From efcf50f1c8e187a0f04f6580783a6a9ee07bc6ef Mon Sep 17 00:00:00 2001
From: "T. van der Sar" <t.vandersar@tudelft.nl>
Date: Wed, 3 Apr 2019 21:02:40 +0000
Subject: [PATCH] Update 13_semiconductors.md - typos

---
 src/13_semiconductors.md | 6 +++---
 1 file changed, 3 insertions(+), 3 deletions(-)

diff --git a/src/13_semiconductors.md b/src/13_semiconductors.md
index 8adbca76..e4898b09 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:
 
-- 
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