Modern Physics

Fermi Energy

EF=ℏ22m(3π2n)2/3E_F = \frac{\hbar^2}{2m}(3\pi^2 n)^{2/3}

The highest occupied energy level at absolute zero for a system of fermions. Crucial for understanding metals, white dwarfs, and neutron stars.

Variables

EF
Fermi energy
J
ℏ
Reduced Planck constant
Allowed values: greater than 0
J·s
m
Particle mass (e.g., electron)
Allowed values: greater than 0
kg
n
Number density of particles
Allowed values: greater than 0
m⁻³

Derivation

Derived from counting quantum states up to the Fermi level in a free electron gas.

Your notes

0/5000

Constants in this formula

ℏ
1.054 571 817… × 10⁻³⁴. Exact by definition.
J·s

Check value

Round inputs run through this page's calculator, with each constant at its tabulated value. Use it to test your own working or a rearranged form. It is a test case, not a real-world example.

  • ℏ = 1.054 571 817… × 10⁻³⁴ J·s (Reduced Planck constant)
  • m = 2 kg (Particle mass (e.g., electron))
  • n = 3 m⁻³ (Number density of particles)

Result: EF = 5.535037 × 10⁻⁶⁸ J

Other modern physics formulas with the same quantities

Where this formula is listed

Common questions

What is the Fermi Energy formula?

Fermi Energy is written E_F = (ℏ²/(2m))(3π²n)^(2/3). The highest occupied energy level at absolute zero for a system of fermions. Crucial for understanding metals, white dwarfs, and neutron stars. It belongs to Modern Physics, under Quantum Mechanics.

What are the units in E_F = (ℏ²/(2m))(3π²n)^(2/3)?

In SI units, E_F is fermi energy, measured in J; ℏ is reduced planck constant, measured in J·s; m is particle mass (e.g., electron), measured in kg; n is number density of particles, measured in m⁻³.

How is the Fermi Energy formula derived?

Derived from counting quantum states up to the Fermi level in a free electron gas. The result is E_F = (ℏ²/(2m))(3π²n)^(2/3).