update reasoning in Q1

src/14_doping_and_devices_solutions.md

@@ -15,6 +15,8 @@ $$ n_e n_h = \frac{1}{2} \left(\frac{k_BT}{\pi\hbar^2}\right)^3
 Charge balance condition:
 $$ n_e - n_h + n_D - n_A = N_D - N_A $$
 ### Subquestion 2
+
+Since $E_G \gg k_B T$, we know that $E_C-E_D < E_G$ and $E_A - E_V < E_G$ and so we are in the regime of full dopant ionization. Therefore, $n_D = n_A = 0$.
  
 $$ n_{e} = \frac{1}{2}(\sqrt{D^2+4n_i^2}+D)$$
 $$ n_{h} = \frac{1}{2}(\sqrt{D^2+4n_i^2}-D)$$