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  • #4242
    axlamelas
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    1.a)
    From the video
    ɣ = µ/(Iħ) = g µ_N / ħ
    Therefore g = µ/(I * µ_N)

    For the ground state of 111Cd: µ/µ_N = -0.5940(3), I = 1/2 => g = -1,18806
    For the 245 KeV excited state of 111Cd: µ/µ_N = -0.766(3), I = 5/2 => g = -0.30652

    1.b)
    For the electron ɣ = µ/(S * ħ) = g µ_B / ħ, therefore g = µ/(S * µ_B) = µ_B/(S * µ_B) =1/S = 2

    1.c)
    µ = g * I * µ_N

    For the neutron I = 1/2, therefore µ = -1.913 nuclear magneton

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