JEE Advanced · 2022 · 28 Aug 2022 · Paper 1 · Modern Physics

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

Answer: (A), (B), (D)

Solution

A proton in the nucleus feels Coulomb repulsion from the other protons, which raises its energy and so lowers its binding energy relative to a neutron; thus

E_{b}^{p} - E_{b}^{n}

is negative, not positive. The magnitude of this difference scales with the number of proton pairs, i.e. with

Z (Z - 1)

, and inversely with the nuclear radius

R = R_{0} A^{1/3}

, i.e. with

A^{- 1/3}

. In positron (

β^{+}

) emission a proton becomes a neutron, so

Z

decreases, the repulsion on the remaining protons eases, and

E_{b}^{p}

increases. Hence the

Z (Z - 1)

dependence, the

A^{- 1/3}

dependence, and the increase of

E_{b}^{p}

β^{+}

decay all hold, while the statement that the difference is positive does not.

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JEE Advanced · 2022 · 28 Aug 2022 · Paper 1 · Modern Physics

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