Question

An atom is located in a magnetic firld of induction `B = 2.50kG`. Find the value of the total splitting of the following terms (expressed in `eV` units):
(a) `.^(1)D`, (b)`.^(3)F_(4)`.

Answer 1

(a) For the `.^(1)D_(2)` term
`g=1+(2xx3+0-2xx3)/(2xx2xx3)=1`
and `Delta E= -mu_(B)M_(J)B`
`M_(J)= -2,-1,0+1,+2`. Thus the spliting is
`deltaE=4mu_(B)`
Substitution gives `deltaE= 57.9mu eV`
(b) For the `.^(3)F_(4)` term `g=1+(4xx5+1xx2-3xx4)/(2xx4xx5)=1+(10)/(40)=(5)/(4)`
and `Delta=-(5)/(4)mu_(B) BM_(J)`
where `M_(J)= -4 t o +4`. Thus
`deltaE=(5)/(4)mu_(B) Bxx8= 10mu_(B)(= 2g Jmu_(B))`
Substitution gives `deltaE=144.7 mueV`