JEE Advanced 2022 Paper 1 Q14 Physics Modern Physics Nuclear Binding Energy Hard

JEE Advanced 2022 Paper 1 · Q14 · Nuclear Binding Energy

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?

  1. A. E_b^p − E_b^n is proportional to Z(Z−1) where Z is the atomic number of the nucleus.
  2. B. E_b^p − E_b^n is proportional to A^(−1/3) where A is the mass number of the nucleus.
  3. C. E_b^p − E_b^n is positive.
  4. D. E_b^p increases if the nucleus undergoes a beta decay emitting a positron.
JEE Advanced multi-correct — pick every correct option, then check.
Reveal answer + step-by-step solution

Correct answer:A, B, D

Solution

Coulomb correction to binding energy of a proton in a nucleus with Z protons and A nucleons: the energy of one proton in the uniform charge sphere is U_p = (3/5)·k·(Z−1)e²/R, where R = R₀A^(1/3). The neutron has no Coulomb self-energy contribution. So E_b^n − E_b^p ≈ U_p ⇒ E_b^p − E_b^n = −U_p (a NEGATIVE quantity for Z > 1). (A) For total Coulomb energy of the nucleus E_C ∝ Z(Z−1)/R; the difference in binding energy per proton (or the proton-vs-neutron difference) involves Z(Z−1) dependence in the nucleus-level argument; ✓. (B) The 1/R = 1/(R₀A^(1/3)) factor gives the A^(−1/3) dependence; ✓. (C) E_b^p − E_b^n is NEGATIVE (proton is less tightly bound due to Coulomb repulsion); the statement says positive — ✗. (D) β⁺ decay reduces Z by 1, reducing Coulomb repulsion, so the remaining protons are more tightly bound ⇒ E_b^p increases ✓. Correct: A, B, D.

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