Mumbai university > FE > SEM 1 > Applied Physics 1

Marks: 3M

Year: May 2013

Given:-

Fermi level, $E_f = 2.1eV$

Temperature, T=300K

Probability of Occupancy

(i) $f(E_1) = 0.99$

(ii) $f(E_1) = 0.01$

To Find:-

Energy levels $E_1$ and $E_2$

Solution:

We know that,

$f(E) = \frac{1}{1 + e^{[(E - E_F)/KT]}}$

$\frac{1}{f(E)} = 1 + e^{[(E-E_F)/KT]}$

$e^{[(E - E_F)/KT]} = \frac{1-f(E)}{f(E)}$

Taking ln on both sides,

$\frac{E - E_F}{kT} = ln[\frac{1-f(E)}{f(E)}]$

$E = E_F + kTln[\frac{1 - f(E)}{f(E)}]$

(i) $f(E_1) = 0.99$

$E_1 = 2.1 + \frac{1.38 × 10^{-23}}{1.6 × 10^{-19}} × 300 ln(\frac{0.01}{0.99})$

$E_1 = 1.98eV$

(ii) $f(E_2) = 0.01$

$E_2 = 2.1+ \frac{1.38 × 10^{-23}}{1.6 × 10^{-19}} × 300 ln(\frac{0.99}{0.01})$

$E_2 = 2.22eV$