From 7ad3ccaf13bf06493b37efbddb3f39e993294641 Mon Sep 17 00:00:00 2001
From: Bowy La Riviere <b.m.lariviere@student.tudelft.nl>
Date: Sun, 14 Mar 2021 14:34:49 +0000
Subject: [PATCH] Apply 11 suggestion(s) to 1 file(s)
---
src/10_xray.md | 23 +++++++++++------------
1 file changed, 11 insertions(+), 12 deletions(-)
diff --git a/src/10_xray.md b/src/10_xray.md
index 847d74fd..4b44ae17 100644
--- a/src/10_xray.md
+++ b/src/10_xray.md
@@ -561,7 +561,7 @@ This improves the chances of fulfilling the Laue condition for a fixed direction
The experiment is illustrated in the figure above.
The result is that the diffracted beam exits the sample via concentric circles at discrete **deflection angles** $2 \theta$.
-To deduce the values of $\theta$ for a specific crystal let us put the Laue condition into a more practical form.
+In order to deduce the values of $\theta$ of a specific crystal, let us put the Laue condition into a more practical form.
We first take the modulus squared of both sides:
$$
@@ -570,7 +570,7 @@ G^2 = 2k^2-2\mathbf{k'} \cdot \mathbf{k},
$$
where we used $|\mathbf{k'}| = |\mathbf{k}|$.
-We then substitute the Laue condition back for $\mathbf{k'}$:
+We then substitute the Laue condition $\mathbf{k'} = \mathbf{k}+\mathbf{G}$:
$$
\begin{align}
@@ -589,30 +589,29 @@ Note, $\phi$ is the angle between the vector $\mathbf{k}$ and $\mathbf{G}$, whic
We are nearly there, but we are left with finding out the relation between $\phi$ and $\theta$.
Last lecture we discussed the concept of Miller planes.
-These were planes designated by Miller indices $(h,k,l)$ which intersect the lattice vectors at $\mathbf{a}_1 / h$, $\mathbf{a}_22 / k$ and $\mathbf{a}_3 / l$.
-It turns out that these miller planes are normal the reciprocal lattice vector $\mathbf{G} = h \mathbf{b}_1 + k \mathbf{b}_2 + l \mathbf{b}_3$ and that the distance between subsequent Miller planes is given by $d_{hkl} = \frac{2 \pi}{\lvert \mathbf{G} \rvert}$ (you will derive this in [exercise ](https://solidstate.quantumtinkerer.tudelft.nl/10_xray/#exercise-2-miller-planes-and-reciprocal-lattice-vectors)).
+These are planes designated by Miller indices $(h,k,l)$ which intersect the lattice vectors at $\mathbf{a}_1 / h$, $\mathbf{a}_22 / k$ and $\mathbf{a}_3 / l$.
+It turns out that these miller planes are normal the reciprocal lattice vector $\mathbf{G} = h \mathbf{b}_1 + k \mathbf{b}_2 + l \mathbf{b}_3$ and that the distance between subsequent Miller planes is given by $d_{hkl} = \frac{2 \pi}{\lvert \mathbf{G} \rvert}$ (you will derive this in [today's exercise](https://solidstate.quantumtinkerer.tudelft.nl/10_xray/#exercise-2-miller-planes-and-reciprocal-lattice-vectors)).
We substitute the expression of $\lvert \mathbf{G} \rvert$ into the equation of the distance:
$$
d_{hkl} \cos (\phi) = \frac{\pi}{\lvert \mathbf{k} \rvert}.
$$
-We know that $\lvert \mathbf{k} \rvert$ is related to the wavelength by $\lvert \mathbf{k} \rvert = \pi/\lambda$.
+We know that $\lvert \mathbf{k} \rvert$ is related to the wavelength by $\lvert \mathbf{k} \rvert = 2\pi/\lambda$.
Therefore, we can write the equation above as
$$
2 d_{hkl} \cos (\phi) = \lambda.
$$
-We substitute $\phi = \theta - \pi/2$.
+Lastly, we express the equation in terms of the deflection angle through the relation $\phi = \theta - \pi/2$.
With this, one can finally derive **Bragg's Law**:
$$
\lambda = 2 d_{hkl} \sin(\theta)
$$
-This equation allows us to obtain the distance $d_{hkl}$ from an experiment.
-Thus we can sue Bragg's Law in a diffraction powder diffraction experiment to obtain the distances between atoms in a crystal structre!
+Bragg's law allows us to obtain atomic distances in the crystal $d_{hkl}$ through powder diffraction experiments!
## Summary
@@ -631,7 +630,7 @@ Thus the band structure can be fully described by considering the 1st Brillouin
1. Why is the amplitude of a scattered wave zero if $\mathbf{k'}-\mathbf{k} \neq \mathbf{G}$?
2. Calculate the structure factor of the triangular lattice using the reciprocal lattice vectors found in the lecture.
Do any intensity peaks dissapear?
-3. Calculate $\mathbf{a}_1 \cdot \mathbf{b}_1$ and $\mathbf{a}_2 \cdot \mathbf{b}_1$ using the definitions of the reciprocal lattice vectors given in the lecture. Is this what was expected?
+3. Calculate $\mathbf{a}_1 \cdot \mathbf{b}_1$ and $\mathbf{a}_2 \cdot \mathbf{b}_1$ using the definitions of the reciprocal lattice vectors given in the lecture. Is the result what you expected?
### Exercise 1: Equivalence of direct and reciprocal lattice
@@ -644,9 +643,9 @@ The volume of a primitive cell of a lattice with lattice vectors $\mathbf{a}_1,
Make use of the vector identity
$$\mathbf{A}\times(\mathbf{B}\times\mathbf{C}) = \mathbf{B}(\mathbf{A}\cdot\mathbf{C}) - \mathbf{C}(\mathbf{A}\cdot\mathbf{B})$$
-3. Write down the primitive lattice vectors of the [BCC lattice](https://solidstate.quantumtinkerer.tudelft.nl/test_builds/lecture_9/9_crystal_structure/#body-centered-cubic-lattice) and calculate its reciprocal lattice vector.
-Which type of lattice is the reciprocal lattice of the real-space BCC lattice?
-4. Determine the shape of the 1st Brillouin zone of its reciprocal lattice.
+3. Write down the primitive lattice vectors of the [BCC lattice](https://solidstate.quantumtinkerer.tudelft.nl/test_builds/lecture_9/9_crystal_structure/#body-centered-cubic-lattice) and calculate its reciprocal lattice vectors.
+Which type of lattice is the reciprocal lattice of a BCC crystal?
+4. Determine the shape of the 1st Brillouin zone.
### Exercise 2: Miller planes and reciprocal lattice vectors
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