Update 14_doping_and_devices_solutions.md

src/14_doping_and_devices_solutions.md

@@ -114,7 +114,7 @@ $$g_h = \frac{4 \pi m_h^{\ast}}{\hbar^2}$$
 L can be found here using previous subquestions.
 Setting $$ E_e - E_h - E_c + E_v = 1 eV = \frac{\hbar^2}{2}(\frac{\pi n}{L}^2+k_x^2+k_y^2)
 (\frac{1}{m_e^{\ast}}+\frac{1}{m_h^{\ast}})$$
-L can be found as $6.85$ nm approx.
+By choosing the correct $n$, $k_x$ and $k_y$, L can be found as $6.85$ nm approx.
 
 
 ### Subquestion 6