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 $(e^2/a_0)$

$$\bgroup\color{black} {e\over{2\mu c}}\vec{B}\cdot\vec{L} / (e... ... \sim {e\over{2\mu c}}B\left(m\hbar\right) / (e^2/a_0) \egroup$$

$$\bgroup\color{black} = m{\alpha eBa_0\over 2} / \left(e^2/a_0\right) = m{\alpha a_0^2\over {2 e}} B \egroup$$

$$\bgroup\color{black} = mB{ \left(0.5\times 10^{-8}\mbox{ cm}\... ...{10}\right) }} = m{B\over {5\times 10^9\mbox{ gauss}}} \egroup$$

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

divide the second term by the third:

$$\bgroup\color{black} \sim {B^2 a^2_0 {e^2\over{8mc^2}} \over {{e\over{2\mu c}}B\hbar}} = \alpha {a^2_0\over{4 e}} B \egroup$$

$$\bgroup\color{black} = { \left(0.5\times 10^{6}\right)^2 \o... ...mes 10^{10}\right) }} = {B\over{10^{10}\mbox{ gauss}}} \egroup$$