JEE Advanced 2022 — Paper 1 — PHY-II solution

JEE Advanced2022Paper 1PHY-II

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?

A
$E_{b}^{p} - E_{b}^{n}$ is proportional to $Z (Z - 1)$ where $Z$ is the atomic number of the nucleus.
B
$E_{b}^{p} - E_{b}^{n}$ is proportional to $A^{- 1/3}$ where $A$ is the mass number of the nucleus.
C
$E_{b}^{p} - E_{b}^{n}$ is positive.
D
$E_{b}^{p}$ increases if the nucleus undergoes a beta decay emitting a positron.

Solution

Neutrons carry no charge, so Coulomb repulsion only affects proton binding. The total Coulomb energy from pairwise proton repulsion scales as the number of proton pairs $(2 Z) = \frac{Z ( Z - 1 )}{2}$ , divided by the characteristic separation $\sim R$ .

$E_{Coul} \propto \frac{Z ( Z - 1 )}{R} \cdot \frac{e ^{2}}{4 π ε _{0}}$ . This Coulomb repulsion reduces proton binding energy compared with neutron binding energy.

(A) $E_{b}^{p} - E_{b}^{n}$ scales with this Coulomb term: proportional to $Z (Z - 1)$ . Correct.

(B) Nuclear radius $R = R_{0} A^{1/3}$ , so $1/ R \propto A^{- 1/3}$ . Correct.

(C) Because Coulomb repulsion weakens proton binding, $E_{b}^{p} < E_{b}^{n}$ , so $E_{b}^{p} - E_{b}^{n}$ is negative. Wrong.

(D) In $β^{+}$ decay a proton converts to a neutron, decreasing $Z$ and reducing the total Coulomb repulsion. With less repulsion per proton, $E_{b}^{p}$ increases. Correct.

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