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Calculate the magnetic moment of an atom (in Bohr magnetons) (a) in .^(1)F state, (b) in .^(2)D_(3//2) state, (c ) in the state in which S=1, L=2, and Lande factor g=4//3. |
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Answer» Solution :(a) For the `(1)F` states `S=0,L=3=J` `g=1+(3xx4-3xx4)/(2xx3xx4)=1` HENCE `mu=SQRT(3xx4) mu_(B)=2sqrt(3)mu_(B)` (b) For the `.^(2)D_(3//2)` state `S=(1)/(2),L=2,J=(3)/(2)` `g=1+((15)/(4)+(3)/(4)-6)/(2xx(15)/(4))=1+(18-24)/(30)=(4)/(5)` Hence `mu=(4)/(5)sqrt(15//4)mu_(B)=(2)/(5)sqrt(15)mu_(B)=2sqrt((3)/(5))mu_(B)` (C ) We have `(4)/(3)=1+(J(J+1)+2-6)/(2J(J+1))` or `(4)/(3)J(J+1)=J(J+1)-4` or `J(J+1)=12implies J=3` Hence `mu=(4)/(3)sqrt(12)mu_(B)=(8)/(sqrt(3))mu_(B)`. |
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