fix math eq bugs

src/14_doping_and_devices.md

@@ -65,9 +65,10 @@ Therefore, the electron's mass is the conduction band's effective mass.
 Furthermore, the interactions between the electron and proton are screened by the lattice.
 As a result, we need to introduce the following substitutions: $m_e \to m_e^*$, $\epsilon_0 \to \epsilon\epsilon_0$.
 We thus estimate the energy of the bound state created by the impurity:
-$$E = -\frac{m_e^*}{m_e \varepsilon^2} R_E = -0.01 \text{eV (in Ge)}$$
-$r_B = 4$ nm (vs $r_B = 0.5$ Å in H)$.
-The electron is very weakly bound to the impurity! At room temperature (0.026 eV), the donor electron is easily thermally excited into the conduction band. 
+$$E = -\frac{m_e^*}{m_e \varepsilon^2} R_E = -0.01 \text{eV (in Ge)},$$
+with Bohr radius $r_B = 4$ nm (vs $r_B = 0.5$ Å in Hydrogen).
+The electron is very weakly bound to the impurity!
+At room temperature (0.026 eV), the donor electron is easily thermally excited into the conduction band. 
 
 On the other hand, we can add a group III element to reduce the average number of electrons in the system.
 Group III elements lacks 1 electron and 1 proton and are therefore known as **acceptors**.
@@ -103,7 +104,7 @@ Therefore, we model the density of states of donors/acceptors as a Dirac delta f
 $$
 g_D(E) = N_D \delta(E- E_D), \quad  g_A(E) = N_A \delta(E-E_A), 
 $$
-where N_D and N_A are donor and acceptor concentrations respectively.
+where $N_D$ and $N_A$ are donor and acceptor concentrations respectively.
 The binding energies of the donor and acceptor are defined as $E_A$ and $E_D$.
 
 How good is this Dirac delta approximation?