Abstract
The effect of finite nucleon size on the Coulomb energy of ${\mathrm{He}}^{3}$ has been investigated. The experimental value of the difference between the binding energies of ${\mathrm{H}}^{3}$ and ${\mathrm{He}}^{3}$ is 0.764 MeV, while the calculated Coulomb energy is approximately equal to or greater than 1.0 MeV if the nuclear force has no repulsive core. We show that, if the finite size of nucleon is taken into consideration, the Coulomb energy of ${\mathrm{He}}^{3}$ is reduced by about 15-20%. The effect of finite charge distribution is determined mainly by the mean square radius. If there is a hard core (with radius $D$), the calculated Coulomb energy (assuming point nucleons) is already smaller, with the values 0.8-0.9 MeV for $D=0.2\ifmmode\times\else\texttimes\fi{}{10}^{\ensuremath{-}13}$ cm, \ensuremath{\sim}0.7 MeV for $D=0.6\ifmmode\times\else\texttimes\fi{}{10}^{\ensuremath{-}13}$ cm. The reduction of Coulomb energy due to the finite size is about 8% and 3%, respectively, for two different models. The Coulomb potential between extended unpolarized nucleons is given in closed form for exponential and Yukawa charge distributions.
Published Version
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