First and second answer

[Himanshu Sharma]

First one :- if the energy is right amount it will change it’s orientation
Second one :-
1) Without hyperfine interaction
△Eo= gμBBo=1×(5.78838 × 10-5)x2= 1.15767636 × 10-4 eV.
1.1577x 10-4 eV
2) With hyperfine term written as
Hnf = A,mrm]
The other a
BUT when a transition has △m=+1 and a fixed nuclear
projection my=±},the hyperfine contribution to the
transition energy is
△Enr=Am△my=A.my·1=±
Numerically:
• A/2=1.05082 ×10-7eV.
Thus the two components of the transition are
△E=△EO±=1.15767636×10-4eV±1.05082×10-7eV.

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