Verified Solution • 2025–26 Session moderate

QJEEAdv22Ph4e5b: JEE Advanced 2022 Physics - Class 12 JEEAdv22Ph4e5b Solution: The binding energy of nucleons in a nucleus can be affected by…

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 EbpE_{b}^{p} and the binding energy of a neutron be EbnE_{b}^{n} in the nucleus.

Which of the following statement(s) is(are) correct?

Multiple Choice Options

  • A Ebp−EbnE_{b}^{p}-E_{b}^{n} is proportional to Z(Z−1)Z(Z-1) where ZZ is the atomic number of the nucleus.
  • B Ebp−EbnE_{b}^{p}-E_{b}^{n} is proportional to A−13A^{-\frac{1}{3}} where AA is the mass number of the nucleus.
  • C Ebp−EbnE_{b}^{p}-E_{b}^{n} is positive.
  • D EbpE_{b}^{p} increases if the nucleus undergoes a beta decay emitting a positron.

Step-by-Step Verified Solution

Total binding energy (without considering repulsions),

Eb=[Zmp+(A−Z)mn−mx]c2{E_b} = [Z{m_p} + (A - Z){m_n} - {m_x}]{c^2}

Where, ZAX_Z^AX is the nuclei under consideration.

Now, considering repulsion :

Number of proton pairs =zC2= {}^z{C_2}

⇒\Rightarrow This repulsion energy ∝ Z(Z−1)2×14π∈0e2R\propto \,{{Z(Z - 1)} \over 2} \times {1 \over {4\pi { \in _0}}}{{{e^2}} \over R}

Where R is the radius of the nucleus

⇒Ebp−Ebn∝Z(Z−1)\Rightarrow E_b^p - E_b^n \propto Z(Z - 1) ∴\therefore there will be no repulsion term for neutrons.

Also, since R=R0A1/3R = {R_0}{A^{1/3}}

⇒Ebp−Ebn∝A−1/3\Rightarrow E_b^p - E_b^n \propto {A^{ - 1/3}}

Because of repulsion among protons,

E_b^p < E_b^n

Since in β+\beta^+ decay, number of protons decrease ⇒\Rightarrow repulsion would decrease

⇒Ebp\Rightarrow E_b^p increases

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