
$F_{hkl} = f [1 + e^{\pi i (h+k+l)}]$
Analysis:
(Using $e^{\pi i N} = (-1)^N$)
If $(h+k+l)$ is EVEN:
$F_{hkl} = f (1 + e^{\pi i (\text{even})}) = f(1 + 1) = 2f$. The reflection is
PRESENT
.
If $(h+k+l)$ is ODD:
$F_{hkl} = f (1 + e^{\pi i (\text{odd})}) = f(1 - 1) = 0$. The reflection is
ABSENT
.
Derivation for Face-Centered Cubic (FCC)
Atoms:
4 identical atoms per cell, at (0, 0, 0), (½, ½, 0), (½, 0, ½), (0, ½, ½).
Structure Factor ($F_{hkl}$):
$F_{hkl} = f [e^{0} + e^{2\pi i (h/2 + k/2)} + e^{2\pi i (h/2 + l/2)} + e^{2\pi i (k/2 + l/2)}]$
$F_{hkl} = f [1 + e^{\pi i (h+k)} + e^{\pi i (h+l)} + e^{\pi i (k+l)}]$
Analysis:
If $h, k, l$ are UNMIXED (all even or all odd):
All exponents $(h+k)$, $(h+l)$, $(k+l)$ are even numbers.
$F_{hkl} = f(1 + 1 + 1 + 1) = 4f$. The reflection is
PRESENT
.
If $h, k, l$ are MIXED (e.g., two even, one odd):
One exponent will be even, two will be odd.
$F_{hkl} = f(1 + 1 - 1 - 1) = 0$. The reflection is
ABSENT
.
Analysis of (100) vs. (110) Reflections
💡
Tutor's Note:
There seems to be a typo in the original question's premise. The (100) reflection is actually
absent
in
both
BCC and FCC.
A much better example is the
(110)
reflection, which is what was likely intended.
(110) in BCC:
$h+k+l = 1+1+0 = 2$
(Even).
$\implies F = 2f$. It is
PRESENT
.
(110) in FCC:
$h,k,l = (1,1,0)$
(Mixed).
$\implies F = 0$. It is
ABSENT
.
4. Structure Factor ($F_{hkl}$)
Atomic Scattering Factor ($f$):
This is the scattering amplitude of a
single, isolated atom
. It represents the efficiency of that atom at scattering X-rays. It depends on the atomic number ($Z$) (more electrons = more scattering) and decreases as the scattering angle ($\theta$) increases (because the atom's electron cloud is "fluffy" and not a point).
Structure Factor ($F_{hkl}$):
This is the
total scattering amplitude from all atoms within ONE unit cell
. It's a vector sum that combines the
amplitude
of each atom's scattering ($f_j$) with its
phase
, which is determined by its position ($u_j, v_j, w_j$) in the unit cell.
Derivation of General Expression
We define the phase of a wave scattered from an atom $j$ at fractional coordinates ($u_j, v_j, w_j$) relative to a wave scattered from the origin (0,0,0).
The phase difference is given by $2\pi (hu_j + kv_j + lw_j)$.
The wave scattered by this atom is its scattering factor $f_j$ multiplied by its phase factor: $f_j e^{2\pi i (hu_j + kv_j + lw_j)}$.
To get the
total
scattering from the unit cell, we sum this expression over all $N$ atoms in the cell:
$$ F_{hkl} = \sum_{j=1}^{N} f_j e^{2\pi i (hu_j + kv_j + lw_j)} $$
Application to BCC
We will prove the BCC extinction rule $h+k+l = \text{even}$.
Atoms:
2 identical atoms ($f$) at (0, 0, 0) and (½, ½, ½).
Structure Factor:
$F_{hkl} = f_1 e^{2\pi i (h\cdot 0 + k\cdot 0 + l\cdot 0)} + f_2 e^{2\pi i (h\cdot \frac{1}{2} + k\cdot \frac{1}{2} + l\cdot \frac{1}{2})}$
$F_{hkl} = f e^{0} + f e^{\pi i (h+k+l)}$
$F_{hkl} = f [1 + e^{\pi i (h+k+l)}]$
Analysis:
If $(h+k+l)$ is EVEN:
$e^{\pi i (\text{even})} = +1$.
$F_{hkl} = f(1 + 1) = 2f$.
Reflection observed.
If $(h+k+l)$ is ODD:
$e^{\pi i (\text{odd})} = -1$.
$F_{hkl} = f(1 - 1) = 0$.
Reflection not observed.
Thus, reflections are only observed for a BCC lattice when $h+k+l$ is an even number.
5. Powder Pattern Indexing
We are given $\lambda = 1.5418$ Å and six $2\theta$ values.
(a) Indexing & (b) Lattice Type
Calculate $\sin^2\theta$:
For a cubic system, $\sin^2\theta = (\frac{\lambda^2}{4a^2}) (h^2+k^2+l^2) = C \times N$, where $N$ is an integer. This means the
ratios
of $\sin^2\theta$ values must be
ratios of integers
.
Create a table:
$2\theta$ (°)
$\theta$ (°)
$\sin^2\theta$
Ratio (divide by smallest)
$\times 3$ (to get integers)
44.4
22.2
0.1427
1.000
3
51.7
25.85
0.1901
1.332
4
76.1
38.05
0.3798
2.661
8
92.7
46.35
0.5236
3.669
11
98.2
49.1
0.5712
4.003
12
121.5
60.75
0.7612
5.334
16
Analyze Ratios:
The ratios 1, 1.33, 2.66... are not simple integers. By multiplying by 3, we get the integer sequence
3, 4, 8, 11, 12, 16
.
Identify Lattice:
This sequence of allowed $N = h^2+k^2+l^2$ values is characteristic of a
Face-Centered Cubic (FCC)
lattice.
$N=3 \implies (111)$
$N=4 \implies (200)$
$N=8 \implies (220)$
$N=11 \implies (311)$
$N=12 \implies (222)$
$N=16 \implies (400)$
(c) Calculate Lattice Parameter 'a'
From the indexing formula: $a^2 = \frac{\lambda^2}{4} \frac{N}{\sin^2\theta}$. Let's calculate $C = \frac{\sin^2\theta}{N}$ for each peak and average it.
Peak 1: $C = 0.1427 / 3 = 0.04757$
Peak 2: $C = 0.1901 / 4 = 0.04753$
Peak 3: $C = 0.3798 / 8 = 0.04748$
... (the others are similar)
The average value for $C$ is approximately
$0.04753$
.
Here is a direct comparison of the three main X-ray diffraction techniques:
Feature
Laue Method
Rotating Crystal Method
Debye-Scherrer (Powder)
Incident X-ray Beam
Polychromatic
(variable $\lambda$)
Monochromatic
(fixed $\lambda$)
Monochromatic
(fixed $\lambda$)
Sample
Single Crystal
Single Crystal
Polycrystalline (Powder)
Sample Motion
Stationary
Rotated (about one axis)
Stationary (or slowly rotated)
Diffraction Pattern
A pattern of
spots
on a flat detector.
A pattern of
spots
arranged in
layer lines
.
A set of
concentric cones
, recorded as
lines
or
peaks
.
Primary Information
Crystal
orientation
and
symmetry
.
Lattice parameters
(unit cell size) from layer line spacing.
Phase identification
(fingerprinting) and precise
lattice parameters
.
7. Miller Indices and Interplanar Spacing
Part A: Determining Miller Indices ($hkl$)
Find Intercepts:
Determine where the crystal plane intercepts the three crystallographic axes (a, b, c). Express these as fractions of the lattice parameters (e.g., $1a$, $\frac{1}{2}b$, $\infty c$).
Take Reciprocals:
Invert these fractional intercepts. An intercept at infinity (for a parallel plane) becomes $1/\infty = 0$.
(e.g., $1/1$, $1/(1/2)$, $1/\infty$
$\implies$
$1, 2, 0$).
Clear Fractions:
Multiply by a common factor to get the smallest possible set of integers $h, k, l$.
Final Form:
Write the integers in parentheses: $(hkl)$. For our example, $(120)$.
Part B: Drawing Planes in a Cubic Cell
(110) Plane:
Intercepts at $1a, 1b, \infty c$. This is a diagonal plane that is parallel to the c-axis.
(111) Plane:
Intercepts at $1a, 1b, 1c$. This plane cuts the corner of the unit cell, passing through three adjacent vertices.
(200) Plane:
Intercepts at $\frac{1}{2}a, \infty b, \infty c$. This plane is parallel to the (100) face but is shifted halfway *inside* the unit cell.
Part C: $d$-spacing for Orthorhombic Crystal
We want to find the perpendicular distance $d$ from the origin (0,0,0) to the plane $(hkl)$.
Equation of the Plane:
The $(hkl)$ plane has intercepts $(a/h, b/k, c/l)$. The equation for this plane is:
$\frac{x}{a/h} + \frac{y}{b/k} + \frac{z}{c/l} = 1$
or
$(\frac{h}{a})x + (\frac{k}{b})y + (\frac{l}{c})z - 1 = 0$
Distance Formula:
The perpendicular distance $D$ from a point $(x_0, y_0, z_0)$ to a plane $Ax+By+Cz+E=0$ is:
$D = \frac{|Ax_0+By_0+Cz_0+E|}{\sqrt{A^2+B^2+C^2}}$
Apply to our system:
Our point is $(0,0,0)$ and our plane has $A=h/a$, $B=k/b$, $C=l/c$, $E=-1$.
$d_{hkl} = \frac{|(h/a)0 + (k/b)0 + (l/c)0 - 1|}{\sqrt{(h/a)^2 + (k/b)^2 + (l/c)^2}}$
Final Formula:
Squaring both sides gives the common form:
$$ \frac{1}{d_{hkl}^2} = \frac{h^2}{a^2} + \frac{k^2}{b^2} + \frac{l^2}{c^2} $$
8. The Laue Method
Part A: Experimental Arrangement
The Laue method is unique in that it uses
polychromatic
(white) X-rays and a
stationary single crystal
.
Source:
A broad-spectrum X-ray tube provides a beam with a wide range of wavelengths ($\lambda$).
Collimator:
A pinhole system creates a narrow beam.
Sample:
A single crystal, mounted in a fixed orientation.
Detector:
A large-area flat detector (e.g., a CCD or photographic plate).
Back-Reflection:
The detector is placed between the source and sample, with a hole for the beam to pass through. It records beams that are diffracted
backward
(angles $2\theta > 90^\circ$). This is used for thick samples.
Transmission:
The detector is placed
behind
the sample.
Part B: Purpose and Function
Primary Purpose:
The Laue method is
not
used to determine lattice parameters (since $\lambda$ is unknown). Its main uses are:
Determining Crystal Orientation:
It's a fast way to see how a single crystal is oriented (e.g., to align it for another experiment).
Assessing Crystal Symmetry:
The
symmetry
of the "Laue pattern" of spots on the detector directly reflects the
point group symmetry
of the crystal along the direction of the X-ray beam. A 4-fold symmetry in the pattern implies the beam is aligned with a 4-fold axis of the crystal.
How it Works:
In Bragg's Law ($2d_{hkl} \sin\theta = n\lambda$), the $d_{hkl}$ and $\theta$ values are fixed by the crystal's structure and orientation. The "white" beam provides a continuous spectrum of $\lambda$ values, so the crystal simply "selects" the wavelength $\lambda$ that satisfies the equation for each $(hkl)$ plane.
Laue Spots & Gnomonic Projection:
Each spot on the detector is a
Laue spot
, corresponding to a specific $(hkl)$ plane. A
gnomonic projection
is a mathematical method used to map the positions of these spots from the 2D detector to a 2D plot that helps in indexing the planes and determining the crystal's precise 3D orientation.
9. Factors Affecting X-ray Intensities
The measured intensity of a diffraction peak,
$I_{hkl}$
, is
not
just the structure factor. It is a product of several factors:
1. Structure Factor ($|F_{hkl}|^2$):
This is the primary factor. It represents the "ideal" intensity from a single unit cell and depends on the
type
(atomic scattering factor $f$) and
position
($u,v,w$) of all atoms in the cell. If $F_{hkl}=0$, the peak is systematically absent.
2. Multiplicity Factor ($p$):
This accounts for the number of
symmetrically equivalent
planes that all contribute to the same peak. In a cubic crystal, the (100) and (010) planes have the same $d$-spacing. The $\{100\}$ family has a multiplicity $p=6$ ( (100), (010), (001), and their negatives). The $\{111\}$ family has $p=8$. A higher multiplicity leads to a more intense peak.
3. Lorentz-Polarization (Lp) Factor:
This combines two geometric corrections:
Polarization:
The incident X-ray beam is unpolarized, but scattering causes it to become partially polarized. This effect is angle-dependent and reduces intensity.
Lorentz:
In a powder or rotating-crystal experiment, this factor accounts for the different amounts of
time
that planes at different $\theta$ angles spend in the correct orientation to diffract.
For a powder diffractometer, the combined $Lp$ factor is $\frac{1+\cos^2(2\theta)}{\sin^2\theta \cos\theta}$.
4. Temperature Factor (Debye-Waller Factor):
Atoms are not static; they vibrate due to thermal energy. This "smears" their electron density, making them less effective scatterers. This effect
reduces all peak intensities
, and the reduction is
stronger at high $2\theta$ angles
.
5. Absorption Factor:
The sample itself absorbs X-rays, both on the way in and on the way out. This
reduces all peak intensities
. The effect is highly dependent on the sample's chemical composition (elements with high $Z$ absorb more) and its physical shape and density.
10. JCPDS (ICDD) Card File
Part A: What is it?
The
JCPDS (Joint Committee on Powder Diffraction Standards)
file, now maintained by the
ICDD (International Centre for Diffraction Data)
, is a massive, standardized database of powder diffraction patterns.
Its primary purpose is
phase identification
. The powder diffraction pattern of a crystalline material is a unique
"fingerprint"
. By matching the experimental pattern of an unknown sample to a pattern in the database, the material can be identified.
Part B: Structure of a JCPDS (PDF) Entry
Each entry in the database (Powder Diffraction File, or PDF) contains:
PDF Number:
A unique reference number (e.g., 00-004-0784 for Aluminum).
Formula & Name:
e.g., Al, Aluminum.
The $d$-$I$ List:
The most critical data. This is a table of all the
$d$-spacings
(in Å) for the material and their
relative intensities
($I/I_1$), where the strongest peak is scaled to 100%.
Indexing:
The $(hkl)$ Miller indices corresponding to the $d$-spacings.
Crystallographic Data:
Crystal system (e.g., Cubic), lattice parameters (e.g., $a = 4.0494$ Å), and space group.
Physical & Optical Data:
Density, color, etc.
Part C: Hanawalt Method (Search-Match)
This is the classic procedure for using the database:
Acquire Data:
Run a powder XRD experiment on your unknown sample.
Process Data:
Create a list of all your peaks, convert their
$2\theta$
positions to
$d$-spacings
, and find their
relative intensities
($I/I_1$).
Search:
Identify the
three strongest peaks
in your pattern. This is your "Hanawalt triplet".
Match:
Search the JCPDS/ICDD database using the $d$-spacings of these three peaks. The database is indexed to allow for rapid searching based on the strongest lines.
Verify:
The search will return a list of "candidate" materials. You must then compare your
entire
experimental pattern (all $d$-spacings and intensities) to the reference patterns of the candidates. A positive match occurs when
all
your peaks (and their relative intensities) match a single reference pattern.
11. Structure Factor of NaCl
We can model the NaCl structure as an
FCC lattice
with a two-ion
basis
:
Basis Atom 1:
Na$^+$ at (0, 0, 0)
Basis Atom 2:
Cl$^-$ at (½, ½, ½)
The total structure factor
$F_{hkl}$
is the product of the
Structure Factor of the Basis
and the
Structure Factor of the Lattice
.
Structure Factor of Basis:
$F_{\text{basis}} = f_{Na^+} e^{2\pi i (h\cdot 0 + ...)} + f_{Cl^-} e^{2\pi i (h\cdot \frac{1}{2} + ...)}$
$F_{\text{basis}} = f_{Na^+} + f_{Cl^-} e^{\pi i (h+k+l)}$
Structure Factor of FCC Lattice:
(Sum over lattice points: (0,0,0), (½,½,0), (½,0,½), (0,½,½))
$F_{\text{lattice}} = [1 + e^{\pi i (h+k)} + e^{\pi i (h+l)} + e^{\pi i (k+l)}]$
Total Structure Factor ($F_{hkl}$):
$F_{hkl} = F_{\text{basis}} \times F_{\text{lattice}}$
$F_{hkl} = [f_{Na^+} + f_{Cl^-} e^{\pi i (h+k+l)}] \times [1 + e^{\pi i (h+k)} + e^{\pi i (h+l)} + e^{\pi i (k+l)}]$
Analysis of Conditions
We first analyze the
$F_{\text{lattice}}$
term (the FCC rule):
If $h,k,l$ are MIXED:
$F_{\text{lattice}} = 0$
. Therefore,
$F_{hkl} = 0$
. These reflections are
ABSENT
.
If $h,k,l$ are UNMIXED (all even or all odd):
$F_{\text{lattice}} = 4$
. The reflection is present, so we must now check the
$F_{\text{basis}}$
term.
$F_{hkl} = 4 \times [f_{Na^+} + f_{Cl^-} e^{\pi i (h+k+l)}]$
Now we analyze this result based on the two unmixed cases:
(a) $h,k,l$ are all EVEN (e.g., (200), (220)):
The sum $h+k+l$ is
even
.
$e^{\pi i (\text{even})} = +1$.
$F_{hkl} = 4 [f_{Na^+} + f_{Cl^-}(+1)] =
4(f_{Na^+} + f_{Cl^-})
$
The scattering factors add up. These are
STRONG
reflections.
(b) $h,k,l$ are all ODD (e.g., (111), (311)):
The sum $h+k+l$ is
odd
.
$e^{\pi i (\text{odd})} = -1$.
$F_{hkl} = 4 [f_{Na^+} + f_{Cl^-}(-1)] =
4(f_{Na^+} - f_{Cl^-})
$
The scattering factors subtract. These are
WEAK
reflections.
Bragg's Law assumes diffraction from a
perfectly infinite
crystal, which would produce infinitely sharp peaks. In reality, crystalline materials are composed of finite-sized domains called
crystallites
.
When the crystallite size ($L$) is very small (typically < 200 nm), there are not enough crystal planes to create perfect destructive interference at angles
just
off the main Bragg angle. This results in the diffraction peaks becoming
broader
. The
smaller
the crystallite, the
broader
the peak.
Part B: The Scherrer Equation
The Scherrer equation is an empirical formula that relates the measured peak broadening to the average crystallite size:
$$ L = \frac{K\lambda}{\beta \cos\theta} $$
$L$
: The volume-averaged
crystallite size
, perpendicular to the $(hkl)$ plane.
$K$
: The
Scherrer constant
, a dimensionless "shape factor" that depends on the crystallite shape (e.g., sphere, cube). It is typically approximated as
$K \approx 0.9$
.
$\lambda$
: The
X-ray wavelength
(e.g., 1.5418 Å for Cu K$\alpha$).
$\beta$
: The
peak broadening
, measured as the Full-Width at Half-Maximum (FWHM).
Crucially
, this is the
instrument-corrected
broadening (i.e., you must subtract the broadening caused by the instrument itself) and it
MUST be in RADIANS
.
$\theta$
: The
Bragg angle
(
not
$2\theta$).
Part C: Use and Limitation
Use:
It is a fundamental tool in
nanoscience
and materials science for quickly estimating the size of nanoparticles, nanograins, or thin-film domains.
Limitation:
The Scherrer equation is an
estimate
. Its main weakness is that it
conflates size broadening with strain broadening
. Microstrain (non-uniform distortion) in the lattice also causes peaks to broaden. To separate these two effects, more advanced methods like a Williamson-Hall plot are required.
13. Reciprocal Lattice and Ewald Sphere
Part A: Reciprocal Lattice
The
Reciprocal Lattice
is a mathematical concept that provides an elegant way to visualize diffraction. It is a 3D lattice in "reciprocal space" (or "k-space") where distances have units of 1/distance (e.g., Å
$^{-1}$
).
It has a direct relationship to the
real lattice
(the crystal's atomic arrangement):
Each
set of planes $(hkl)$
in the real lattice (with spacing $d_{hkl}$) is represented by a
single point $(hkl)$
in the reciprocal lattice.
The vector $\vec{G}_{hkl}$ from the origin (000) to the point $(hkl)$ has two properties:
Direction:
It is
perpendicular
to the $(hkl)$ planes in the real lattice.
Magnitude:
Its length is $1 / d_{hkl}$.
Part B: Ewald Sphere Construction
The Ewald Sphere is a brilliant geometric construction that shows exactly when Bragg's Law is satisfied.
Draw the Reciprocal Lattice
of the crystal.
Draw the Incident Beam ($k_{\text{in}}$):
This is a vector of length
$1/\lambda$
. Draw it so that its
tip
points to the
origin (000)
of the reciprocal lattice.
Draw the Ewald Sphere:
Draw a sphere of radius
$1/\lambda$
, centered at the
tail
of the $k_{\text{in}}$ vector. By this construction, the origin (000) is guaranteed to be on the sphere's surface.
The Diffraction Condition:
A diffraction event occurs
if and only if
another reciprocal lattice point
$(hkl)$
lies
exactly on the surface
of the Ewald Sphere.
Proof of Bragg's Law:
If a point $(hkl)$ is on the sphere, we can draw a triangle.
The vector from the sphere's center to the origin (000) is $k_{\text{in}}$ (length $1/\lambda$).
The vector from the sphere's center to the point $(hkl)$ is the diffracted beam $k_{\text{out}}$ (also length $1/\lambda$).
The vector from the origin (000) to $(hkl)$ is the reciprocal lattice vector $\vec{G}_{hkl}$ (length $1/d_{hkl}$).
This is an isosceles triangle. The angle between $k_{\text{in}}$ and $k_{\text{out}}$ is $2\theta$. If we bisect this triangle, we get two right-angled triangles where $\sin\theta = (\frac{1}{2}|\vec{G}_{hkl}|) / |k_{\text{in}}| = (\frac{1}{2}d_{hkl}) / (1/\lambda)$.
This simplifies to $\sin\theta = \frac{\lambda}{2d_{hkl}}$, or
$2d_{hkl}\sin\theta = \lambda$
.
14. Density Calculation from XRD Data
Given: Mo (Molybdenum), BCC,
$(hkl) = (211)$
,
$2\theta = 73.6^\circ$
,
$\lambda = 1.54$
Å, M = 95.96 g/mol.
Structure Factor ($F_{hkl}$):
This is a
theoretical amplitude
. It's a complex number that represents the ideal scattering from
one single unit cell
. Its magnitude squared,
$|F_{hkl}|^2$
, is the "ideal intensity" from that cell, determined only by the atom types and positions.
X-ray Intensity ($I_{hkl}$):
This is the
actual, experimentally measured intensity
of a peak. It is
proportional
to
$|F_{hkl}|^2$
, but is
also
multiplied by all the other factors we discussed in Question 9: the Multiplicity ($p$), the Lorentz-Polarization ($Lp$) factor, the Temperature factor, and the Absorption factor.
Part B: Calculation of $I_{111} / I_{200}$ for Copper (FCC)
Given: Cu, FCC, $a=3.61$ Å, $\lambda = 1.54$ Å.
The ratio of intensities is:
$\frac{I_{111}}{I_{200}} = \frac{p_{111} \times (Lp)_{111} \times |F_{111}|^2}{p_{200} \times (Lp)_{200} \times |F_{200}|^2}$
Structure Factor ($|F|^2$):
For FCC, $F = 4f$ for unmixed $hkl$.
$|F_{111}|^2 = (4f_{111})^2 = 16 f_{111}^2$
$|F_{200}|^2 = (4f_{200})^2 = 16 f_{200}^2$
(Note: $f$ depends on $\theta$, so $f_{111} \neq f_{200}$)
Multiplicity ($p$):
$p_{111}$: The $\{111\}$ family has 8 planes. $p_{111} = 8$.
$p_{200}$: The $\{200\}$ family has 6 planes. $p_{200} = 6$.
Atomic Scattering ($f$):
We must use tabulated values for $f_{Cu}$ at the respective angles.
$\sin\theta/\lambda$ for (111) is $0.3695/1.54 \approx 0.24$. $f_{111} \approx 18.5$
$\sin\theta/\lambda$ for (200) is $0.4266/1.54 \approx 0.28$. $f_{200} \approx 17.4$
$|F_{111}|^2 = 16 \times (18.5)^2 = 5476$
$|F_{200}|^2 = 16 \times (17.4)^2 = 4843$