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The electrostatic energy of Z protons uniformly distributed in a nucleus of radius R is E = 3*Z*(Z-1)*e² / (5*4*pi*epsilon₀*R). Given: mass of neutron = 1.008665 u, mass of H-1 = 1.007825 u, mass of N-15 = 15.000109 u, mass of O-15 = 15.003065 u, 1u = 931.5 MeV/c², e²/(4*pi*epsilon₀) = 1.44 MeV*fm. Assuming the difference in binding energies of N-15 and O-15 arises purely from electrostatic energy, find the common nuclear radius R.

1. 2.85 fm
2. 3.03 fm
3. 3.42 fm
4. 3.80 fm

Correct answer: 3.42 fm

Solution

Delta_BE = (mₙ - m_H + M_O - M_N)*931.5 = (0.000840 + 0.002956)*931.5 = 3.54 MeV. Coulomb difference = 3*1.44*(56-42)/(5R) = 3*1.44*14/(5R) = 12.10/R. Setting equal: R = 12.10/3.54 = 3.42 fm.

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