Question

(a) Assume nuclei are spherical in shape, show that its radius \((r)\) is proportional to the cube root of mass number \((A) .\) (b) In general, the radius of a nucleus is given by \(r=r_{0} A^{\frac{1}{3}},\) where \(r_{0},\) the proportionality constant, is given by \(1.2 \times 10^{-15} \mathrm{~m}\). Calculate the volume of the \({ }^{238} \mathrm{U}\) nucleus.

Short answer

The volume of a Uranium-238 nucleus is obtained by substituting the calculated radius into the formula for the volume of a sphere.

Step-by-step solution

Proving Proportionality

Let's assume the nucleus is a sphere. The mass of a sphere is given by \(M = \rho \frac{4}{3}\pi r^{3}\), where \(\rho\) is the density and \(r\) is the radius. Atomic mass is evenly distributed, hence density will remain constant. Therefore, we can say that \(M \propto r^{3}\). Taking cube root on both sides, it implies \(r \propto A^{\frac{1}{3}}\), which proves our assumption.

Given Formula for Radius

In the next part we are given that the radius of a nucleus is represented by \(r=r_{0} A^{\frac{1}{3}}\), where \(A\) is the mass number and \(r_{0}\) is a constant equal to \(1.2 \times 10^{-15} \mathrm{~m}\). In our example \(A\) is equal to \(238\) for the Uranium-238 nucleus.

Calculate Radius of Uranium-238 Nucleus

We plug in the value of \(A\) and \(r_{0}\) into our formula from Step 2. Hence, \(r = 1.2 \times 10^{-15} \times 238^{\frac{1}{3}}\). This gives us the radius of the Uranium-238 nucleus.

Calculate Volume of Uranium-238 Nucleus

The volume of a sphere is given by the formula \(V = \frac{4}{3}\pi r^{3}\). Using the radius calculated in step 3, we substitute the value into this volume formula to receive the volume of the Uranium-238 nucleus.

Key concepts

Mass Number

Understanding the concept of mass number is crucial when we talk about atomic structures and nuclear physics. The mass number, denoted as \(A\), refers to the total number of protons and neutrons in an atom's nucleus. Since protons and neutrons make up nearly all the mass of an atom, the mass number essentially tells us the 'heft' of an atom.

Why is this important? Well, knowing the mass number is critical for identifying isotopes of the same element, which have the same number of protons (therefore the same place in the periodic table) but different numbers of neutrons. This has implications in fields ranging from chemistry to astrophysics. For example, Carbon-12 and Carbon-14 are both carbon isotopes, but Carbon-14 has two more neutrons, giving it a higher mass number and different properties.

Nucleus Volume Calculation

When we picture the nucleus of an atom, it helps to visualize it as a tiny, incredibly dense sphere nestled at the center of an atom. To calculate the volume of this sphere, we must first find its radius. The formula for the volume \(V\) is given by \(V = \frac{4}{3}\pi r^{3}\), where \(r\) is the radius of the nucleus.

This might seem straightforward, but there's a catch - how do we find the nucleus's radius? This is where the proportionality constant \(r_0\), known as the nuclear radius constant, comes into play. By using the formula \(r = r_0 A^{\frac{1}{3}}\) - with \(A\) being the mass number and \(r_0\) typically given as \(1.2 \times 10^{-15} \mathrm{~m}\) - we can calculate the radius for any nucleus, which then allows us to calculate the volume.

Proportionality of Radius and Mass Number

The size of the nucleus (radius) is not directly proportional to the mass number (\(A\)), because if it were, larger nuclei would have much greater volumes than they actually do. Instead, the relationship between the radius of a nucleus and its mass number is a cube root one, signified by the equation \(r = r_0 A^{\frac{1}{3}}\). What this means is that as the mass number increases, the radius increases as well, but at a much slower rate, as it’s related to the cube root of the mass number.

This cube root relationship reveals an important aspect of nuclear density: it tends to remain relatively constant across different nuclei, since as the mass number increases, the volume increases to a lesser degree than if it were a direct proportion. This interplay largely influences nuclear stability and the properties of different isotopes, vital concepts in nuclear chemistry and physics.