| Line 30: | Line 30: | ||
E = \pm\alpha\sqrt{k_x^x+k_y^2} | E = \pm\alpha\sqrt{k_x^x+k_y^2} | ||
</math> | </math> | ||
| + | |||
Let ; <math>k=\pm\sqrt{k_x^x+k_y^2}</math> | Let ; <math>k=\pm\sqrt{k_x^x+k_y^2}</math> | ||
| + | |||
<math> | <math> | ||
\therefore E = \alpha k | \therefore E = \alpha k | ||
</math> | </math> | ||
| + | |||
<math> | <math> | ||
\begin{align*} | \begin{align*} | ||
| Line 51: | Line 54: | ||
\end{align*} | \end{align*} | ||
</math> | </math> | ||
| + | |||
Taking <math>E_c = 0</math> | Taking <math>E_c = 0</math> | ||
| + | |||
<math> | <math> | ||
n = \frac{E_F^2}{4\alpha^2\pi} | n = \frac{E_F^2}{4\alpha^2\pi} | ||
</math> | </math> | ||
| − | [[Image:MN1_2007_1.png|Alt text| | + | [[Image:MN1_2007_1.png|Alt text|400x400px]] |
------------------------------------------------------------------------------------ | ------------------------------------------------------------------------------------ | ||
| Line 68: | Line 73: | ||
</math> | </math> | ||
| − | [[Image:MN1_2007_2.png]] | + | [[Image:MN1_2007_2.png|Alt text|400x400px]] |
<math> | <math> | ||
Revision as of 10:56, 6 August 2017
MICROELECTRONICS and NANOTECHNOLOGY (MN)
Question 1: Semiconductor Fundamentals
August 2007
Questions
All questions are in this link
Solutions of all questions
1)
$ E = \pm\alpha\sqrt{k_x^x+k_y^2} $
Let ; $ k=\pm\sqrt{k_x^x+k_y^2} $
$ \therefore E = \alpha k $
$ \begin{align*} D(E) &= \frac{1}{2\pi}k\frac{dk}{dE}\\ &= \frac{1}{2\pi}\cdot\frac{E}{\alpha}\cdot\frac{1}{\alpha} = \frac{E}{2\pi\alpha^2} \end{align*} $
------------------------------------------------------------------------------------
2)
$ \begin{align*} n&=\int_{E_c}^{E_f}D(E)dE\\ &=\int_{E_c}^{E_f}\frac{E}{2\alpha^2}\cdot dE\\ &=\frac{1}{4\alpha^2}(E_F^2 - E_c^2) \end{align*} $
Taking $ E_c = 0 $
$ n = \frac{E_F^2}{4\alpha^2\pi} $
------------------------------------------------------------------------------------ 3)
$ \begin{align*} F &= -qE =qE_x \hspace{0.5cm} [\because E = -\hat{x}E_x]\\ F&= \frac{d(\hslash k)}{dt} \end{align*} $
$ \implies \int_0^{k_x}dk = \frac{1}{\hslash}\int_0^t Fdt $ $ \implies k_x = \frac{qE_xt}{\hslash} $ $ E = \alpha k $ $ V = \frac{1}{\hslash}\frac{dE}{dk} = \frac{\alpha}{\hslash} $ $ x = \int_0^tVdt = \frac{\alpha}{\hslash}t $
