Commit 4bd0c2e8 authored by Isidora Araya's avatar Isidora Araya
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Update 14_doping_and_devices_solutions.md

parent c8e9c7f7
......@@ -102,8 +102,8 @@ $$-\frac{\hbar^2}{2m_h^{\ast}} \nabla^2 \Psi_h = (E_v-E_h)\Psi_h $$
### Subquestion 3
$$E_e = E_c + \frac{\hbar^2 (k_x^2+k_y^2)}{2m_e^{\ast}}$$
$$E_h = E_v - \frac{\hbar^2 (k_x^2+k_y^2)}{2m_h^{\ast}}$$
$$E_e = E_c + \frac{\hbar^2}{2m_e^{\ast}} (\frac{\pi n}{L}^2+k_x^2+k_y^2+)}$$
$$E_h = E_v - \frac{\hbar^2}{2m_h^{\ast}} (\frac{\pi n}{L}^2+k_x^2+k_y^2+)}$$
### Subquestion 4
This is a 2D electron/hole gas. Apply 2D density of states (constant).
......@@ -111,10 +111,10 @@ $$g_e = \frac{4 \pi m_e^{\ast}}{\hbar^2}$$
$$g_h = \frac{4 \pi m_h^{\ast}}{\hbar^2}$$
### Subquestion 5
L can be found here for $k_x$ and $k_y$ using previous subquestions.
Setting $$ E_e - E_h - E_c + E_v = 1 eV = \frac{\hbar^2 (k_x^2+k_y^2)}{2}
L can be found here using previous subquestions.
Setting $$ 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}})$$
Taking $k_x^2+k_y^2=(\frac{\pi}{L})^2$, L can be found as $6.85$ nm approx.
L can be found as $6.85$ nm approx.
### Subquestion 6
......
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