Tried to fix the numbered list. Manually inserted the numbered list, so it should work now

src/6_bonds_and_spectra_solutions.md

@@ -2,13 +2,8 @@
 
 ### Exercise 1: linear triatomic molecule 
 
-1. 
-
-In 1D, there are two normal modes and in 3D there are 4 normal modes
-
-2. 
-
-$$
+1.  In 1D, there are two normal modes and in 3D there are 4 normal modes
+2.  $$
 \begin{cases}
     m \ddot{x}_1 & = -\kappa(x_1-x_2)\\
     M \ddot{x}_2 & = -\kappa(2x_2-x_1-x_3)\\ 
@@ -16,38 +11,29 @@ $$
 \end{cases}
 $$
 Where $m$ is the mass of the oxygen atoms and $M$ the mass of the carbon atom.
-3.
-$$
+3.  $$
 \omega = \frac{\kappa}{m}
 $$
 4.
 $$
 \frac{x_1}{x_2} = \frac{M}{2m}
 $$
-5.
-$$
+5.  $$
 \omega = \sqrt{\frac{\kappa(2m+M)}{mM}}
 $$
 
-### Exercise 2: Lennard-Jones potential
-
-1. 
-
-See lecture slides/internet
 
-2. 
-
-The equilibrium position is $r_0 = 2^{1/6}\sigma$. The energy at the inter atomic distance $r_0$ is given by:
+### Exercise 2: Lennard-Jones potential
+1.  See lecture+internet
+2.  The equilibrium position is $r_0 = 2^{1/6}\sigma$. The energy at the inter atomic distance $r_0$ is given by:
 $$
 U(r_0) = -\epsilon
 $$
-3.
-$$
+3.  $$
 U(r) = -\epsilon + \frac{\kappa}{2}(r-r_0)^2
 $$
 Where $\kappa = \frac{72\epsilon}{2^{1/3}\sigma^2}$
-4.
-The ground state energy is given by
+4.  The ground state energy is given by
 $$
 E_0 = -\epsilon+\frac{1}{2}\sqrt{\frac{2\kappa}{m}}
 $$
@@ -55,8 +41,7 @@ And the breaking energy is given by
 $$
 E_{break} = \epsilon - \frac{1}{2}\sqrt{\frac{2\kappa}{m}}
 $$
-5.
-Distance from which $U(r)$ becomes anharmonic:
+5.  Distance from which $U(r)$ becomes anharmonic:
 $$
 r_{anharmonic} = \frac{6}{7}2^{1/6}\sigma
 $$
@@ -65,6 +50,7 @@ $$
 n = \frac{36}{49}\frac{\epsilon}{\hbar\omega}-\frac{1}{2}
 $$
 
+
 ### Exercise 3: Numerical simulation
 1.