Question

Prove that the density of nuclear matter is same for all nuclei.

Hint:

Take mass = A u and volume = (4/3) pi R0^3 A using R = R0 A^(1/3). A cancels in the ratio.

Answer 1

Step 1: Understanding the Concept:
Density is mass divided by volume. We write the mass and the volume of a nucleus of mass number \(A\), and see whether \(A\) cancels.

Step 2: Mass of the nucleus:
Each nucleon has a mass of about one atomic mass unit, \(u = 1.66\times 10^{-27}\) kg. A nucleus with \(A\) nucleons has mass
\[ M \approx A u \]

Step 3: Volume of the nucleus:
Experiments show that the nuclear radius is \(R = R_0 A^{1/3}\), with \(R_0 \approx 1.2\times 10^{-15}\) m. The nucleus is nearly a sphere, so
\[ V = \frac{4}{3}\pi R^3 = \frac{4}{3}\pi R_0^3 A \]

Step 4: Density:
\[ \rho = \frac{M}{V} = \frac{A u}{\frac{4}{3}\pi R_0^3 A} = \frac{3u}{4\pi R_0^3} \]
The mass number \(A\) cancels out. The density depends only on the constants \(u\) and \(R_0\), so it is the same for all nuclei.

Step 5: Numerical value:
\[ \rho = \frac{3 \times 1.66\times 10^{-27}}{4\pi (1.2\times 10^{-15})^3} \approx 2.3\times 10^{17}\ \text{kg m}^{-3} \]
This is the same for light and heavy nuclei alike, and it is far larger than the density of ordinary matter.

Final Answer:
Both mass and volume of a nucleus are proportional to A, so A cancels and the density is the same for all nuclei. \[ \boxed{\rho = \frac{3u}{4\pi R_0^3} \approx 2.3\times 10^{17}\ \text{kg m}^{-3}} \]