The ratio of the orbital dipole moment to the orbital angular moment is given by

The orbital dipole moment is given by M = 0.5 x eVangx r2 and the orbital angular moment is given by Ma = m x Vangx r2. Their ratio M/Ma is given by –e/2m, the negative sign indicates the charge of electron.

For a multi-electron atom $$l$$, L and S specify the one electron orbital angular momentum, total orbital angular momentum and total spin angular momentum respectively. The selection rules for electric dipole transition between the two electronic energy levels, specified by $$l$$, L and S are

ΔL = 0, ±1; ΔS = 0; Δ$$l$$ = 0, ±1

ΔL = 0, ±1; ΔS = 0; Δ$$l$$ = ±1

ΔL = 0, ±1; ΔS = ±1; Δ$$l$$ = 0, ±1

ΔL = 0, ±1; ΔS = ±1; Δ$$l$$ = ±1

A magnetic dipole of dipole moment $$\overrightarrow {\bf{m}} $$ is placed in a non-uniform magnetic field $$\overrightarrow {\bf{B}} .$$ If the position vector of the dipole is $$\overrightarrow {\bf{r}} ,$$ the torque acting on the dipole about the origin is

$$\overrightarrow {\bf{r}} \times \left( {\overrightarrow {\bf{m}} \times \overrightarrow {\bf{B}} } \right)$$

$$\overrightarrow {\bf{r}} \times \overrightarrow \nabla \left( {\overrightarrow {\bf{m}} .\overrightarrow {\bf{B}} } \right)$$

$$\overrightarrow {\bf{m}} \times \overrightarrow {\bf{B}} $$

$$\overrightarrow {\bf{m}} \times \overrightarrow {\bf{B}} + \overrightarrow {\bf{r}} \times \overrightarrow \nabla \left( {\overrightarrow {\bf{m}} .\overrightarrow {\bf{B}} } \right)$$