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    Manon
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    a) we can read from the table the following:
    Ground state: I=1/2 and mu=-0,5940mu_N. g = (-0,5940)hbar/(1/2) = -1.2528 • 10^-34 m^2 kg s^-1
    245 keV: analoguesly g=-0.32 • 10^-34 m^2 kg s^-1

    b) g = 2mu_B hbar/mu_N and mu_B is about 2000mu_N, which means g=4000hbar=4,218•10^-31

    c) neutron has spin 0 thus mu=0 J•T

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