Fermi-Dirac function and 1.5kT probability

The Fermi-Dirac distribution function $f(E)$ gives the probability that an available energy state at energy $E$ is occupied by an electron at temperature $T$:

$$f(E) = \frac{1}{1 + \exp\left(\dfrac{E - E_F}{kT}\right)}$$

where $E_F$ is the Fermi energy and $k$ is Boltzmann's constant. It follows from the Pauli exclusion principle (at most one electron per state).

Key features:

• At $T = 0$: $f = 1$ for $E < E_F$ and $f = 0$ for $E > E_F$ (a step).
• At any $T > 0$: $f(E_F) = 1/2$.
• The change from 1 to 0 happens over a few $kT$ around $E_F$.

 f(E)
 1.0 |------\
     |       \   T > 0
 0.5 |- - - - *
     |         \
 0.0 |__________\------> E
               EF

Probability at 1.5kT above the Fermi level

Put $E - E_F = 1.5kT$:

$$\begin{aligned} f(E_F + 1.5kT) &= \frac{1}{1 + e^{1.5kT/kT}} = \frac{1}{1 + e^{1.5}} \\ &= \frac{1}{1 + 4.4817} = 0.182 \end{aligned}$$

The result does not depend on temperature or on the material: at any $T$, a state lying $1.5kT$ above $E_F$ is occupied with probability about 18%, and empty with probability $1 - 0.182 = 0.818$.

Answer: $f = 1/(1+e^{1.5}) = 0.182$, i.e. about 18.2%.