---
title: "Temperature for 30% occupation 0.75 eV above"
subject: "Electrical Engineering Material"
chapter: 1
key: "semester_3/electrical_engineering_material:fermi-temperature-30-percent"
source: https://www.bctengineer.com/q/semester_3/electrical_engineering_material/fermi-temperature-30-percent
site: BCT Engineering
---

# Temperature for 30% occupation 0.75 eV above

Electrical Engineering Material · Chapter 1: Theory of Metals

- Asked in: 2081 Chaitra (new course), 4 marks
- 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-30-percent
- Whole chapter: https://www.bctengineer.com/md/semester_3/electrical_engineering_material/chapter-1

## Question

Find the temperature at which the probability of occupation of the energy state 0.75 eV above the Fermi energy is 30%.

## Short answer

From f=0.30 and E−EF=0.75 eV in f=1/[1+exp((E−EF)/kT)], we get exp((E−EF)/kT)=7/3, so x=ln(7/3)=0.8473. Solving T = (E−EF)/(k x) gives T ≈ 1.03×10⁴ K (about 10,300 K).

## Answer

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

## Check yourself

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

1. With f = 0.30, which gives e^x where x = (E-E_F)/kT?
   A. e^x = 1/0.30 - 1 = 2.333
   B. e^x = 0.30
   C. e^x = 1/0.30 + 1
   D. e^x = 0.70
   Answer: A. 1 + e^x = 1/0.3 = 3.333, so e^x = 2.333. (Key point: Rearranging f)

2. What is x?
   A. 0.847
   B. 2.333
   C. 1.20
   D. 0.30
   Answer: A. x = ln 2.333 = 0.8473. (Key point: Value of x)

3. Which expression gives T?
   A. T = (E-E_F) / (k x)
   B. T = k x / (E-E_F)
   C. T = (E-E_F) x / k
   D. T = k / (x (E-E_F))
   Answer: A. x = (E-E_F)/kT gives T = (E-E_F)/(kx), about 1.03e4 K. (Key point: Solving for T)

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