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/cm3. What would be the radius of a sphere of this material that has a critical mass?

Verified step by step guidance

1. Calculate the volume of neptunium-237 required for the critical mass. Use the formula for mass and density, \( V = \frac{m}{\rho} \), where \( m = 60 \, \text{kg} \) (convert this mass into grams) and \( \rho = 19.5 \, \text{g/cm}^3 \).
2. Convert the mass from kilograms to grams by multiplying the mass in kilograms by 1000, since 1 kg = 1000 g.
3. Substitute the values into the volume formula to find the volume in cubic centimeters.
4. Calculate the radius of the sphere using the volume of a sphere formula, \( V = \frac{4}{3} \pi r^3 \). Solve for \( r \) (the radius).
5. Take the cube root of the value obtained in the previous step to find the radius of the sphere in centimeters.

Key Concepts

Critical Mass

Critical mass is the minimum amount of fissile material needed to maintain a nuclear chain reaction. When a sufficient quantity of material is present, the neutrons released from fission events can induce further fission in nearby nuclei, leading to a self-sustaining reaction. Understanding critical mass is essential for nuclear physics and safety in nuclear reactions.

Density

Density is defined as mass per unit volume and is a crucial property in determining how much material can fit into a given space. In this context, the density of neptunium-237 (19.5 g/cm³) allows us to calculate the volume of the material needed to achieve the critical mass. This relationship between mass, volume, and density is fundamental in physics and engineering.

Volume of a Sphere

The volume of a sphere is calculated using the formula V = (4/3)πr³, where r is the radius. This formula is essential for determining the size of a sphere that would contain the critical mass of neptunium-237. By rearranging this formula, one can find the radius when the volume is known, which is necessary for solving the given problem.