Given below are two statements : one is labelled as Assertion (A) and the other is labelled as Reason (R). Assertion (A) : The density of the copper (6429Cu) nucleus is greater than that of the carbon (126C) nucleus. Reason (R) : The nucleus of mass number A has a radius proportional to A1/3. In the light of the above statements, choose the most appropriate answer from the options given below :

The correct answer is Option 4. Full mathematical solution steps and hints are accessible in the interactive StemCET workspace.

Physics Q44 – JEE-MAIN 2025

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Home Physics NucleiQuestion #374

Given below are two statements : one is labelled as Assertion (A) and the other is labelled as Reason (R). Assertion (A) : The density of the copper (

_{29}^{64} Cu

) nucleus is greater than that of the carbon (

_{6}^{12} C

) nucleus. Reason (R) : The nucleus of mass number A has a radius proportional to

A^{1 / 3}

. In the light of the above statements, choose the most appropriate answer from the options given below :

Clue 1 of 3: Nuclear Properties

Consider the fundamental properties of atomic nuclei, specifically how their size and mass relate to their density.

🧠 Full Solution Path

Step 1: Evaluate Reason (R)✦ Active

Reason (R) states that the nucleus of mass number $A$ has a radius proportional to $A^{1 / 3}$ . This is a well-established empirical relation in nuclear physics, given by $R = R_{0} A^{1 / 3}$ , where $R_{0} \approx 1.2 \times 10^{- 15}$ m (or 1.2 fm) is a constant. Therefore, Reason (R) is correct.

Step 2: Evaluate Assertion (A)○ Expand

To evaluate Assertion (A), we need to determine the nuclear density. The mass of a nucleus is approximately $M = A \times m_{p}$ , where $m_{p}$ is the average mass of a nucleon (proton or neutron). The volume of the nucleus is $V = \frac{4}{3} π R^{3}$ . Using the relation $R = R_{0} A^{1 / 3}$ from Reason (R), the volume becomes:

V = \frac{4}{3} π (R_{0} A^{1 / 3})^{3} = \frac{4}{3} π R_{0}^{3} A

Now, the nuclear density $ρ$ is:

ρ = \frac{M}{V} = \frac{A \times m_{p}}{\frac{4}{3} π R_{0}^{3} A} = \frac{m_{p}}{\frac{4}{3} π R_{0}^{3}}

This expression for nuclear density is independent of the mass number $A$ . This implies that the nuclear density is approximately constant for all nuclei, regardless of whether it's a light nucleus like carbon ( $_{6}^{12} C$ ) or a heavier one like copper ( $_{29}^{64} Cu$ ). Therefore, the density of the copper nucleus is not greater than that of the carbon nucleus; they are approximately equal. Thus, Assertion (A) is incorrect.

💡 Teacher's Secret Hint

Remember that nuclear density is remarkably constant across different nuclei, a key characteristic of nuclear matter.

Step 3: Conclusion○ Expand

Based on the analysis, Assertion (A) is incorrect, but Reason (R) is correct. This corresponds to option 4.

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Physics Question 44 – JEE-MAIN 2025

Clue 1 of 3: Nuclear Properties

🧠 Full Solution Path