# g-factors

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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

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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