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What is the probability of an electron being thermally promoted to conduction band in diamond at 27°C, if bandgap is 5.6 eV wide.
1 Answer
written 3.5 years ago by |
f(Ec) = $\dfrac{1\ }{ 1+e^{\big(\dfrac{E_c-E_v}{RT}\big)}}$
K = Boltzman constant = 1.38 ${ \times}$10-23 J/K
In eV it is given by
K(in eV) = ${1.38\times10^{-23} \over 1.6\times 10^{-6}}$ = 86.25 $ {\times}$ 10-6 eV
Also in intrinsic semiconductor
EC - EV = $\dfrac{E_g }{2}$
Ec - Ev = $\dfrac{5.6}{2}$ = 2.8
$\dfrac{E_c - E_v\ }{KT}$ = $\dfrac{2.8}{86.25\times10^{-6}\times(27+273)} $ = 108.212
f(Ec) = $\dfrac{1 }{1+e^{108.212}}$
f(EC) = $1 \times10^{-47}$
It implies that it is insulator.