Question

If all the nuclei in a pure sample of uranium-235 were to fission, yielding about 200 MeV each. What is the kinetic energy yield in joules per kilogram of fuel?

Short answer

Kinetic energy is$E=832\times {10}^{11}J/kg$

Step-by-step solution

Finding N is the number of uranium atoms.

We need to find the kinetic energy per kilogram of uranium-235 that is used as a fuel. It has about 200 MeV energy.

The total kinetic energy E in this fission process is calculated using the formula:

$E=NQ$

Here, Q is the energy released in the process, and N is the number of uranium atoms.

First, we will find N is the number of uranium atoms:

$\frac{m}{M}=N$

Here, m is carbon mass, and M is molecule mass.

By substituting numerical values, we have:

$$\begin{gathered} N=\frac{m}{M} \\ {N}_0=\frac{1kg}{3.9\times {10}^{-23}kg} \\ {N}_0=2.6\times {10}^{24} \end{gathered}$$

By converting 200 MeV into J.

$\begin{array}{rcl}200MeV& =& 200\times 1.6\times {10}^{-13}J\\ & =& 320\times {10}^{-13}J\\ & & \end{array}$

Finally, we can calculate the kinetic energy as:

role="math" localid="1658420259863" $$\begin{gathered} E=320\times {10}^{-13}J\times 2.6\times {10}^{24} \\ E=832\times {10}^{11}J/kg \end{gathered}$$

Therefore, Kinetic energy is$E=832\times {10}^{11}J/kg$