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Homework Statement
a) Calculate the difference in binding energy between the mirror nuclide ##^{15}O## and ##^{15}N##.
b) Calculate the radius of both nuclide’s assuming the difference in binding energy exclusively depends on difference in Coulomb energy.
Homework Equations
Mean nuclear radius
##R = R_0A^{1/3}##.
Binding energy
##B = \left[ Zm(^1H)+Nm_n-m(^AX)\right]c^2##.
Binding energy due to Coulomb repulsion between protons
##E_c = -\frac{3}{5}\frac{e^2}{4\pi \epsilon_0 R_0} Z(Z-1)A^{-1/3}##.
Constants:
##m(^1H) = 1.007825u##
##m_n = 1.00866501u##
##m(^{15}O) = 15.003065u##
##m(^{15}N)=15.000109u##
##1 u = 931.502MeV/c^2##
##\frac{e^2}{4\pi \epsilon_0} = 1.439976MeV \cdot fm##.
The Attempt at a Solution
I think I solved part a) but I'm having trouble with part b).
The difference in binding energy should be
##\Delta B = \left[ m_n-m(^1H)+m(^{15}O)-m(^{15}N\right] = (0.00379601u)/c^2 = 3.536MeV##. (The answer says ##3.532## but I guess I worked with more decimals)
The difference in Coulomb energy should be (since ##8\cdot 7 - 7\cdot 6 = 14##)
##\Delta B= 3/5 \frac{e^2}{4\pi \epsilon_0R_0} 14A^{-1/3}##
##R = R_0A^{1/3} = \frac{3\cdot 14 \cdot 1.439976}{5\cdot 3.536}fm = 3.420752fm##
The answer however says ##R = 3.67fm## so I did something wrong (using 3.532 instead doesn't change anything)
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