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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.

Solution :

Step 1: Theoretical Derivation of Nuclear Density

To demonstrate that nuclear density is constant, we define the following variables:

  • Nuclear radius: $r = r_0 A^{1/3}$
  • Mass number: $A$
  • Nuclear mass: $M_{nuc} \approx A \cdot m_p$, where $m_p \approx 1.67 \times 10^{-27} \, \text{kg}$ [Assuming mass of nucleon $\approx$ mass of proton].
  • Nuclear volume: $V = \frac{4}{3} \pi r^3$

Substituting the expression for $r$ into the volume formula:

$V = \frac{4}{3} \pi (r_0 A^{1/3})^3 = \frac{4}{3} \pi r_0^3 A$ [By substitution of radius empirical relation]

Now, define the nuclear density ($\rho_{nuc}$):

$\rho_{nuc} = \frac{\text{Mass}}{\text{Volume}} = \frac{A \cdot m_p}{\frac{4}{3} \pi r_0^3 A} = \frac{3 m_p}{4 \pi r_0^3}$

Since $m_p$, $\pi$, and $r_0$ are constants, $\rho_{nuc}$ is independent of the mass number $A$. Thus, nuclear density is constant for all nuclei.

Step 2: Estimation of Sodium Nucleus Density

Given values:

  • $r_0 = 1.2 \, \text{f} = 1.2 \times 10^{-15} \, \text{m}$
  • $m_p \approx 1.67 \times 10^{-27} \, \text{kg}$

Calculating the density:

$\rho_{nuc} = \frac{3 \times 1.67 \times 10^{-27}}{4 \times 3.14159 \times (1.2 \times 10^{-15})^3}$ $\rho_{nuc} = \frac{5.01 \times 10^{-27}}{4 \times 3.14159 \times 1.728 \times 10^{-45}}$ $\rho_{nuc} = \frac{5.01 \times 10^{-27}}{2.171 \times 10^{-44}} \approx 2.3 \times 10^{17} \, \text{kg/m}^3$

Step 3: Comparison with Sodium Atom Density

From Exercise 2.27, the radius of a sodium atom is approximately $R_a \approx 1.9 \times 10^{-10} \, \text{m}$ and the mass is $M_a \approx 3.8 \times 10^{-26} \, \text{kg}$.

Density of sodium atom ($\rho_{atom}$):

$\rho_{atom} = \frac{M_a}{\frac{4}{3} \pi R_a^3} = \frac{3.8 \times 10^{-26}}{\frac{4}{3} \times 3.14159 \times (1.9 \times 10^{-10})^3}$ $\rho_{atom} = \frac{3.8 \times 10^{-26}}{2.87 \times 10^{-27}} \approx 1.3 \times 10^3 \, \text{kg/m}^3$
System Density ($\text{kg/m}^3$)
Sodium Nucleus $\approx 2.3 \times 10^{17}$
Sodium Atom $\approx 1.3 \times 10^{3}$

Geometric Representation of Relative Scales

Nucleus (Radius $\approx 10^{-15}$m) Atom (Radius $\approx 10^{-10}$m) r

Final Solution: The derivation shows that nuclear density is constant because $A$ cancels out in the ratio of mass to volume. The estimated density of the sodium nucleus is $2.3 \times 10^{17} \, \text{kg/m}^3$, which is approximately $10^{14}$ times greater than the density of the sodium atom, illustrating that the mass of an atom is concentrated almost entirely in its nucleus while the volume is dominated by electron orbits.


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