Understanding Nuclear Density and Mass Number

The question asks about the ratio of densities of two nuclei with given mass numbers, A and B. To determine this ratio, we need to understand how the density of an atomic nucleus is related to its mass number.

Relationship Between Nuclear Properties and Mass Number

Atomic nuclei are composed of protons and neutrons, collectively called nucleons. The mass number (A) is the total number of nucleons in a nucleus.

• Nuclear Mass: The mass of a nucleus is approximately proportional to the total number of nucleons. If 'm' is the average mass of a nucleon, the mass of a nucleus with mass number A is approximately $M \approx m \times A$.
• Nuclear Radius: Experimental observations show that the radius (R) of a nucleus is approximately proportional to the cube root of its mass number. This relationship is given by the empirical formula: $R \approx R_0 A^{1/3}$, where $R_0$ is a constant (approximately 1.2 femtometers or $1.2 \times 10^{-15}$ meters).
• Nuclear Volume: Assuming a nucleus is spherical, its volume (V) is given by the formula for the volume of a sphere: $V = \frac{4}{3}\pi R^3$. Substituting the expression for R, we get:
  $V = \frac{4}{3}\pi (R_0 A^{1/3})^3 = \frac{4}{3}\pi R_0^3 A$

Deriving Nuclear Density

Density ($\rho$) is defined as mass per unit volume ($\rho = \frac{M}{V}$). Using the approximate expressions for nuclear mass and volume:

$\rho = \frac{m A}{\frac{4}{3}\pi R_0^3 A}$

We can see that the mass number 'A' appears in both the numerator and the denominator, so it cancels out:

$\rho = \frac{m}{\frac{4}{3}\pi R_0^3}$

This formula shows that the density of a nucleus depends on the average mass of a nucleon (m) and fundamental constants ($\frac{4}{3}\pi$ and $R_0$). It does not depend on the mass number (A).

Comparing Densities of Two Nuclei

Let's consider two nuclei, one with mass number A and the other with mass number B.

• Density of the first nucleus ($\rho_A$): $\rho_A \approx \frac{m}{\frac{4}{3}\pi R_0^3}$
• Density of the second nucleus ($\rho_B$): $\rho_B \approx \frac{m}{\frac{4}{3}\pi R_0^3}$

Since the expression for density is the same for both nuclei, regardless of their mass numbers A and B, their densities are approximately equal.

The ratio of their densities is:

$\frac{\rho_A}{\rho_B} = \frac{\frac{m}{\frac{4}{3}\pi R_0^3}}{\frac{m}{\frac{4}{3}\pi R_0^3}} = 1$

So, the ratio $\rho_A : \rho_B$ is $1 : 1$.

This indicates that nuclear matter is extremely dense and its density is nearly constant across all nuclei, irrespective of their size (mass number).

Summary of Density Calculation

Since the density is independent of the mass number, the ratio of densities for any two nuclei is always 1:1.

Revision Table: Key Nuclear Concepts

Additional Information: Significance of Constant Nuclear Density

The fact that nuclear density is roughly constant ($\approx 2.3 \times 10^{17}$ kg/m$^3$) is a significant property of nuclear matter. It suggests that nucleons inside the nucleus are packed together in a way that the addition of more nucleons (increasing A) primarily increases the volume proportionally, maintaining a constant average density.

• This constant density is much higher than the density of normal matter, even the densest elements.
• This property is consistent with the idea that nuclear forces have a very short range and exhibit saturation, meaning each nucleon interacts strongly only with its immediate neighbors.
• Comparing densities: Water $\approx 10^3$ kg/m$^3$, Iron $\approx 7.8 \times 10^3$ kg/m$^3$, Nuclear matter $\approx 2.3 \times 10^{17}$ kg/m$^3$. Nuclear matter is about $10^{14}$ times denser than water!
• This extreme density is observed in neutron stars, which are essentially giant nuclei held together by gravity.

Therefore, for any two nuclei, regardless of their mass numbers A and B, their densities are approximately the same, leading to a density ratio of 1:1.