updates numbered lists

src/4_sommerfeld_model_solutions.md

@@ -64,17 +64,14 @@ $$
 $$
 2g(k)dk = g(E)dE
 $$
-
 $$
 g(E)=2g(k(E))\frac{dk}{dE}
 $$
-
 $$
 g(E) = \frac{2}{\Gamma(\frac{n}{2})}\left(\frac{L}{\hbar}\sqrt{\frac{m_e}{2\pi}}\right)^n (E)^{\frac{n}{2}-1}
 $$
 
 6. 
-
 $$
 N = \int_{0}^{\infty}g(E)f(E)dE = \frac{2}{\Gamma(\frac{n}{2})}\left(\frac{L}{\hbar}\sqrt{\frac{m_e}{2\pi}}\right)^n\int_{0}^{\infty}\frac{(E)^{\frac{n}{2}-1}}{e^{\frac{E-\mu}{k_BT}}+1}dE
 $$
@@ -140,7 +137,7 @@ Check the source code written in python for solving integral using midpoint rule
 
 ### Exercise 4: graphene
 
-1. 
+1.
 
 ```python
 import numpy as np
@@ -166,7 +163,7 @@ ax.yaxis.set_label_coords(0.5,1)
 ax.xaxis.set_label_coords(1.0, 0.49)
 ```
 
-2. The DOS for the positive energies is given by
+2.The DOS for the positive energies is given by
 $$
 g(\epsilon) = (2_s + 2_v) 2 \pi \left(\frac{L}{2 \pi}\right)^2 \frac{\epsilon}{c^2},
 $$
@@ -176,7 +173,7 @@ $$
 g(\epsilon) = (2_s + 2_v) 2 \pi \left(\frac{L}{2 \pi}\right)^2 \frac{|\epsilon|}{c^2}.
 $$
 
-3. $g(\epsilon)$ vs $\epsilon$ is a linear plot. Here, the region marked by $-k_B T$ is a triangle whose area gives the number of electrons that can be excited:
+3.$g(\epsilon)$ vs $\epsilon$ is a linear plot. Here, the region marked by $-k_B T$ is a triangle whose area gives the number of electrons that can be excited:
 $$
 \begin{align}
 n_{ex} &= \frac{1}{2} g(-k_B T) k_B T\\
@@ -188,7 +185,7 @@ $$
 E(T) - E_0 = \frac{L^2 k_B^3T^3}{\pi c^2}.
 $$
 
-4. 
+4.
 $$
 C_v(T) = \frac{\partial E}{\partial T} = \frac{3L^2k_B^3T^2}{\pi c^2}
 $$