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*}

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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 $E_c = 0$

$n = \frac{E_F^2}{4\alpha^2\pi}$


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3)

\begin{align*} F &= -qE =qE_x \hspace{0.5cm} [\because E = -\hat{x}E_x]\\ F&= \frac{d(\hslash k)}{dt} \end{align*}


$\implies \int_0^{k_x}dk = \frac{1}{\hslash}\int_0^t Fdt$ $\implies k_x = \frac{qE_xt}{\hslash}$ $E = \alpha k$ $V = \frac{1}{\hslash}\frac{dE}{dk} = \frac{\alpha}{\hslash}$ $x = \int_0^tVdt = \frac{\alpha}{\hslash}t$

## Alumni Liaison

Abstract algebra continues the conceptual developments of linear algebra, on an even grander scale.

Dr. Paul Garrett