JEE Advanced 2022 · Paper 1 · Physics Q14

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PhysicsMultiple correct+4 / -2Question 32 of 54

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?

E_b^p - E_b^n

is proportional to

Z(Z-1)

where

Z

is the atomic number of the nucleus.

E_b^p - E_b^n

is proportional to

A^{-1/3}

where

A

is the mass number of the nucleus.

E_b^p - E_b^n

is positive.

E_b^p

increases if the nucleus undergoes a beta decay emitting a positron.

Answer and worked solution

Official answer: A, B, D

Coulomb term. The Coulomb energy of a uniformly charged nucleus is
$E_C\propto\frac{Z(Z-1)}{R},\qquad R\propto A^{1/3}.$
Only protons feel it, and it reduces their binding:
$E_b^p-E_b^n\propto-\frac{Z(Z-1)}{A^{1/3}}.$

Options.

(A) and (B) are true: the difference is proportional to $Z(Z-1)$ and to $A^{-1/3}$ .
(C) is false: the difference is negative.
(D) is true: positron emission lowers $Z$ , which reduces the Coulomb repulsion, so $E_b^p$ increases.

Topics:Mass defect and binding energy Nuclear composition and size Radioactive decay

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JEE Advanced 2022 · Paper 1, Physics question 14