Typical atomic lengths, wave numbers, and areas in physics can be related to the reduced Compton wavelength for the electron ({{tmath|1= \textstyle \lambda\!\!\!\bar{}_\text{e} \equiv \tfrac{\lambda_\text{e} }{2\pi}\simeq \mathrm{386~fm} }}) and the electromagnetic
fine-structure constant {{nowrap|({{tmath|1= \alpha\simeq\tfrac{1}{137} }}).}} The
classical electron radius is about 3 times larger than the
proton radius, and is written: r_\text{e} = \alpha\lambda\!\!\!\bar{}_\text{e} \simeq 2.82~\textrm{fm} The
Bohr radius is related to the Compton wavelength by: a_0 = \frac{\lambda\!\!\!\bar{}_\text{e}}{\alpha} \simeq 5.29\times 10^4~\textrm{fm} The
angular wavenumber of a photon with one
hartree (the
atomic unit of energy {{tmath|1= E_\text{h} = 4\pi\hbar cR_\infty }}, where is the
Rydberg constant), being (approximately) the negative potential energy of the electron in the hydrogen atom, and twice the energy needed to ionize it, is: \frac{\hbar c}{E_\text{h}} = \frac{1}{4\pi R_\infty} = \frac{\lambda\!\!\!\bar{}_\text{e}}{\alpha^2} \simeq 7.25~\textrm{nm} This yields the sequence: \alpha^{-1} r_{\text{e}} = \lambda\!\!\!\bar{}_{\text{e}} = \alpha a_0 = \alpha^2 {\hbar c/E_\text{h}}. For
fermions, the classical (electromagnetic) radius sets the cross-section of electromagnetic interactions of a particle. For example, the cross-section for
Thomson scattering of a photon from an electron is equal to \sigma_\mathrm{e} = \frac{8\pi}{3}r_\text{e}{}^2 \simeq \mathrm{66.5~fm^2} , which is roughly the same as the cross-sectional area of an iron-56 nucleus. == Geometrical interpretation ==