11 - University Physics with Modern Physics

Question

Question- Neptunium. 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?

Step-by-Step Solution

Answer

The radius of sphere is 0.0270 m.

Calculation of Volume of 60 kg material

Given data:

The mas of neptunium-237 is $m=60 Kg$.

The density of neptunium -237 is $\rho =19.5 g/c{m}^3$.

The volume of the sphere can be calculated as

$$\begin{gathered} V=\frac{m}{\rho } \\ V=\frac{60 kg}{19.5 g/c{m}^3\times {\displaystyle \frac{1000 kg/m^3}{1 g/c{m}^3}}} \\ V=3.07\times {10}^{-3}{m}^3 \end{gathered}$$

Calculating the radius of sphere

The radius of sphere will be,

$$\begin{gathered} V=\frac{4}{3}\pi R^3 \\ 3.07\times {10}^{-3}{m}^3=\frac{4}{3}\pi R^3 \\ R^3=7.33\times {10}^{-4}{m}^3 \\ R=0.0270 m \end{gathered}$$

 

Thus, the radius of the sphere is 0.0270 m.