Question

(a) In calclum at room temperature, what is the electron energy at which the Fermi-Dlrac distributlon function has the value 0.13? (Glve thls energy to at least three declmal places. Take the temperature to be room temperature, 293 K.) (b) Over what energy range ?? does the Fermi-Dirac distribution function for calcium drop from 0.95 to 0.13? ?? ev

Answer 1

Fermi distribution function is given by:

$f(E) = \frac{1}{1+e^{\frac{E-E_f}{kT}}}$

Fermi energy of Calcium is: E_f = 4.69 eV = 7.504 x 10^-19 J

so, $0.13 = \frac{1}{1+e^{\frac{E-7.504\times 10^{-19}}{k(293)}}}$

=> $e^{\frac{E-7.504\times 10^{-19}}{k(293)}} = \frac{1}{0.13}-1$

taking natural log on both sides gives:

$\frac{E-7.504\times 10^{-19}}{k(293)} = ln(\frac{1}{0.13}-1)$

=> $E= ln(\frac{1}{0.13}-1)k(293) + 7.504\times 10^{-19} = 7.581\times 10^{-19}J=4.738eV$

This is the energy at 293K.

b]

Repeat the above procedure for f(E) = 0.95 to obtain E'

The energy range will then be: $\Delta E = |E'-E|$ .