Temperature for 98% occupation probability

The occupation probability of a state is given by the Fermi–Dirac function

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

Given: $f(E) = 0.98$, $E - E_F = -0.3$ eV (state is below $E_F$), $k = 8.617\times10^{-5}$ eV/K ($= 1.38\times10^{-23}$ J/K).

Solve for T

$$\begin{aligned} 0.98 &= \frac{1}{1 + e^{-0.3/kT}} \\ 1 + e^{-0.3/kT} &= \frac{1}{0.98} = 1.020408 \\ e^{-0.3/kT} &= 0.020408 \\ \frac{0.3}{kT} &= \ln\frac{1}{0.020408} = \ln 49 = 3.892 \\ kT &= \frac{0.3}{3.892} = 0.07708\ \text{eV} \end{aligned}$$ $$T = \frac{0.07708\times1.6\times10^{-19}}{1.38\times10^{-23}} = 894\ \text{K}$$

Answer: $T \approx 894$ K (about 621 °C).

At this temperature $kT = 0.077$ eV. Below 894 K the probability of occupation of that state is higher than 98 % (it tends to 100 % as $T \to 0$); above 894 K it falls below 98 %, because more electrons are thermally excited above $E_F$, leaving states below $E_F$ empty.