written 7.9 years ago by | • modified 7.9 years ago |
Mumbai University > First Year Engineering > Sem 1 > Applied Physics 1
Marks: 3M
Year: May 2016
written 7.9 years ago by | • modified 7.9 years ago |
Mumbai University > First Year Engineering > Sem 1 > Applied Physics 1
Marks: 3M
Year: May 2016
written 7.9 years ago by |
Given:
$T_1= 27^0 C = 300K$
$E_R = 5.6 eV=5.6 x 1.6 x 10^{-19} J$
Find:
$f (E_c)$
Solution:
The probability that an electron being thermally promoted to the conduction band is given as
$f(E_c) = \frac{1}{1 + exp \bigg(\frac{E_g}{2KT}\bigg)}$
where k is the Boltzmann constant.
Substituting the given data for $E_R$, T, k into above equation, we get
$f (E_c) = \frac{1}{1 + exp\bigg(\frac{5.6 \times 1.6 \times 10^{-19}}{2 \times 1.38 \times 10^{-23}\times 300}\bigg)}$
$f (E_c) = 1.7 \times 10^{-47}$
Since the probability $f (E_c)$ is extremely small, electrons cannot be thermally promoted to the conduction band of diamond at $27^0 C$.