Question

Demonstrate that if the average nucleus is assumed to be spherical with a radius (r=r₀A^{1/3}) and mass (A) u, then its density is independent of (A).

a) The density is independent of (A).
b) The density depends on (A^{1/3}).
c) The density increases with (A).
d) The density decreases with (A).

Answer 1

The density of a nucleus is proven to be independent of the mass number A by showing that while the volume of a nucleus increases with A, the mass also increases in the same proportion, leading to a constant density. This is demonstrated using the formula for the volume of a sphere and the expression for the nuclear radius in terms of A.

Step-by-step explanation:

Let's address the student's question regarding the density of a nucleus and its independence from mass number A. By assuming an average nucleus is spherical, its density can be computed using its mass and volume.

First, we express the radius r in terms of the mass number A: r = ro A1/3, where ro is a constant with a value of approximately 1.2 femtometers (fm). The volume V of a sphere is given by V = (4/3)πr3. Substituting the expression for r, we get V = (4/3)π(ro A1/3)3 = (4/3)πro3A. This shows that the volume of a nucleus is directly proportional to A. Since the mass of a nucleus is essentially A atomic mass units (u), the density ρ is mass per unit volume: ρ = m/V = A u / ((4/3)πro3A) = 3u / (4πro3). Hence, the density of a nucleus is independent of A and depends only on the constant ro.

To calculate the density for specific elements such as gold (Au) and compare it with iron (Fe), one could simply apply the formula for density given above. Both elements would yield the same density value, supporting the claim that nuclear density is indeed independent of A.