Commit ce166090 authored by Isidora Araya's avatar Isidora Araya
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Update 14_doping_and_devices_solutions.md

parent 4bd0c2e8
......@@ -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
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