Problem 11 - University Physics with Modern Physics

Question

Neptunium. In the fall of \(2002,\) a group of scientists at Los Alamos National Laboratory determined the critical mass of neptunium- 237 is about 60 \(\mathrm{kg}\) . The critical mass of a fissionable material is the minimum amount that must be brought together to start a chain reaction. This element has a density of 19.5 \(\mathrm{g} / \mathrm{cm}^{3}\) . What would be the radius of a sphere of this material that has a critical mass?

Step-by-Step Solution

Answer

The radius is approximately 9.01 cm.

Step 1: Convert Mass from Kilograms to Grams

The critical mass is provided in kilograms. First, we need to convert this into grams because density is in grams per cubic centimeter. Since 1 kg equals 1000 grams, multiply 60 kg by 1000.\[ 60 \text{ kg} \times 1000 \text{ grams/kg} = 60000 \text{ grams} \]

Step 2: Volume Calculation Using Density Formula

We can calculate the volume of the neptunium required using the formula:\[ \text{Density} = \frac{\text{Mass}}{\text{Volume}} \implies \text{Volume} = \frac{\text{Mass}}{\text{Density}} \]Substitute the known values:\[ \text{Volume} = \frac{60000 \text{ grams}}{19.5 \text{ grams/cm}^3} \approx 3077 \text{ cm}^3 \]

Step 3: Determine the Radius from the Volume

Assume the mass is in the form of a sphere. The formula for the volume of a sphere is:\[ V = \frac{4}{3} \pi r^3 \]Set this equal to the volume calculated in Step 2:\[ \frac{4}{3} \pi r^3 = 3077 \]Solve for \(r^3\) by isolating it:\[ r^3 = \frac{3077 \times 3}{4 \pi} \approx 733.23 \]Now, take the cube root to find \(r\):\[ r \approx \, \sqrt[3]{733.23} \approx 9.01 \text{ cm} \]

Key Concepts

Critical Mass

Critical mass is a foundational concept in nuclear physics and chemistry. It refers to the minimum quantity of a fissionable material necessary to maintain a self-sustaining nuclear chain reaction.
For neptunium-237, as stated, the critical mass is approximately 60 kg. Achieving this mass in a controlled setting allows scientists to harness the energy produced for various applications, including nuclear reactors and research.
When scientists discuss critical mass, they focus on creating the conditions necessary for a continuous sequence of reactions.

• The material's properties, such as density and isotopic composition, impact the critical mass.
• Shape and arrangement, especially in spherical configurations, can influence it significantly.
• Environmental factors, such as the presence of neutron reflectors, may also alter what is considered critical mass.

Understanding these facets is crucial for ensuring safe and efficient use or stabilization of fissionable materials like neptunium.

Density Calculation

Density is a measure of mass per unit volume, typically expressed in grams per cubic centimeter (g/cm³) for solid materials.
In the exercise, the density of neptunium is 19.5 g/cm³. This parameter is crucial when calculating how much volume a specific mass of a material will occupy.
To find the volume, we use the formula:
\[ \text{Volume} = \frac{\text{Mass}}{\text{Density}} \]Substituting the critical mass of 60000 grams and the density, the calculation gives approximately 3077 cm³.

• This step highlights the interrelation of mass, volume, and density, crucial in various scientific domains.
• Such calculations are vital for creating models for the storage, transport, and application of nuclear materials like neptunium.

Having these precise measurements ensures that materials used in sensitive applications are managed safely and effectively.

Volume of a Sphere

The volume of a sphere is an essential concept in geometry, often applied across scientific disciplines, including physics and engineering.
It is calculated using the formula:
\[ V = \frac{4}{3} \pi r^3 \]where \(V\) is volume and \(r\) is the radius. For neptunium with a specified volume, we derive the radius by rearranging this formula.
In this case, setting the known volume of 3077 cm³ equal to the formula and solving for \(r\) involves:

• Isolating \(r^3\):\[ r^3 = \frac{3077 \times 3}{4 \pi} \approx 733.23 \]
• Finding the cube root:\[ r \approx \sqrt[3]{733.23} \approx 9.01 \text{ cm} \]

This calculation provides the radius of a sphere of neptunium with a critical mass, highlighting how geometry aids in practical applications, such as determining containment dimensions in nuclear safety.