Update 13_semiconductors_solutions.md

src/13_semiconductors_solutions.md

@@ -106,18 +106,27 @@ This approximation indicates the chemical potential is "well bellow" the conduct
 the valence band. This way, we only have a few electrons and a few holes per band, which allows us to
 approximate Fermi statistics by Boltzmann's.
 
+Assume $t_{cb}$ and $t_{vb}$ positive.
+
 For electrons, do a taylor expansion around $k=0$ (min) of $E_{cb} = E_G - 2 t_{cb} [\cos(ka)-1],$.
 For holes, do a taylor expansion around $k=0$ (max) of $E_{vb} = 2 t_{vb} [\cos(ka)-1]$.
-
-Assume $t_{cb}$ and $t_{vb}$ positive.
+Near $k=0$
+$$ E_{cb} = E_G - t_{cb}(ka)^2$$
+$$ E_{vb} = t_{vb}(ka)^2 $$
 
 ### Subquestion 3
 
+Now, $g_e$ and $g_h$ can be calculated.
+??? hint "How?"
+    The approximated bands are quadratic! We've seen these before during the lecture.
+
 
 ### Subquestion 4
+
 $$n_e = \int_{E_{cb}}^{E_{cb}+4t_{cb}} f(\varepsilon)g_c(\varepsilon)d\varepsilon$$
 $$n_h = \int_{-4t_{vb}}^{0} (1-f(\varepsilon))g_h(\varepsilon)d\varepsilon$$
 
+
 ### Subquestion 5
 For an intrinsic semiconductor the number of electrons excited into the conduction band must be equal
 to the number of holes left behind in the valence band, so $p = n$. Impose this condition with the results of