Update 13_semiconductors_solutions.md

src/13_semiconductors_solutions.md

@@ -80,13 +80,17 @@ dispersion(10, 2, 8)
 ```
 
 ### Subquestion 1
-Apply the following to max an min of valence and conduction band, respectively:
+Apply the following to valence and conduction band, respectively:
 $$v=\hbar^{-1}\partial E(k)/\partial k$$
 $$m_{eff} = \hbar^2\left(d^2 E(k)/dk^2\right)^{-1}$$
 
-
 Be careful with hole calculations.
-These results assume $t_{cb}$ and $t_{vb}$ positive.
+
+$$v_e = \frac{2at_{cb}}{\hbar}sin(ka) $$
+$$m_e =  \frac{\hbar^2}{2a^2t_{cb}cos(ka)}$$
+
+$$v_h = -\frac{2at_{vb}}{\hbar}sin(ka) $$
+$$m_h =  \frac{\hbar^2}{2a^2t_{vb}cos(ka)}$$
 
 ### Subquestion 2
 This approximation indicates the chemical potential is "well bellow" the conduction band and "well above"
@@ -96,12 +100,14 @@ approximate Fermi statistics by Boltzmann's.
 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.
+
 ### Subquestion 3
 
 
 ### Subquestion 4
-$$n = \int_{E_{cb}}^{E_{cb}+4t_{cb}} f(\varepsilon)g_c(\varepsilon)d\varepsilon$$
-$$p = \int_{-4t_{vb}}^{0} (1-f(\varepsilon))g_h(\varepsilon)d\varepsilon$$
+$$n_e = \int_{E_{cb}}^{E_{cb}+4t_{cb}} f(\varepsilon)g_c(\varepsilon)d\varepsilon$$
+$$n_h = \int_{-4t_{vb}}^{0} (1-f(\varepsilon))g_h(\varepsilon)d\varepsilon$$
 
 ### Subquestion 5
 For an intrinsic semiconductor the number of electrons excited into the conduction band must be equal