Update 14_doping_and_devices_solutions.md

src/14_doping_and_devices_solutions.md

@@ -102,7 +102,7 @@ $$g_h = \frac{4 \pi m_h^{\ast}}{\hbar^2}$$
 ### Subquestion 5
 L can be found here for $k_x$ and $k_y$ using previous subquestions.
 Setting $$ E_e - E_h - E_c + E_v = 1 eV = \frac{\hbar^2 (k_x^2+k_y^2)}{2}
-(\frac{1}{m_e^{\ast}}-\frac{1}{m_h^{\ast}})$$
+(\frac{1}{m_e^{\ast}}+\frac{1}{m_h^{\ast}})$$
 
 ### Subquestion 6