Question

The probability of finding an electron at energy level of 0.05 eV above the fermi level in a semiconductor is_____.
(Assume kT = 0.026 eV)

• A. 0.127

Answer 1

The correct option is A 0.127
We know that,
$f\left(E\right)=\frac{1}{1+e^{\left(\frac{E-E_f}{kT}\right)}}$
Given, $E=E_F+0.05$
$\therefore f\left(E\right)=\frac{1}{1+e^{\frac{E_f+0.05-E_f}{kT}}}=\frac{1}{1+e^{\frac{0.05}{0.026}}}$
$f\left(E\right)=0.127$