Question

How does the size of a nucleus depend on its mass number A ? Hence prove that the density of nucleus is a constant, independent of A, for all nuclei.

Hint:

The fact that nuclear density remains staggeringly constant strongly implies that nucleons act like incompressible fluid drops; adding more nucleons purely adds more volume without squishing the existing ones tighter.

Answer 1

Concept:
• Extensive scattering experiments have historically revealed that atomic nuclei are approximately spherical physical objects.
• The physical volume of any specific nucleus is directly and strictly proportional to its total contained number of nucleons (which is its mass number $A$).
• Consequently, the geometric radius $R$ aggressively scales mathematically with the cube root of the given mass number $A$.

Step 1: State the dependency of nuclear size on mass number
Based securely on experimental observations, the physical radius $R$ of any spherical nucleus is mathematically related to its mass number $A$ strictly by the established empirical formula:
\[ R = R_0 A^{1/3} \]
where $R_0$ is a universal empirical constant roughly equal to $1.2 \times 10^{-15} \text{ m}$ (or $1.2 \text{ fm}$). This directly demonstrates that size (radius) grows precisely as the cube root of the atomic mass number.

Step 2: Formulate the physical Volume and Mass of the nucleus
Assuming the stable nucleus is a perfectly rigid sphere, its physical spatial volume $V$ is mathematically calculated as:
\[ V = \frac{4}{3} \pi R^3 \]
Substitute our previously defined radius dependency explicitly into this volume equation:
\[ V = \frac{4}{3} \pi (R_0 A^{1/3})^3 \]
\[ V = \frac{4}{3} \pi R_0^3 A \]
This vividly proves that nuclear volume $V$ is strictly and linearly proportional to $A$.
Next, the total physical mass $M$ of the entire nucleus is approximately cleanly equal to the total mass number $A$ explicitly multiplied by the average baseline mass of a single isolated nucleon ($m \approx 1.66 \times 10^{-27} \text{ kg}$):
\[ M \approx A \cdot m \]

Step 3: Derive the density expression to prove constancy
Nuclear density $\rho$ is definitively calculated by strictly dividing the total nuclear mass by its total spatial volume:
\[ \rho = \frac{\text{Mass}}{\text{Volume}} = \frac{M}{V} \]
Substitute the robust expressions meticulously established in Step 2 securely into the density formula:
\[ \rho = \frac{A \cdot m}{\frac{4}{3} \pi R_0^3 A} \]
The mass number variable $A$ wonderfully and completely cancels out from both the numerator and the massive denominator:
\[ \rho = \frac{m}{\frac{4}{3} \pi R_0^3} \]
\[ \rho = \frac{3m}{4 \pi R_0^3} \]

Step 4: Conclusion
In this final, rigorously derived expression, every single remaining term ($m$, $\pi$, $R_0$) is a completely fixed universal constant. Because the variable $A$ has entirely vanished, it explicitly proves that the physical density of nuclear matter is a universal constant (roughly $2.3 \times 10^{17} \text{ kg/m}^3$), completely independent of the mass number $A$ for all atoms.