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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. |
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Answer» Solution :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. |
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