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## Q. 13.6

Calculate the excitation energy of the compound nuclei produced when ${ }^{235} U \text { and }{ }^{238} U$ absorb thermal neutrons.

Strategy The two reactions are

\begin{aligned}&n+{ }^{235} U \rightarrow{ }^{236} U ^{*} \\&n+{ }^{238} U \rightarrow{ }^{239} U ^{*}\end{aligned}

As we did in Example 13.4, we find the excitation energy from the atomic masses using Equation (13.11). A thermal neutron has a negligible kinetic energy (about 0.03 eV).

$E\left( CN ^{*}\right)=M_{x} c^{2}+M_{X} c^{2}-M_{C N} c^{2}$ (13.11)

## Verified Solution

We have from Equation (13.11)

\begin{aligned}E\left({ }^{236} U ^{*}\right) &=\left[m(n)+M\left({ }^{235} U \right)-M\left({ }^{236} U \right)\right] c^{2} \\&=[1.0087 u +235.0439 u -236.0456 u ] c^{2} \\&=\left(0.0070 c^{2} \cdot u \right)\left(\frac{931.5 MeV }{c^{2} \cdot u }\right)=6.5 MeV\end{aligned}

\begin{aligned}E\left({ }^{239} U ^{*}\right) &=\left[m(n)+M\left({ }^{238} U \right)-M\left({ }^{239} U \right)\right] c^{2} \\&=[1.0087 u +238.0508 u -239.0543 u ] c^{2} \\&=\left(0.0056 c^{2} \cdot u \right)\left(\frac{931.5 MeV }{c^{2} \cdot u }\right)=4.8 MeV\end{aligned}

Thus ${ }^{236} U *$ has almost 2 MeV more excitation energy than ${ }^{239} U *$ when both are produced by thermal neutron absorption. This helps explain why ${ }^{235} U$ more easily undergoes thermal neutron fission.