Finite size

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    Johanna Waltenspiel
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    If no electrons penetrate the nucleus, the Fermi contact correction and the Bohr-Weisskopf correction would be zero.

    The other energy corrections (such as spin dipolar and the orbital contribution) would not be affected. They are however not due to the finite size of the nucleus.

    I should think that the fact that an area of space (nucleus) is ‘forbidden’ for the electron would give rise to a needed correction. What the exact nature of this contribution would be is not obvious to me.

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