finite size corrections

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    I would say if all electrons stay outside the nucleus the Fermi contact potential would vanish, but the Bohr-Weiskampf effect would definitely be there. It is said the nucleus is spherically symmetric but that does not mean it has a uniform magnetic moment, so the different contituents of the nucleus would react differently with the magnetic field caused by the hyperfine interactions so that exactly because of the finite size of the nucleus there will be e a small correction.

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