Update 13_semiconductors.md - typo

src/13_semiconductors.md

@@ -175,9 +175,9 @@ $$ g(E_h) = (2m_h)^{3/2}\sqrt{E_h}/2\pi^2\hbar^3$$
 
 Applying the algorithm:
 
-$$n_h = \int_0^\infty f_h(E+E_F)g_h(E)dE = \int_0^\infty\frac{(2m_h)^{3/2}}{2\pi^2\hbar^3} \sqrt{E}\frac{1}{e^{(E+E_F)/kT}+1}dE$$
+$$n_h = \int_0^\infty f(E+E_F)g_h(E)dE = \int_0^\infty\frac{(2m_h)^{3/2}}{2\pi^2\hbar^3} \sqrt{E}\frac{1}{e^{(E+E_F)/kT}+1}dE$$
 
-$$n_e = \int_{E_G}^\infty f_h(E-E_F)g_e(E)dE = \int_{E_G}^\infty\frac{(2m_e)^{3/2}}{2\pi^2\hbar^3} \sqrt{E-E_G}\frac{1}{e^{(E-E_F)/kT}+1}dE$$
+$$n_e = \int_{E_G}^\infty f(E-E_F)g_e(E)dE = \int_{E_G}^\infty\frac{(2m_e)^{3/2}}{2\pi^2\hbar^3} \sqrt{E-E_G}\frac{1}{e^{(E-E_F)/kT}+1}dE$$
 
 We need to solve $n_e = n_h$