Question

Show that the probability that an energy level which is at an energy ∆E above the
Fermi level is occupied is equal to the probability that an energy level ∆E below the
Fermi level is not occupied.

Answer 1

Answer on Question #61853-Physics-Solid State Physics

Show that the probability that an energy level which is at an energy $\Delta E$ above the Fermi level is occupied is equal to the probability that an energy level $\Delta E$ below the Fermi level is not occupied.

Solution

Let the energy above the Fermi energy $E_{F}$ be $E_{1}$ . Then $\Delta E = E_{1} - E_{F}$ , and the probability of occupancy $f(E_{1})$ of the level $E_{1}$ is given by the FD distribution function, i.e.,

$$f (E _ {1}) = \frac {1}{1 + \exp \left(\frac {E _ {1} - E _ {F}}{k T}\right)} = \frac {1}{1 + \exp \left(\frac {\Delta E}{k T}\right)}$$

The probability of vacancy of the level $E_{1}$ is

$$1 - f (E _ {1}) = 1 - \frac {1}{1 + \exp \left(\frac {\Delta E}{k T}\right)} = \frac {\exp \left(\frac {\Delta E}{k T}\right)}{1 + \exp \left(\frac {\Delta E}{k T}\right)}$$

The probability of occupancy of an energy level $E_{2}$ below $E_{F}$ , where $E_{F} - E_{2} = \Delta E$ , is

$$f (E _ {2}) = \frac {1}{1 + \exp \left(\frac {E _ {2} - E _ {F}}{k T}\right)} = \frac {1}{1 + \exp \left(- \frac {\Delta E}{k T}\right)} = \frac {\exp \left(\frac {\Delta E}{k T}\right)}{1 + \exp \left(\frac {\Delta E}{k T}\right)} = 1 - f (E _ {1}).$$

which proves the desired result.

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