Chapter 11: Q64E (page 520)

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?

Expert verified

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

Step by step solution

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:

$E = N Q$

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:

$N = \frac{m}{M} N_{0} = \frac{1 k g}{3.9 \times 10^{- 23} k g} N_{0} = 2.6 \times 10^{24}$

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

Finally, we can calculate the kinetic energy as:

role="math" localid="1658420259863" $E = 320 \times 10^{- 13} J \times 2.6 \times 10^{24} E = 832 \times 10^{11} J / k g$

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

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