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  • #3637 Score: 0

    In the notes we can find the following two formula:
    vec{mû}_I = (mu/I hbar) vec{Î}
    vec{mû}_I = (g mu_N /hbar) vec{Î}
    From these we can find a relation for g:
    (mu/I hbar) = (g mu_N /hbar)
    <=> g = mu / (I mu_N)

    a) Let’s consider the groundstate of 111Cd with mu = -0.5940 mu_N (the paper giving info on the Nuclear data base mentions that mu is given in units nuclear magnetons) and I=1/2.
    So we have:
    g = -0.5940 x 2 = -1.188.
    Now for the 245 keV level: mu = -0.766 mu_N and I = 5/2:
    g = -0.766 x 2 / 5 = -0.3064

    b) For a free electron the relation between the magnetic moment and its spin is determined by the g-factor and here the Bohr magneton instead of the nuclear magneton, giving finally:
    g = mu / (S mu_B)
    It is given that the free electron has mu = mu_B, and S=1/2. So that gives us, in a approximation:
    g = 2

    c) In the case of the free neutron we have again
    g = mu / (I mu_N)
    We are give that g = -3.826. So we have:
    -3.826 = mu / (I mu_N) <=> mu = -3.826 I mu_N
    We need to know the nuclear spin to calculate this further.

    #3638 Score: 0

    I realise, after reading some other answers that I read ‘neutron’ as ‘nucleus’ for some reason.

    Let me try to correct then my last answer:
    So we have the 2 formulae:
    vec{mû}_S = (mu/S hbar) vec{\Hat{S}}
    vec{mû}_S = (g e/2m) vec{\Hat{S}}

    For an eletron, the last formula is:
    vec{mû}_S = (g u_B/ hbar) vec{\Hat{S}}
    A neutron is around 1836 times heavier than an electron, and thus for the neutron:
    vec{mû}_S = (g u_B/1836 hbar) vec{\Hat{S}}

    Equating the 2 formulae we have:
    mu/S hbar =g mu_B/1836 hbar <=> mu = g mu_B S /1836
    We are give that g = -3.826, and have S=1/2. So we have:
    mu = -0.00104 mu_B

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