Calculate the density of the nucleus of 47 Ag^107 assuming rnucleus is 1.4 A^1 / 3 ×10^-13 cm

step 1 Hi everyone. So in this question, they ask, calculate the density of the nucleus of 47 silver 10

Calculate the density of the nucleus of 47 Ag^107 assuming...

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 $10.5 \mathrm{~g} \mathrm{~cm}^{-3}$.

Key Concepts

Nuclear Density Calculation

Nuclear density is obtained by dividing the mass of the nucleus by its volume. In practice, the mass is approximated using the mass number A multiplied by the mass of a nucleon. Since the volume scales with A and the mass scales linearly with A, this results in a density that is nearly constant across different nuclei, revealing the tightly packed structure of nuclear matter.

Comparison of Densities

This concept involves contrasting the extremely high density of nuclear matter with the much lower densities found in everyday materials like metals. While metallic densities are on the order of a few grams per cubic centimeter, nuclear densities are many orders of magnitude greater, underscoring the extreme compactness and energy concentration within atomic nuclei.

Volume of a Sphere

Assuming the nucleus is spherical, its volume is calculated using the formula V = (4/3)?R³. This geometrical principle is essential because it allows for converting the nuclear radius into a volume, which is a key component when determining the density of nuclear matter.

Nuclear Radius Formula

This concept involves the empirical relation R = r? A^(1/3), where R is the radius of a nucleus, A is the mass number, and r? is a constant (usually around 1.2-1.4 × 10?¹³ cm). It is based on the observation that a nucleus's volume is proportional to the number of nucleons, implying that the size of the nucleus increases roughly as the cube root of A.

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