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a) From the formula in the slides we get g = (\mu*\hbar)/(I*\mu_N).
For the 111Cd ground state g = (-0,594\mu_N*\hbar)/((1/2)*\hbar*\mu_N) = -1,188
For the 111Cd 245 eV state g = (-0,766\mu_N*\hbar)/((5/2)*\hbar*\mu_N) = -0,3064b) For the free electron the spin 1/2 and we have g = (\mu*\hbar)/(S*\mu_N). This leads to g = 2(\mu/\mu_N)
c) We use again the same formula from the slides to get
\mu = (g*I*\mu_N)/\hbar = -3,826*(1/2)*\mu_N = -1,913\mu_N
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