The binding energy of nucleons in a nucleus can be affected by the pairwise Coulomb repulsion. Assume that all nucleons are uniformly distributed inside the nucleus. Let the binding energy of a proton

Question

The binding energy of nucleons in a nucleus can be affected by the pairwise Coulomb repulsion. Assume that all nucleons are uniformly distributed inside the nucleus. Let the binding energy of a proton be $E_{b}^{p}$ and the binding energy of a neutron be $E_{b}^{n}$ in the nucleus.
Which of the following statement(s) is(are) correct?

TestHub · Accepted Answer

Total binding energy (without considering repulsions),

$E_{b} = [Z m_{p} + (A - Z) m_{n} - m_{x}] c^{2}$

Where, ${}_{Z}^{A}X$ is the nuclei under consideration.

Now, considering repulsion :

Number of proton pairs $= {}^{Z}C_{2}$

$\Rightarrow$ Thus repulsion energy $\propto \frac{Z (Z - 1)}{2} \times \frac{1}{4 π ϵ_{0}} \frac{e^{2}}{R}$

Where $R$ is the radius of the nucleus

$\Rightarrow E_{b}^{p} - E_{b}^{n} \propto Z (Z - 1) ∴$ there will be no repulsion term for neutrons.

Also, since $R = R_{0} A^{\frac{1}{3}}$

$\Rightarrow E_{b}^{p} - E_{b}^{n} \propto A^{- \frac{1}{3}}$

Because of repulsion among protons,

$E_{b}^{p} < E_{b}^{n}$

Since in $β^{+}$ decay, number of protons decrease $\Rightarrow$ repulsion would decrease

$\Rightarrow E_{b}^{p}$ increases

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