An atom in the state `.^(2)P_(3//2)` is located in the external magnetic field on induction `B=1.0kG`. In terms of the vector model find the angular precession velocity of the total angular momentum of that atom.

Explanation

If `vec(M)` is the total angular momentum vector if he atom then there is magnetic moment
`vec(mu)_(m)=g mu_(B)vec(M)//ħ`
associated with it, here `g` is the Lande factor. In a magnetic field of induction `vec(B)`, an energy
`H'= -g mu_(B)vec(M).vec(B)/ħ`
is associated with it. This interaction term corresponds to a presession of the angular momentum vector beacuse if leads to an equation of motion of hte angular momentum vectro of the form
`(dvec(M))/(dt)=vec(Omega)xxM`
where `vec(Omega)=(g mu_(B)vec(B))/(ħ)`
Using Gaussian unit expression of `mu_(B)=0.927xx10^(-20) erg//gauss, B=10^(3)gauss ħ= 1.054xx10^(-27)` erg sec and for the `.^(2)P_(3//2)` state
`g=1+((3)/(2)xx(5)/(2)+(1)/(2)xx(3)/(2)-1xx2)/(2xx(3)/(2)xx(5)/(2))=1+(1)/(3)=(4)/(3)`
and `Omega=1.17xx10^(10)rad//s`
The same formula is valid in `MKS` units also But `mu_(B)= 0.972xx10^(-23)A/m^(2),B=10^(-1)T` and `ħ=1.054xx10^(-34)` Joul e sec. The answer is the same.