Chapter 1: Problem 11
In the fall of 2002, scientists at Los Alamos National Laboratory determined that the critical mass of neptunium-237 is about 60 kg. The critical mass of a fissionable material is the minimum amount that must be brought together to start a nuclear chain reaction. Neptunium-237 has a density of 19.5 g/cm\(^3\). What would be the radius of a sphere of this material that has a critical mass?
Short Answer
Step by step solution
Convert Mass to Grams
Calculate Volume using Density Formula
Use Volume of a Sphere Formula
Solve for the Radius
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Nuclear Chain Reaction
These additional neutrons can then induce fission in nearby nuclei, propagating the reaction. For a chain reaction to be sustained, a critical mass of the fissionable material is needed. This is the minimum amount required so that the neutrons produced can effectively cause further fissions without escaping the system or being absorbed without causing fission.
Key aspects to consider in a nuclear chain reaction include:
- The type of fissionable material used (e.g., uranium-235, plutonium-239, neptunium-237).
- The presence of a moderator to slow down neutrons, enhancing their probability of causing additional fissions.
- The geometry and arrangement of the material to ensure sufficient neutron capture.
Density Formula
\[ \rho = \frac{m}{V} \]
This formula helps determine how much space a material's mass occupies. It's especially useful when calculating the volume required to achieve a specific mass when the density is known.
In the context of the exercise, the density of neptunium-237 was crucial for determining the volume of the sphere at critical mass. By rearranging the formula, we find the volume (\( V \)) when the mass (\( m \)) and density (\( \rho \)) are given:
\[ V = \frac{m}{\rho} \]
This calculation provides the sphere's volume through which the critical mass of the material can be achieved.
Volume of a Sphere
\[ V = \frac{4}{3} \pi r^3 \]
This equation calculates the three-dimensional space enclosed within the sphere. Understanding this concept allows us to determine the required radius when the volume is known, as in this exercise.
After calculating the volume using the density formula, we can rearrange the sphere volume formula to find the radius:
\[ r^3 = \frac{3V}{4\pi} \]
Finally, by taking the cube root of both sides, we arrive at the radius. This step is critical for determining the physical dimensions necessary to achieve a particular volume. The formula is applicable whenever one needs to find the sphere's radius using a given volume.
Fissionable Material
Examples of fissionable materials include uranium-235, plutonium-239, and neptunium-237. In the exercise, neptunium-237's capability to reach a critical mass and maintain a nuclear chain reaction was analyzed. Key properties of fissionable materials include:
- Their ability to absorb neutrons and undergo fission.
- The critical mass, which is the smallest amount needed to maintain a chain reaction.
- The energy released during the fission process, contributing to nuclear power or weaponry.