Update 13_semiconductors.md

src/13_semiconductors.md

@@ -207,11 +207,11 @@ Now we can calculate $n_e$ and $n_h$:
 $$n_h \approx \frac{(2m_h)^{3/2}}{2\pi^2\hbar^3}e^{-E_F/kT} \int_{-E_v}^\infty\sqrt{E+E_v}e^{-E/kT}dE =
 N_V e^{E_v-E_F/kT},$$
 
-with
+where we used $\int_0^\infty \sqrt{x}e^{-x}dx=\sqrt{\pi}/2$ and we defined 
 
 $$N_V = 2\left(\frac{2\pi m_h kT}{h^2}\right)^{3/2}$$
 
-where we used $\int_0^\infty \sqrt{x}e^{-x}dx=\sqrt{\pi}/2$. Note that $N_V$ represents the density of holes with energy $E<kT$ (compare with the rule above).
+We see that holes are exponentially activated into the valence band. 
 
 ??? question "how large is $N_V$ at room temperature? (hard question)"
     If $kT \sim 1\textrm{eV}$ (the typical energy size of a band), then electrons in the whole band may be excited and $N_V \sim 1$ per unit cell. On the other hand, $N_V \sim T^{3/2}$ Therefore $N_V \sim (kT/1 \textrm{eV})^{3/2}\sim 1\%$.