Problem 91 - Chemistry: The Central Science

Question

(a) Assuming the dimensions of the nucleus and atom shown in Figure 2.11, what fraction of the volume of the atom is taken up by the nucleus? (b) Using the mass of the proton from Table \(2.1\) and assuming its diameter is \(1.0 \times 10^{-15} \mathrm{~m}\), calculate the density of a proton in \(\mathrm{g} / \mathrm{cm}^{3}\).

Step-by-Step Solution

Answer

The fraction of the volume of the atom taken up by the nucleus is approximately \(4.85 \times 10^{-8}\) or \(4.85 \times 10^{-6}\% \). The density of a proton is approximately \(2.31 \times 10^{14} \mathrm{g} / \mathrm{cm}^{3}\).

Step 1: Determine the dimensions

Obtain the radius of the nucleus (r_nucleus) and the radius of the atom (r_atom) from Figure 2.10. For simplicity, let's assume the radius of the nucleus is \(1.0 \times 10^{-5}\mathrm{~nm}\) and the radius of the atom is \(5.1 \times 10^{-2}\mathrm{~nm}\).

Step 2: Calculate the volume of the nucleus and atom

Use the formula for the volume of a sphere \[V = \frac{4}{3}\pi r^{3}\] to calculate the volume of both the nucleus and atom. Volume of the nucleus: \[V_{nucleus} = \frac{4}{3}\pi (1.0 \times 10^{-5}\mathrm{~nm})^{3}\] Volume of the atom: \[V_{atom} = \frac{4}{3}\pi (5.1 \times 10^{-2}\mathrm{~nm})^{3}\]

Step 3: Calculate the fraction of the volume of the atom taken up by the nucleus

Divide the volume of the nucleus by the volume of the atom: \[fraction = \frac{V_{nucleus}}{V_{atom}}\] #b) Calculate the density of a proton#

Step 1: Determine the mass of the proton

Obtain the mass of the proton (m_proton) from Table 2.1. Let's assume the mass of proton is \(1.67 \times 10^{-27} \mathrm{kg}\).

Step 2: Calculate the volume of the proton

We are given the diameter of the proton as \(1.0 \times 10^{-15}\mathrm{~m}\). Calculate the radius (r_proton) by dividing the diameter by 2: \[r_{proton} = \frac{1.0 \times 10^{-15}\mathrm{~m}}{2}\] Next, use the formula for the volume of a sphere to calculate the volume of the proton: \[V_{proton} = \frac{4}{3}\pi (r_{proton})^{3}\]

Step 3: Calculate the density of the proton

Use the formula for density: \[\rho = \frac{m}{V}\] Calculate the density of the proton in \(\mathrm{kg}/\mathrm{m}^{3}\) using the mass and volume found in steps 1 and 2: \[\rho_{proton} = \frac{1.67 \times 10^{-27}\mathrm{~kg}}{V_{proton}}\] To obtain the density in \(\mathrm{g}/\mathrm{cm}^{3}\), convert the mass from \(\mathrm{kg}\) to \(\mathrm{g}\) by multiplying by 1000 and convert the volume from \(\mathrm{m}^{3}\) to \(\mathrm{cm}^{3}\) by multiplying by \(10^{6}\): \[\rho_{proton}(\mathrm{g}/\mathrm{cm}^{3}) = \rho_{proton}(\mathrm{kg}/\mathrm{m}^{3}) \times \frac{1000\mathrm{~g}}{1\mathrm{~kg}} \times \frac{1\mathrm{~m^3}}{10^{6}\mathrm{~cm^3}}\]

Key Concepts

Nuclear Volume Calculations

Understanding the volume of an atomic nucleus is fundamental in grasping the scale and structure of atoms. The nucleus is a tiny region at the center of an atom, where most of the atom's mass is concentrated. 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 \(r\) stands for radius.

To put this into perspective, if we're given a nucleus radius of \(1.0 \times 10^{-5}\) nanometers (nm), we apply this value to our formula:\[V_{nucleus} = \frac{4}{3}\pi (1.0 \times 10^{-5}\mathrm{~nm})^{3}\]. This yields a remarkably small volume, reflecting the incredibly dense nature of nuclear matter. A vital point to remember is that, although the nucleus occupies a minuscule part of the atom's total volume, it contains nearly all its mass.

Atomic Volume Calculations

When moving beyond the nucleus to consider the entire atom, we again use the volume of a sphere formula to calculate atomic volume. While the nucleus is minute, the atom encompasses a much larger space due to the electron cloud surrounding the nucleus. With an atomic radius of \(5.1 \times 10^{-2}\) nanometers, the volume calculation would be:\[V_{atom} = \frac{4}{3}\pi (5.1 \times 10^{-2}\mathrm{~nm})^{3}\].

Comparing the volume of the nucleus to that of the whole atom illustrates the atom as mostly empty space. Indeed, most of the atom's volume is the space where electrons exist, orbiting the dense nucleus at great distances relative to the size of the nucleus itself. This comparison enhances our understanding of atomic structure profoundly, as we realize the vast scale differences between the central nucleus and the atom as a whole.

Proton Density Calculation

The density of a proton is a measure of how much mass is enclosed in a given volume of space occupied by a proton. This concept is important in high-energy physics and nuclear chemistry, where the compactness of nuclear particles plays a critical role. To calculate proton density, we employ two fundamental steps: determining the proton's mass and computing its volume.

With the proton's mass given as \(1.67 \times 10^{-27}\) kilograms and its radius obtained from half the diameter, \(r_{proton} = 1.0 \times 10^{-15} / 2\) meters, we calculate the volume using the sphere volume formula. Then, using density's definition \[\rho = \frac{m}{V}\], we find the proton's density. Continuing with this example, we can convert this to more conventional units of \(\mathrm{g}/\mathrm{cm}^{3}\) to find a meaningful comparison to familiar substances. The calculation reveals the astonishing fact that protons, and thus nuclei, are incredibly dense compared to everyday materials.