Q11E - University Physics with Modern Physics

Question

Neptunium. In the fall of 2002, scientists at Los Alamos National Laboratory determined that 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 nuclear chain reaction. Neptunium-237 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

Volume of $60 \mathrm{kg}$ material $=3.076\times {10}^3{ \mathrm{cm}}^3$.

Radius of sphere of material $=9.021 \mathrm{cm}$.

Given information

Neptunium mass $=60\mathrm{Kg}$

Neptunium density $=19.5 \mathrm{g}/{\mathrm{cm}}^3$

Calculation of Volume of $60 \mathrm{~kg}$ material

The Volume is calculated as

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

Herel

$\mathrm{V}$ is volume

$M$ is mass

$\rho =$ density

$$\begin{gathered} V=60\mathrm{Kg}/\left(19.5 \mathrm{g}/{\mathrm{cm}}^3\right)\left(\frac{{10}^3 \mathrm{kg}}{1 \mathrm{g}}\right) \\ V=3.076\times {10}^3{ \mathrm{cm}}^3 \end{gathered}$$

Calculating Radius of sphere of material

The Sphere radius is given as

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

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

Substituting the values as

$R=\sqrt[a]{\frac{3\times 3/076\times {10}^3{ \mathrm{cm}}^3}{4\pi }}$

$\mathrm{R}=9.021 \mathrm{cm}$