Question

In the fall of 2002, a group of scientists at Los Alamos National Laboratory determined that the critical mass of neptunium-237 is about 60.0kg. The critical mass of a fissionable material is the minimum amount that must be brought together to start a chain reaction. Neptunium-237 has a density of 19.5g/cm^3.What would be the radius r of a sphere of neptunium-237 that has a critical mass?

Answer 1

The concepts required to solve this problem are those related to density, as a function of mass and volume. In turn, we will use the geometric concept defined for the volume.

The relationship between volume, density and mass is given under the function

$\rho = \frac{m}{V}$

Here,

m = Mass

V = Velocity

Rearranging for the Volume,

$V = \frac{m}{\rho}$

With our information the volume is

$V = \frac{60kg}{19.5g/cm^3 (\frac{10^3kg/m^3}{1g/cm^3})}$

$V = 3.0769*10^{-3}m^3$

Now the volume of sphere is expressed as

$V = \frac{4}{3} \pi r^3$

Here r is the radius of Sphere, then rearranging to find the radius we have

$r = \sqrt[3]{\frac{3V}{4\pi}}$

$r = \sqrt[3]{\frac{3(3.0769*10^{-3})}{4\pi}}$

$r = 0.0902m$

Therefore the radius of a sphere made of this material that has a critical mass is 9.02cm