Temperature for 30% occupation 0.75 eV above

The probability of occupation is given by the Fermi-Dirac function:

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

Given: $E - E_F = 0.75\ \text{eV} = 0.75 \times 1.602 \times 10^{-19} = 1.2015 \times 10^{-19}\ \text{J}$, $f = 0.30$, $k = 1.381 \times 10^{-23}\ \text{J/K}$.

Solve for T

$$\begin{aligned} 0.30 &= \frac{1}{1 + e^{x}}, \quad x = \frac{E - E_F}{kT} \\ 1 + e^{x} &= \frac{1}{0.30} = 3.333 \\ e^{x} &= 2.333 = \frac{7}{3} \\ x &= \ln(2.333) = 0.8473 \end{aligned}$$

So

$$T = \frac{E - E_F}{k \times 0.8473} = \frac{1.2015 \times 10^{-19}}{1.381 \times 10^{-23} \times 0.8473} = 1.03 \times 10^{4}\ \text{K}$$

Such a high temperature is needed because a state 0.75 eV above $E_F$ is far from the Fermi level compared with $kT$ at room temperature (0.026 eV), where its occupation would be almost zero.

Answer: $T \approx 10\,268\ \text{K} \approx 1.03 \times 10^{4}\ \text{K}$.