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I cannot get the correct units for the magnetic susceptibility and the Fermi energy for a free electron gas!

According to "Introduction to Solid State Physics", 8th edition by Charles Kittel, the total magnetization of a free electron gas is (page 317):

$M=\dfrac{N\mu_B^2}{\epsilon_F}B$

where:

  • $M$ is the magnetization (SI units: A/m)
  • $N$ is the number of electrons,
  • $\mu_B$ is the Bohr magneton constant (SI units: J/T),
  • $\epsilon_F$ is the Fermi energy (SI units: J), and
  • $B$ is the magnetic field strength (SI units: A/m).

The Fermi energy $\epsilon_F$ of the electron gas is given by:

$\epsilon_F=\dfrac{\hbar^2}{2m_e}\left(\dfrac{3\pi^2N}{V}\right)^{2/3}$

where:

  • $\hbar$ is the reduced Planck constant (SI units: Js),
  • $m_e$ the rest mass of each electron (SI units: kg/(number of electrons)), and
  • $V$ the volume (SI units: m$^3$).

According to the above equation the Fermi energy is expressed in Joules$\times$(number of electrons)$^{5/3}$ instead of only Joules. But most importantly, the magnetic susceptibility, $\chi = \dfrac{N\mu_B^2}{\epsilon_F}$, is expressed in units of $\dfrac{\text{Joules}}{\text{Tesla}^2\text{(number of electrons)}^{2/3}}$ instead of being dimensionless!

Where is my understanding wrong? Thank you.

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1 Answer 1

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$\epsilon_F=\dfrac{\hbar^2}{2m_e}\left(\dfrac{3\pi^2N}{V}\right)^{2/3}$

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According to the above equation the Fermi energy is expressed in Joules$\times$(number of electrons)$^{5/3}$ instead of only Joules.

No, $\epsilon_F$ is not expressed in the units you suggest. It has units of energy (Joules).

Notice that $1/V$ has units of inverse length cubed. So, $(1/V)^{(2/3)}$ has units of inverse length squared. So, $\hbar^2 V^{-2/3}$ has units of momentum squared. So, $\epsilon_F$ has units of momentum squared divided by mass, i.e., units of kinetic energy, i.e., units of energy.

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