Mumbai university > FE > SEM 1 > Applied Physics 1

Marks: 5M

Year: May 2013

Given:-

Fermi level lies 0.4eV below conduction band

$E_{(F_1 )} = E_C - 0.4$

Concentration is doubled

$n_2 = 2n_1$

To find:-

New position of Fermi level $E_{(F_2 )}$

Solution:

We know that,

$n = N_C e^{-[(E_C-E_F)/kT]}$

$\frac{n}{N_C} = e^{[(E_F-E_C)/kT]}$

$ln(\frac{n}{N_C}) = \frac{E_F - E_C}{kT}$

$E_F - E_C = kT ln(\frac{n}{N_C})$

Initially,

$E_{(F_1 )} - E_C = kT ln(\frac{n_1}{N_C }) ------(i)$

After doubling the concentration,

$E_{(F_2 )} - E_C = kT ln(\frac{n_2}{N_C }) ------(ii)$

Subtracting (i) from (ii),

$E_{(F_2 )} - E_{(F_1 )} = kT [ln(\frac{n_2}{N_C }) - ln(\frac{n_1}{N_C })]$

$E_{(F_2 )} - E_{(F_1 )} = - kTln(\frac{n_1}{n_2})$

$ E_{(F_2 )} - E_C – 0.4 = - kT ln(1/2) \ \ \ [ n_2 = 2n_1]$

Considering room temperature i.e T=300K

$E_{(F_2 )} = E_C - 0.4 - \frac{1.38 × 10^{-23}}{1.6 × 10^{-19}} × 300 ln(1/2)$

$E_{(F_2 )} = E_C – 0.38eV$

Fermi level is below conduction band by 0.38eV