Q2.28:

The unit of length convenient on the nuclear scale is a fermi : 1 f = $10^{–15}$ m. Nuclear sizes obey roughly the following empirical relation : $r = r_0 A^{1/3}$ where $r$ is the radius of the nucleus, $A$ its mass number, and $r_o$ is a constant equal to about, 1.2 f. Show that the rule implies that nuclear mass density is nearly constant for different nuclei. Estimate the mass density of sodium nucleus. Compare it with the average mass density of a sodium atom obtained in Exercise. 2.27.

Step 1: Theoretical Proof of Constant Nuclear Density

To prove that the nuclear mass density $\rho_n$ is constant regardless of the mass number $A$, we define the mass and volume of a nucleus:

• Let $m_p$ be the average mass of a nucleon (proton or neutron), $m_p \approx 1.67 \times 10^{-27}\text{ kg}$.
• Total nuclear mass $M = A \cdot m_p$. [Assumption: Neglecting binding energy mass defect for density estimation].
• Nuclear radius $r = r_0 A^{1/3}$.
• Nuclear volume $V = \frac{4}{3}\pi r^3 = \frac{4}{3}\pi (r_0 A^{1/3})^3 = \frac{4}{3}\pi r_0^3 A$.

Now, calculate density $\rho_n$:

$\rho_n = \frac{M}{V} = \frac{A \cdot m_p}{\frac{4}{3}\pi r_0^3 A}$

Since the term $A$ cancels out in the numerator and denominator, we are left with:

$\rho_n = \frac{m_p}{\frac{4}{3}\pi r_0^3}$ [Result: Density is independent of $A$].

Step 2: Calculation of Nuclear Density for Sodium ($A=23$)

Given $r_0 = 1.2 \text{ f} = 1.2 \times 10^{-15} \text{ m}$:

$\rho_n = \frac{1.67 \times 10^{-27} \text{ kg}}{\frac{4}{3} \cdot 3.14 \cdot (1.2 \times 10^{-15} \text{ m})^3}$

$\rho_n = \frac{1.67 \times 10^{-27}}{4.188 \cdot 1.728 \times 10^{-45}} \approx \frac{1.67 \times 10^{-27}}{7.239 \times 10^{-45}}$

$\rho_n \approx 2.3 \times 10^{17} \text{ kg/m}^3$

Step 3: Comparative Analysis of Nuclear vs. Atomic Density

From the previously referenced Exercise 2.27, the atomic radius of Sodium is $R_a \approx 1.9 \text{ \AA} = 1.9 \times 10^{-10} \text{ m}$. The mass of a sodium atom $M_a \approx 23 \times 1.67 \times 10^{-27} \text{ kg} = 3.84 \times 10^{-26} \text{ kg}$.

$\rho_a = \frac{M_a}{\frac{4}{3}\pi R_a^3} = \frac{3.84 \times 10^{-26}}{\frac{4}{3} \cdot 3.14 \cdot (1.9 \times 10^{-10})^3} \approx 2.24 \times 10^3 \text{ kg/m}^3$

Step 4: Summary Table of Findings

The ratio $\frac{\rho_n}{\rho_a} \approx 10^{14}$, demonstrating that nuclear matter is significantly more dense than atomic matter due to the vast empty space between the nucleus and orbital electrons.

Final Solution: The nuclear mass density is approximately $2.3 \times 10^{17} \text{ kg/m}^3$, which is roughly $10^{14}$ times greater than the average atomic mass density of $2.3 \times 10^3 \text{ kg/m}^3$.