The Size of the B field Terms in Atoms

In the equation

$$\bgroup\color{black}{-\hbar^2\over{2\mu}}\nabla^2\psi + {e\o... ...{B}\right)^2\right]\psi = \left(E + e \phi\right)\psi. \egroup$$

the second term divided by $(e^2/a_0)$

$$\begin{eqnarray*} {e\over{2\mu c}}\vec{B}\cdot\vec{L} / (e^2/a_0) & \sim & {e\ov... ...imes 10^{-10}\right) }} = m{B\over {5\times 10^9\mbox{ gauss}}} \end{eqnarray*}$$

$$\bgroup\color{black}\left(\alpha = {e^2\over {\hbar c}} \qquad a_0={\hbar\over{\alpha m c}}\right)\egroup$$

Divide the third term by the second:

$$\bgroup\color{black}{B^2 a^2_0 {e^2\over{8mc^2}} \over {{e\... ...es 10^{-10}\right) }} = {B\over{10^{10}\mbox{ gauss}}} \egroup$$