# Two isobars they differ in binding energy, will they differ in radius as well?

It is a problem from introduction to nuclear physics- by krane, chapter: nuclear properties, problem 7. It goes like this: From the known masses of $_{~~8} ^{15}\rm O$ and $_{~~7}^{15}\rm N$, compute the difference in binding energy. Assuming this difference to arise from the difference in coulomb energy, compute the nuclear radius of $_{~~8} ^{15}\rm O$ and $_{~~7} ^{15}\rm N$. I computed the difference in binding energy $$BE(_{~~7} ^{15}\mathrm N)-BE(_{~~8} ^{15}\mathrm O)= 3.536489\,\rm MeV$$ I can't relate it with the radius. From the formula $$R= R_{o} A^{ \frac{1}{3} }$$ my intuition tells me isobars should have same radius as $A$ is same for both of them. How can I solve this?

In other words my question is suppose you have two nucleus with same mass number but they have different binding energy. What would be the formula for their radius? How much they would differ in their radius?

• Comment to the question (v2): the precision you imply for the binding energy (about $1\rm\,eV$) is unrealistically aggressive. – rob May 22 '15 at 8:15
• Any idea? @phonon – Self-Made Man May 22 '15 at 9:17

One approach: the liquid-drop model says they should have the same $R\propto A^{1/3}$, but different Coulomb energies proportional to $Q^2/R = Z^2e^2/R$.
To wit: the self-energy of a uniformly-charged sphere with radius $R$ and charge $Q=Ze$ is $$U = \frac35 \cdot \frac1{4\pi\epsilon_0} \frac{Q^2}{R} = \frac35 \cdot \alpha\hbar c \frac{Z^2}{R}.$$ (Remember that $\alpha\hbar c = e^2/4\pi\epsilon_0.$) So the difference in the Coulomb energy between two nuclei with the same (unknown) $R$ but $Z=7$ and $Z=8$ would be \begin{align} U_8-U_7 &= \frac 35 \frac{\alpha\hbar c}{R} \left( 8^2 - 7^2 \right) = 9 \frac{\alpha\hbar c}R \\ R &= \frac{9\alpha\hbar c}{U_8-U_7} \\ R &= \frac{9}{137} \frac{\rm 197\,MeV\,fm}{\rm 3.54\,MeV} = 3.39\rm\,fm \end{align} If we compare that to the liquid drop radius model \begin{align} R &= R_0 A^{1/3} \\ 3.39\,\mathrm{fm} &= R_0 15^{1/3} \end{align} we get a nucleon radius $R_0 = 1.37\,\mathrm{fm}$. Not far from the commonly cited $R_0=1.2\,\mathrm{fm}$, especially if you take the implied (im)precision in that value seriously.