Question

In an assembly of fissionable material, the larger the sur. face area per fissioning nucleus (i.e., per unit volume), the more likely is the escape of valuable neutrons. (a) What is the surface-to-volume ratio of a sphere of radius \(r_{0} ?\) (b) What is the surf ace-to-volume ratio of a cube of the same volume? (c) What is the surface-to. volume ratio of a sphere of twice the volume?

Short answer

The surface to volume ratio for: (a) a sphere of radius \(r_{0}\) is \(\frac{3}{r_{0}}\) ; (b) a cube of equivalent volume is \(\frac{6}{\sqrt[3]{\frac{4}{3}\pi r_{0}^{3}}}\) ; and (c) a sphere of twice the volume is \(\frac{3}{r_{0}\sqrt[3]{2}}\).

Step-by-step solution

Calculate Surface-to-Volume Ratio for a Sphere

For a sphere of radius \(r_{0}\), the surface area \(A\) is given by the formula \(A = 4\pi r_{0}^{2}\), and the volume \(V\) is given by the formula \(V = \frac{4}{3}\pi r_{0}^{3}\). The surface-to-volume (S/V) ratio is obtained by dividing the surface area by the volume. So, for a sphere, \(S/V = \frac{4\pi r_{0}^{2}}{\frac{4}{3}\pi r_{0}^{3}} = \frac{3}{r_{0}}\).

Calculate Surface-to-Volume Ratio for a Cube

Assuming that the cube has the same volume as the sphere, the side length of the cube \(a\) can be obtained by finding the cube root of the sphere's volume, i.e., \(a = \sqrt[3]{\frac{4}{3}\pi r_{0}^{3}}\). The surface area of the cube \(A\) is given by \(A = 6a^{2}\) and the volume \(V\) of the cube is \(a^{3}\). The surface-to-volume ratio of the cube is then given by \(S/V = \frac{6a^{2}}{a^{3}} = \frac{6}{a}\). Substituting \(a\) from above, \(S/V_{cube} = \frac{6}{\sqrt[3]{\frac{4}{3}\pi r_{0}^{3}}}\).

Calculate Surface-to-Volume Ratio for a Sphere of Twice the Volume

For a sphere of twice the volume, the new radius \(r_{1}\) can be found using the formula for the volume of a sphere, \(V = \frac{8}{3}\pi r_{1}^{3}\). Solving for \(r_{1}\), we have \(r_{1} = r_{0}\sqrt[3]{2}\). The surface-to-volume ratio for this larger sphere would then be \(S/V = \frac{3}{r_{1}} = \frac{3}{r_{0}\sqrt[3]{2}}\).

Key concepts

Nuclear Fission Material

Understanding the importance of surface-to-volume ratio in the context of nuclear fission materials is crucial to grasping the fundamentals of nuclear physics. In nuclear reactions, specifically fission, the aim is to split the nuclei of atoms to release energy. However, the efficiency of this process can be affected by how easily neutrons, which facilitate the reaction, can escape the material.

When dealing with fissionable material, such as uranium or plutonium, a high surface area relative to the volume is advantageous because it increases the likelihood of neutron emission into the surrounding environment, rather than being captured back into the material. This behavior is paramount in reactor design and understanding the geometric considerations of the fuel's shape, whether it's in the form of rods, spheres, or pellets.

Geometry in Physics

Geometry plays a vital role in various physical processes, and in the case of nuclear materials, the geometry determines how substances interact with each other. In our example, comparing different shapes and their surface-to-volume ratios illustrates this concept perfectly.

A sphere, with its symmetrical properties, offers a unique S/V ratio that is distinct compared to other shapes like a cube. These geometric comparisons not only serve as mathematical exercises but also provide valuable insights for applications in science and engineering, such as maximizing the efficiency of reactions in nuclear reactors.

Mathematical Problem-Solving

The step-by-step solutions to these problems exemplify mathematical problem-solving in physics. Firstly, by using the formulae for the surface area and volume of a sphere, we can establish the S/V ratio for a sphere of a given radius. Moreover, using cube roots and algebraic manipulation, we extend this approach to compare a cube of the same volume and a sphere with twice the volume.

These exercises not only test one's ability to perform algebraic calculations but also reinforce understanding of three-dimensional geometry and its application in the real world. An essential takeaway for students is the link between mathematical problem-solving and practical applications, specifically in scientific contexts like nuclear physics, where precise calculations lead to a greater understanding and optimization of physical properties and interactions.