Calculate the density of the nucleus of \(_{47} \mathrm{Ag}^{107}\) assuming \(r_{\text {nucleus }}\) is \(1.4 A^{1 / 3} \times 10^{-13} \mathrm{~cm} .\) Where \(A\) is mass number of nucleus. Compare its density with density of metallic silver \(10.5 \mathrm{~g} \mathrm{~cm}^{-3}\).

Understanding the volume of an atomic nucleus is crucial for numerous calculations in nuclear physics. Given that nucleons (protons and neutrons) are packed into a sphere, the volume of the nucleus can be calculated using the formula for the volume of a sphere, which is \( V = \frac{4}{3} \pi r^3 \), where \( r \) represents the radius of the sphere.

To calculate the volume of a nucleus, we apply the aforementioned formula with a specific way to find the nuclear radius. The radius is proportional to the cube root of the atomic mass number \( A \) and can be expressed as \( r_{\text{nucleus}} = 1.4 A^{1/3} \times 10^{-13} \text{ cm} \). The mass number for an isotope of silver, \( _{47}\mathrm{Ag}^{107} \), is 107, so this value is used to calculate the radius first, and then to find the volume. The compactness of the nucleus is reflected in the small volume we find with this calculation, emphasizing the density of matter at the nucleon level.

The mass number, denoted as \( A \), is the sum of protons and neutrons in an atomic nucleus. It's a whole number that gives an estimate of the atomic mass and plays a significant role in nuclear calculations. Given that the mass of the nucleus is approximately equal to its mass number times the mass of one nucleon, we can determine the mass of a nucleus if we know its mass number and the approximate mass of a nucleon.

In the case of \( _{47}\mathrm{Ag}^{107} \), the mass number is 107, which means there are 107 protons and neutrons combined within the nucleus. Since each nucleon has a mass of roughly \( 1.67 \times 10^{-24} \text{ g} \), we multiply this value by the mass number to get the total nuclear mass. This approach simplifies calculations and allows for coherent comparison between the actual mass and the theoretically derived mass from the number of nucleons.

To understand the concept of nuclear density, it’s helpful to compare it with a substance we can relate to in our everyday life. Calculating the density of a nucleus involves dividing its mass (obtained from the mass number and mass of a nucleon) by its calculated volume. This yields an extremely high value because the mass is concentrated in a minuscule space.

Comparatively, the density of metallic silver is \( 10.5 \text{ g cm}^{-3} \), which is much lower than the density of an atomic nucleus. For example, once we calculate the density of the silver nucleus, we can contrast it with the density of metallic silver to see just how dense nuclear matter is. This comparison highlights the stark density differences between nuclear matter and the same elements in bulk form. Students often find it astonishing that the dense core of an atom could differ so greatly from the element's density in tangible objects like silverware, emphasizing the unique properties of nuclear physics.