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The same spectral line undergoing anomalous Zeeman splitting is observed in direction1 and, after reflection from the mirror M (figure), in direction 2. How many Zeeman components are observed in both direction if the spectral lines is caused by transition (a).^(2)P_(3//2) rarr .^(2)S_(1//2), (b) .^(3)P_(2) rarr .^(3)S_(1) ? |
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Answer» Solution :The difference arise because of different selection rules in the TWO cases. In (1) the line is emitted perpendicular to the field. The selection rules are then `Delta M_(Ƶ)=0 +- 1` In (2) the light is emotted along the direction of the field. Then the selection rules are `DeltaM_(Ƶ)=+- 1` `DeltaM_(Ƶ)=0` is forbidden (a)In the transition `.^(2)P_(3//2) rarr .^(2)S_(1//2)` This has been considered above. In (1) we get all the SIX lines shown in hte problem above In (2) the line corresponding to `(1)/(2) rarr(1)/(2)` and `-(1)/(2) rarr -(1)/(2)` is forbidden. Then we get four lines (b) `.^(3)P_(2)` level `g=1+(2xx3+1xx2-1xx2)/(2xx2xx3)=(3)/(2)` so the energies of the sublevels are `E'(M_(Ƶ))=E'_(0)-(3)/(2)mu_(B) BM_(Ƶ)` where `M_(Ƶ)= +-2, +-1,0` For the `.^(3)S_(1)` line `g=2` and the enrgies of the sublevels are `E(M_(Ƶ))= E_(0)-2mu_(B)BM_(Ƶ)` where `M_(Ƶ)= +-1, 0`. The lines are `DeltaM_(Ƶ)=M_(Ƶ)-M_(Ƶ)= +1 : -2 rarr -1,-1 rarr 0` and `0 rarr 1` `DeltaM_(Ƶ)= 0-1 rarr -1,0 rarr0,1 rarr 1` `DeltaM_(Ƶ)= 1, rarr 1,1 rarr0, 0 rarr -1` All energy difference are unequal beacuse the two `g` values are unequal. There are then nine lines if viewed along (1) and six lines if viewed along (2) |
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