typo fixes

src/5_atoms_and_lcao_solutions.md

@@ -7,22 +7,24 @@
 2. The atomic number of Tungsten is 74:
 
 $$
-1s^22s^22p^63s^23p^64s^23d^{10}4p^65s^24d^{10}5p^66s^24f^{14}5d^4	
+1s^22s^22p^63s^23p^64s^23d^{10}4p^65s^24d^{10}5p^66s^24f^{14}5d^4
 $$
 
 3.
 $$
-        Cu &= [Ar]4s^23d^9
-		Pd &= [Kr]5s^24d^8
-		Ag &= [Kr]5s^24d^9
-		Au &= [Xe]6s^24f^145d^9
+\begin{align}
+\textrm{Cu} &= [\textrm{Ar}]4s^23d^9
+\textrm{Pd} &= [\textrm{Kr}]5s^24d^8
+\textrm{Ag} &= [\textrm{Kr}]5s^24d^9
+\textrm{Au} &= [\textrm{Xe}]6s^24f^145d^9
+\end{align}
 $$
 
 ### Question 2
-1. 
+1.
 
 $$
-    \psi(x) = 
+    \psi(x) =
     \begin{cases}
         &\sqrt{\kappa}e^{\kappa(x-x_1)}, x<x_1\\
         &\sqrt{\kappa}e^{-\kappa(x-x_1)}, x>x_1
@@ -31,9 +33,9 @@ $$
 
 Where $\kappa = \sqrt{\frac{-2mE}{\hbar^2}} = \frac{mV_0}{\hbar^2}$.
 
-The energy is given by $\epsilon1 = \epsilon2 = -\frac{mV_0}{\hbar^2}$
+The energy is given by $\epsilon_1 = \epsilon_2 = -\frac{mV_0}{\hbar^2}$
 
-The wavefunction of a single delta peak is given by
+The wave function of a single delta peak is given by
 
 $$
     \psi_1(x) = \frac{\sqrt{mV_0}}{\hbar}e^{-\frac{mV_0}{\hbar^2}|x-x_1|}
@@ -41,18 +43,18 @@ $$
 
 $\psi_2(x)$ can be found by replacing $x_1$ by $x_2$
 
-2. 
+2.
 
 $$
     H = -\frac{mV_0^2}{\hbar^2}\begin{pmatrix}
-        1/2+\exp(-\frac{mV_0}{\hbar^2}(x_2-x_1)) & 
+        1/2+\exp(-\frac{mV_0}{\hbar^2}(x_2-x_1)) &
         \exp(\frac{mV_0}{\hbar^2}(x_2-x_1))\\
-        \exp(-\frac{mV_0}{\hbar^2}(x_2-x_1)) & 
+        \exp(-\frac{mV_0}{\hbar^2}(x_2-x_1)) &
         1/2+\exp(+\frac{mV_0}{\hbar^2}(x_2-x_1))
     \end{pmatrix}
 $$
 
-3. 
+3.
 
 $$
     \epsilon_{\pm} = \beta(3/2+\cosh{2\alpha}+2\cosh{\alpha}\pm \cosh{\alpha})
@@ -104,4 +106,3 @@ $$
 $$
     P = -\frac{2\gamma^2}{\mathcal{E}}(\frac{1}{\sqrt{\gamma^2+t^2}})
 $$
-