---
title: "Fermi-Dirac function and 1.5kT probability"
subject: "Electrical Engineering Material"
chapter: 1
key: "semester_3/electrical_engineering_material:fermi-dirac-1-5kt"
source: https://www.bctengineer.com/q/semester_3/electrical_engineering_material/fermi-dirac-1-5kt
site: BCT Engineering
---

# Fermi-Dirac function and 1.5kT probability

Electrical Engineering Material · Chapter 1: Theory of Metals

- Asked in: 2069 Chaitra, 4 marks
- Video: [Fermi Function Explained](https://www.youtube.com/watch?v=cBd5Kaqy7EU) — Jordan Edmunds Chetty
- 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-dirac-1-5kt
- Whole chapter: https://www.bctengineer.com/md/semester_3/electrical_engineering_material/chapter-1

## Question

What is fermi-Dirac Distribution function? Prove that probability of finding electron 1.5 KT above the fermi level.

## Short answer

The Fermi–Dirac distribution f(E) = 1/{1+exp[(E−E_F)/kT]} gives the occupation probability of an electron state. Substituting E = E_F + 1.5kT yields f = 1/(1+e^{1.5}) ≈ 0.182; the probability is about 18.2%, independent of temperature and material.

## Answer

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 $E_F$.

```text
 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%.

## Check yourself

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

1. What does the Fermi-Dirac function give?
   A. Number of states at E
   B. Probability that a state at energy E is occupied at temperature T
   C. Average speed of electrons
   D. Fermi energy
   Answer: B. f(E) is the occupation probability, from Pauli exclusion. (Key point: Meaning of f(E))

2. Putting E - E_F = 1.5kT, what does f become?
   A. 1/(1 + e^1.5)
   B. 1/(1 + e^(1.5 kT))
   C. 1/(1 - e^1.5)
   D. e^(-1.5)
   Answer: A. The kT cancels in the exponent, leaving 1/(1 + e^1.5). (Key point: Substitution)

3. What value does the probability take?
   A. 0.182
   B. 0.5
   C. 0.818
   D. 0.269
   Answer: A. 1/(1 + 4.4817) = 0.182. (Key point: Numerical result)

4. How does this result depend on T and material?
   A. Increases with T
   B. Independent of both
   C. Depends on E_F
   D. Depends on material
   Answer: B. Since the offset is expressed in units of kT, the result is the same for any T and material. (Key point: Temperature independence)

## Related in Electrical Engineering Material

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