Answer to Question #106752 in Atomic and Nuclear Physics for Sri L Basuki

Question

(a) From the known masses of 15O and 15N, compute the difference in binding energy. (b) Assuming this difference to arise from the difference in Coulomb energy, compute the nuclear radius of 15O and 15N.

Answer 1

(a) Compute the binding energy in each atom:

$E_b(^{15}\text{O, MeV})=931.494[Zm_p+Nm_n-M(\text{O})]=\\ =931.494[8\cdot1.0072765+7\cdot1.0086654-15.003066]=\\ =107.87\text{ MeV}.\\ \space\\ E_b(^{15}\text{N, MeV})=931.494[Zm_p+Nm_n-M(\text{N})]=\\ =931.494[7\cdot1.0072765+8\cdot1.0086654-15.003066]=\\ =111.92\text{ MeV}.$

The difference is

$\Delta E_b=4.05\text{ MeV}.$

(b) Compute the nuclear radius of each isotope, denote it $R_C$.

The empirical nuclei radius is

$R=r_0A^\frac{1}{3},$

where

$r_0=1.2\cdot10^{-15}\text{ m}$ is an empirical constant (takes values (1.2...1.5) fm),

A - mass number.

The Coulomb energy is

$E_C=\frac{3}{5}\cdot\frac{1}{4\pi\epsilon_0}\cdot\frac{Q^2}{R},$

where $Q=Ze.$

On the other hand, the Coulomb energy is

$E_C\approx\alpha_C\frac{Z(Z-1)}{A^{{1}/{3}}},$

where

$\alpha_C=\frac{3e^2}{20\pi\epsilon_0r_0}$

is called electrostatic Coulomb constant.

Therefore, the nuclear radius will be

$E_C=\frac{3}{5}\cdot\frac{1}{4\pi\epsilon_0}\cdot\frac{(eZ)^2}{R_C},\\ \space\\ R_C=\frac{3e^2Z^2}{20\pi\epsilon_0\cdot E_C}=\\ \space\\ =\frac{3e^2Z^2}{20\pi\epsilon_0\cdot \alpha_C\frac{Z(Z-1)}{A^{{1}/{3}}}}=\\ \space\\ =\frac{3e^2Z^2}{20\pi\epsilon_0\cdot \frac{3e^2}{20\pi\epsilon_0r_0}\frac{Z(Z-1)}{A^{{1}/{3}}}}=\\ \space\\ =\frac{Z}{Z-1}r_0A^{{1}/{3}}.$

As we can notice, this expression gives a bit higher results than the empirical nuclei radius R, which would give equal results for oxygen and nitrogen. Calculate it and the radii R_C obtained from Coulomb's energy:

$R(\text{O})=R(\text{O})=3\cdot10^{-15}\text{ m}.\\ R_C(\text{O})=3.3\cdot10^{-15}\text{ m},\\ R_C(\text{N})=3.5\cdot10^{-15}\text{ m}.$