Hartree

${\displaystyle E_{\mathrm {h} }={\hbar ^{2} \over {m_{\mathrm {e} }a_{0}^{2}}}=m_{\mathrm {e} }\left({\frac {e^{2}}{4\pi \epsilon _{0}\hbar }}\right)^{2}=m_{\mathrm {e} }c^{2}\alpha ^{2}={\hbar c\alpha \over {a_{0}}}}$

= 2 Ry

≜ 27.211386245988(53) eV

≜ 4.3597447222071(85)×10⁻¹⁸ J

≜ 4.3597447222071(85)×10⁻¹¹ erg

≜ 2625.4996394799(50) kJ/mol

≜ 627.5094740631(12) kcal/mol

≜ 219474.63136320(43) cm⁻¹

≜ 6579.683920502(13) THz

≜ 315775.02480407(61) K

where:

ħ is the reduced Planck constant,

m_e is the electron rest mass,

e is the elementary charge,

a₀ is the Bohr radius,

ε₀ is the electric constant,

c is the speed of light in vacuum, and

α is the fine structure constant.

Note that since the Bohr radius ${\displaystyle a_{0}}$ is defined as ${\displaystyle a_{0}={\frac {4\pi \varepsilon _{0}\hbar ^{2}}{m_{\mathrm {e} }e^{2}}}={\frac {\hbar }{m_{\mathrm {e} }c\alpha }}}$, one may write the Hartree energy as ${\displaystyle E_{\mathrm {h} }=e^{2}/a_{0}}$ in Gaussian units where ${\displaystyle 4\pi \varepsilon _{0}=1}$. Effective Hartree units are used in semiconductor physics where ${\displaystyle e^{2}}$ is replaced by ${\displaystyle e^{2}/\epsilon }$ and ${\displaystyle \epsilon }$ is the static dielectric constant. Also, the electron mass is replaced by the effective band mass ${\displaystyle m^{*}}$. The effective Hartree in semiconductors becomes small enough to be measured in millielectronvolts (meV).[3]