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Calculate the total splitting Delta omega of the spectral line .^(3)D_(3) rarr .^(3)P_(2) in a weak magnetic field with inductionB= 3.4KG. |
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Answer» Solution :For the TWO levels `E'+(0)=E_(0)-g'mu'_(B)M'_(Ƶ)B` `E_(0)=E_(0)0g mu_(B)M_(Ƶ)B` and HENCE the shift of the component is the value of `Delta OMEGA =(mu_(B)B)/( ħ)[g'M'_(Ƶ)-gM_(Ƶ)]` subject to the selection rule `Delta M_(Ƶ)=0, +- 1` For `(3)D_(3)` `g'1+(3xx4+1xx2-2xx3)/(2xx3xx4)= 1+(8)/(24)=(4)/(3)` For `.^(3)P_(2)`, `g=1+(2xx3+1xx2-1xx2)/(2xx2xx3)=(3)/(2)` Thus `Delta omega=(mu_(B)B)/( ħ)|(4)/(3)M'_(Ƶ)-(3)/(2)M_(Ƶ)|` For the different transition we have the following table `M_(Ƶ)g'-M_(Ƶ)g` `3 rarr 2 mu_(B)B 0 rarr1 -(3)/(2)mu_(B)B` `2 rarr2 -(1)/(3)mu_(B)B 0 rarr0 0` `2 rarr7//6mu_(B)B 0 rarr -1 3//2mu_(B)B` `1 rarr 2 -5//3mu_(B) B -1 rarr0 -4//3 mu_(B) B` `1 rarr1 -1//6mu_(B)B -1 rarr-1 1//6//6mu_(B)B` `1 rarr 0 4//3mu_(B)B -1 rarr -2 rarr 5//3mu_(B)B` `2 rarr -1rarr -7//6mu_(B)B` `-2 rarr-2 rarr 1//3mu_(B)B` `-3 rarr -2 rarr -mu_(B)B` There are 15 lines in all. The lines farthest out are `1 rarr 2` and `-1 rarr -2` The splitting between them is the total splitting. It is `Delta omega +(10)/(3) mu_(B) B//ħ` Substitution gives `Delta omega= 7.8xx10^(10) rad//sec`. |
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