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 $(^ {64}_{29} \text{Cu})$ nucleus is greater than that of the carbon $(^ {12}_{6} \text{C})$ 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 :

<p>To evaluate the assertion and reason provided, we need to consider the formula for the density of a nucleus. The density $ \rho $ can be expressed as follows:</p> <p>$ \rho = \frac{M}{V} = \frac{m_n \times A}{\frac{4}{3} \pi R^3} $</p> <p>Here, $ m_n $ represents the nucleon mass, $ A $ is the mass number, and $ R $ is the radius of the nucleus. According to the formula for the radius:</p> <p>$ R = R_0 A^{1/3} $</p> <p>This indicates that the radius $ R $ is proportional to $ A^{1/3} $. By substituting the expression for $ R $ in terms of $ A $ back into the formula for density, we get:</p> <p>$ \rho = \frac{m_n \times A}{\frac{4}{3} \pi (A^{1/3} R_0)^3} $</p> <p>Simplifying the equation shows that the $ A $ terms cancel out:</p> <p>$ \rho = \frac{m_n}{\frac{4}{3} \pi R_0^3} $</p> <p>This reveals that the nuclear density $ \rho $ is approximately constant and independent of the mass number $ A $. Therefore, the densities of copper $(^{64}_{29} \text{Cu})$ and carbon $(^{12}_{6} \text{C})$ nuclei are effectively the same because the density formula results in a similar value regardless of the specific mass number.</p>