Finite size correction

[Tibo]

There will not be an energy correction due to the finite size of this nucleus as in the calculations the r = 0 conditions arose from the fact that only in this case the term delta^2(1/r) is different from zero. So even for a finite size nucleus there will be no energy correction.

[Tibo]

Upon closer inspection of the answers of other I have realised that I might have interpreted the question incorrectly. There will not be energy corrections due to overlap, as the electrons do not but there will be corrections due to the magnetic dipole spin and orbital contributions.

[Author]

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