diff --git a/src/13_semiconductors.md b/src/13_semiconductors.md
index 8adbca769378b0b20b986a211b6169d70ad5cec1..e4898b0971ea6e17c88837309be9d4f097e87c5b 100644
--- a/src/13_semiconductors.md
+++ b/src/13_semiconductors.md
@@ -190,17 +190,17 @@ $$f(E\pm E_F) \approx e^{-(E\pm E_F)/kT}$$
 
 Now we can calculate $n_e$ and $n_h$:
 
-$$n_h \approx \frac{V(2m_h)^{3/2}}{2\pi^2\hbar^3}e^{-E_F/kT} \int_0^\infty\sqrt{E}e^{-E/kT}dE =
+$$n_h \approx \frac{(2m_h)^{3/2}}{2\pi^2\hbar^3}e^{-E_F/kT} \int_0^\infty\sqrt{E}e^{-E/kT}dE =
 N_V e^{-E_F/kT},$$
 
 with
 
 $$N_V = 2\left(\frac{2\pi m_h kT}{h^2}\right)^{3/2}$$
 
-the number of holes with energy $E<T$ (compare with the rule above).
+the density of holes with energy $E<kT$ (compare with the rule above).
 
 ??? 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 $N_V \sim 1$. On the other hand, $N_V \sim T^{3/2}$ Therefore $N_V \sim (kT/1 \textrm{eV})^{3/2}\sim 1\%$.
+    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 $N_V \sim (kT/1 \textrm{eV})^{3/2}\sim 1\%$.
 
 Similarly for electrons: