Finite size

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    The Fermi contact term will of course be 0, as it only gives a correction if there are electrons with a spin imbalance in the nucleus, and we assumed there are no electrons within the nucleus. Maybe we can’t really speak about an energy correction here, because the Fermi contact contribution is most likely the largest contribution to the hyperfine field.
    The Bohr-Weisskopf effect does depend on on the size of the nucleus, as we have to sum the intrinsic magnetic moments of the protons and neutrons which make up the nucleus. Also, a magnetic field will be induced by the movement of the protons within the nucleus, so that hints there is a dependence on the spatial distribution of the nucleons in the nucleus, as the hypefine field will be space dependent. So obtain the interaction energy, we need to integrate the integral the dot product of the space dependend magnetic hyperfine field and the local magnetic moments.
    So: yes, due to the Bohr-Weisskopf interaction, there will be energy corrections (small effects) due to the size of the nucleus.

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