Mumbai university > FE > SEM 1 > Applied Physics 1

Marks: 3M

Year: Dec 2012

Given:-

$Temperature T= 20^oC = 293 K$

Boltzmann Constant, $k = 1.38 ×10^{-23}$

Band Gap Energy, (Eg) =1.12eV

To Find:

Probability of an electron being excited i.e. $F(E_C)$

Solution:

For an intrinsic semiconductor,

$E_C - E_V = \frac{E_Q}{2} = \frac{1.12}{2}$

$E_C - E_V = 0.56eV$

Now,

$k = 1.38 × 10^-23 J/K$

$ = \frac{1.35 ×10^{-23}}{1.6 × 10^{-19}} eV/K$

$ k = 8.625 × 10^{-5} eV/K$

We know that,

$F(E_C) = \frac{1}{1 + e^{[(E_C - E_V)/KT]}}$

$= \frac{1}{1+ e^{[(0.56 / 8.625 × 293 × 10^{-25})]}}$

$F(E_C)=0.4936$