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?

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?

Quantrex Academy · Accepted Answer

Correct answer: A. E _ b ^ p - E _ b ^ n is proportional to Z(Z - 1) where Z is the atomic number of the nucleus.

Total binding energy (without considering repulsions),

${E}_{b} = \left[Z{m}_{p} + \left(A - Z\right){m}_{n} - {m}_{x}\right]{c}^{2}$

Where, $X ZA$ is the nuclei under consideration.

Now, considering repulsion :

Number of proton pairs $= C2 Z$

$⇒$ Thus repulsion energy $∝ \dfrac{Z\left(Z - 1\right)}{2} \times \dfrac{1}{4\pi{ϵ}_{0}}\dfrac{{e}^{2}}{R}$

Where $R$ is the radius of the nucleus

$⇒ {E}_{b}^{p} - {E}_{b}^{n} ∝ Z\left(Z - 1\right) ∴$ there will be no repulsion term for neutrons.

Also, since $R = {R}_{0}A\dfrac{1}{3}$

$⇒ {E}_{b}^{p} - {E}_{b}^{n} ∝ {A}^{- \dfrac{1}{3}}$

Because of repulsion among protons,

${E}_{b}^{p} < {E}_{b}^{n}$

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

$⇒ {E}_{b}^{p}$ increases

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