Magnetic moments

[RubenVanderBorght]

mu_N = 3.152*10^-8 eV/T the nuclear magneton
mu_B = 5.788*10^-5 eV/T the Bohr magneton

a) We have g = mu/(mu_N*I) for the g-factor (no hbar contained in I). For 111^Cd,0 keV, the spin is I = 1/2 and mu = -0.595 mu_N. The g-factor is g = -1.190 mu_N
For 245 keV, it follows that I = 5/2 and mu=-0.766 mu_N. Then g = -0.306 mu_N
b) I = 1/2 and mu = 1 mµ_B, then g = 2 since mu=g*mu_b*I (g-factor electron)
c) I = 1/2 and g = -3.826, so mu = -1.913 mu_N

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