---
title: "Temperature for 98% occupation probability"
subject: "Electrical Engineering Material"
chapter: 1
key: "semester_3/electrical_engineering_material:fermi-temperature-98-percent"
source: https://www.bctengineer.com/q/semester_3/electrical_engineering_material/fermi-temperature-98-percent
site: BCT Engineering
---

# Temperature for 98% occupation probability

Electrical Engineering Material · Chapter 1: Theory of Metals

- Asked in: 2074 Chaitra, 4 marks
- Video: [Fermi Dirac Distribution in Semiconductor (Basics, Statement, Function, Proof & Case Study)](https://www.youtube.com/watch?v=IEGzFze1y6g) — Engineering Funda
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_3/electrical_engineering_material/fermi-temperature-98-percent
- Whole chapter: https://www.bctengineer.com/md/semester_3/electrical_engineering_material/chapter-1

## Question

Calculate the temperature at which there is 98% probability that a state 0.3 eV below the fermi energy level will be occupied by an electron.

## Short answer

Solve f(E)=0.98 with E−E F = −0.3 eV: 0.98 = 1/(1+e^{-0.3/kT}) => e^{-0.3/kT}=0.020408 => 0.3/kT = ln 49 = 3.892 => kT = 0.07708 eV, T = 894 K.

## Answer

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.

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. For a state 0.3 eV below E_F, E - E_F in the Fermi function is:
   A. +0.3 eV
   B. -0.3 eV
   C. 0
   D. kT
   Answer: B. The state is below E_F so E - E_F is negative. (Key point: Sign of exponent)

2. Setting f = 0.98 gives e^(-0.3/kT) equal to:
   A. 0.98
   B. 0.02
   C. 0.020408 (= 1/49)
   D. 49
   Answer: C. 1/0.98 - 1 = 0.020408, so 0.3/kT = ln 49 = 3.892. (Key point: Algebra)

3. kT in this problem is about:
   A. 0.0771 eV
   B. 0.771 eV
   C. 0.0259 eV
   D. 3.892 eV
   Answer: A. kT = 0.3/3.892 = 0.0771 eV. (Key point: kT value)

4. The temperature is about:
   A. 300 K
   B. 894 K
   C. 89.4 K
   D. 2000 K
   Answer: B. T = 0.0771/(8.617x10^-5) is about 894 K. (Key point: Temperature)

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