updates some solutions

src/9_crystal_structure_solutions.md

@@ -20,7 +20,7 @@ $$
 \mathbf{a}_3 &= \frac{a}{2}(\mathbf{\hat{y}} + \mathbf{\hat{z}}).
 \end{align}
 $$
-With respect to the conventional unit cell, the basis in fractional coordinates is $\bigcirc(0,0,0)$, $\bigcirc(1,0,0)$, $\bigcirc(0,1,0)$ and $\bigcirc(0,0,1)$.
+With respect to the conventional unit cell, the basis in fractional coordinates is $\bigcirc(1/2,1/2,0)$, $\bigcirc(1/2,0,1/2)$, $\bigcirc(0,1/2,1/2)$ and $\bigcirc(0,0,0)$.
 With respect to the primitive unit cell, the basis is $\bigcirc(0,0,0)$.
 Let us now consider the BCC lattice.
 The primitive lattice vectors are
@@ -31,7 +31,7 @@ $$
 \mathbf{a}_3 &= a\mathbf{\hat{y}}.
 \end{align}
 $$
-The basis of the conventional unit cell is $\bigcirc(0,0,0)$ and $\bigcirc(1,0,0)$.
+The basis of the conventional unit cell is $\bigcirc(0,0,0)$ and $\bigcirc(1/2,1/2,1/2)$.
 For the primitive unit cell the basis is $\bigcirc(0,0,0)$.
 
 3.
@@ -77,17 +77,17 @@ $$
 \mathbf{a_1} = a \hat{\mathbf{x}}, \quad \mathbf{a_2} = a \hat{\mathbf{y}}.
 $$
 With respect to the primitive lattice vectors, the basis is
-$$
-\Huge \bullet \normalsize(0,0), \quad \bigcirc(\frac{1}{2},\frac{1}{2}).
-$$
+$
+\huge \bullet \normalsize(0,0), \quad \bigcirc(\frac{1}{2},\frac{1}{2}).
+$
 
 4.
 The lattice is a cubic lattice. 
 The basis of the crystal is
 
-$$
-\Huge \bullet \normalsize = (0,0,0), \quad \bigcirc = (\frac{1}{2},\frac{1}{2},\frac{1}{2}).
-$$
+$
+\huge \bullet \normalsize(0,0,0), \quad \bigcirc(\frac{1}{2},\frac{1}{2},\frac{1}{2}).
+$
 
 An example of such a material is Cesium Chloride (CsCl)
 
@@ -106,9 +106,11 @@ $$
 The conevtional unit cell of diamond consists out of two shifted fcc lattices.
 One set of primitive lattice vectors is
 $$
-\mathbf{a_1} = \frac{a}{2} \left(\hat{\mathbf{x}}+\hat{\mathbf{y}} \right) \\
-\mathbf{a_2} = \frac{a}{2} \left(\hat{\mathbf{x}}+\hat{\mathbf{z}} \right) \\
-\mathbf{a_3} = \frac{a}{2} \left(\hat{\mathbf{y}}+\hat{\mathbf{z}} \right).
+\begin{align}
+\mathbf{a_1} &= \frac{a}{2} \left(\hat{\mathbf{x}}+\hat{\mathbf{y}} \right) \\
+\mathbf{a_2} &= \frac{a}{2} \left(\hat{\mathbf{x}}+\hat{\mathbf{z}} \right) \\
+\mathbf{a_3} &= \frac{a}{2} \left(\hat{\mathbf{y}}+\hat{\mathbf{z}} \right).
+\end{align}
 $$
 The volume of the primitive unit cell is
 $$
@@ -120,7 +122,7 @@ The primitive unit cell contains 2 atoms.
 With respect to the set of primitive lattcie vectors, the basis is $ \mathrm{C}(0,0,0)$ and $\mathrm{C}(\frac{1}{4},\frac{1}{4},\frac{1}{4})$.
 
 3.
-One FCC lattice has 4 atoms. 
+One FCC lattice contains 4 atom. 
 Because the diamond conventional unit cell contains two shifted FCC lattices, it will contain 8 atoms.
 The volume of the conventional unit cell is $V = a^3$.