From 598e14b27841fc262e34f0b241f83b686b866b22 Mon Sep 17 00:00:00 2001
From: Kostas Vilkelis <kostasvilkelis@gmail.com>
Date: Wed, 18 Mar 2020 18:09:17 +0000
Subject: [PATCH] show solution 10
---
mkdocs.yml | 2 +-
src/10_xray_solutions.md | 18 ++++++------------
2 files changed, 7 insertions(+), 13 deletions(-)
diff --git a/mkdocs.yml b/mkdocs.yml
index e9132bb9..f6c5f293 100644
--- a/mkdocs.yml
+++ b/mkdocs.yml
@@ -44,7 +44,7 @@ nav:
- Tight-binding model: '7_tight_binding_model_solutions.md'
- Many atoms per unit cell: '8_many_atoms_solutions.md'
- Crystal structure: '9_crystal_structure_solutions.md'
- # - X-ray diffraction: '10_xray_solutions.md'
+ - X-ray diffraction: '10_xray_solutions.md'
# - Nearly free electron model: '11_nearly_free_electron_model_solutions.md'
# - Band structures in 2D: '12_band_structures_in_higher_dimensions_solutions.md'
# - Basic principles: '13_semiconductors_solutions.md'
diff --git a/src/10_xray_solutions.md b/src/10_xray_solutions.md
index 031ee8f2..3f6a4d65 100644
--- a/src/10_xray_solutions.md
+++ b/src/10_xray_solutions.md
@@ -77,7 +77,7 @@ Since $\rho=d / V$, we must maximize $d$. To do that, we must minimize $|G|$ (Su
## Exercise 3: X-ray scattering in 2D
### Subquestion 1
-```
+```python
def reciprocal_lattice(N = 7, lim = 40):
y = np.repeat(np.linspace(-18.4*(N//2),18.4*(N//2),N),N)
x = np.tile(np.linspace(-13.4*(N//2),13.4*(N//2),N),N)
@@ -103,7 +103,7 @@ $k = \frac{2 \pi}{\lambda} = 37.9 nm^{-1}$
Note that $|k| = |k'| = k $ since elastic scatering
-```
+```python
reciprocal_lattice()
# G vector
plt.arrow(0,0,13.4*2,18.4,color='r',zorder=10,head_width=2,length_includes_head=True)
@@ -144,15 +144,9 @@ $$
### Subquestion 4
-For FCC, the structure factor is the following:
+Due to bcc systematic absences, the peaks from lowest to largest angle are:
+$(110),(200),(211), (220), (310)$
+### Subquestion 5
-$S_\mathbf{G} = $
-$$
-\begin{cases}
- 4f & \text{if $h,k,l$ are all odd or even}\\
- 0 & \text{if otherwise}
-\end{cases}
-$$
-
-Since $(110)$ have mixed odd and even indices, no diffraction peak will be observed on FCC. For BCC, however, $(110)$ gives $1+1+0 = 2$ even number, so there will be diffraction.
+$a = 2.9100 \unicode{xC5}$
--
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