JEE Advanced2022PhysicsNuclear PhysicsActual
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?
Options
- AE b p - E b n is proportional to Z Z - 1 where Z is the atomic number of the nucleus.
- BE b p - E b n is proportional to A - 1 3 where A is the mass number of the nucleus.
- CE b p - E b n is positive.
- DE b p increases if the nucleus undergoes a beta decay emitting a positron.
Correct answer
A. E b p - E b n is proportional to Z Z - 1 where Z is the atomic number of the nucleus.
Step-by-step solution
Total binding energy (without considering repulsions), E b = Z m p + A - Z m n - m x c 2 Where, X Z A is the nuclei under consideration. Now, considering repulsion : Number of proton pairs = C 2 Z ⇒ Thus repulsion energy ∝ Z Z - 1 2 × 1 4 π ϵ 0 e 2 R Where R is the radius of the nucleus ⇒ E b p - E b n ∝ Z Z - 1    ∴ there will be no repulsion term for neutrons. Also, since R = R 0 A 1 3 ⇒   E b p - E b n ∝ A - 1 3 Because of repulsion among p