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## Q. 13.5

Calculate the ground state Q value of the induced fission reaction in Equation (13.14) if the neutron is thermal. A neutron is said to be thermal when it is in thermal equilibrium with its environment; it then has an average kinetic energy given by $\frac{3}{2} k T$.

$n+{ }_{92}^{235} U \rightarrow{ }_{92}^{236} U ^{*} \rightarrow{ }_{40}^{99} Zr +{ }_{52}^{134} Te +3 n$ (13.14)

Strategy Because the kinetic energy of a thermal neutron is so small, its kinetic energy can be neglected; even for a temperature of $10^{6}$ K, the thermal energy is only 130 eV. The Q value is given by Equation (13.7).

$Q=M_{x} c^{2}+M_{X} c^{2}-\left(M_{y} c^{2}+M_{Y} c^{2}\right)=K_{y}+K_{Y}-K_{x}$ (13.7)

## Verified Solution

We look up the atomic masses in Appendix 8 and determine the Q value to be

\begin{aligned}Q=&\left\{M\left({ }^{235} U \right)+m(n)-\left[M\left({ }^{99} Zr \right)\right.\right.\\&\left.\left.+M\left({ }^{134} Te \right)+3 m(n)\right]\right\} c^{2}\end{aligned}

\begin{aligned}Q=&[235.0439 u -98.9165 u -133.9115 u \\&-2(1.0087 u )] c^{2} \\=& 0.1985 c^{2} \cdot u \\=&\left(0.1985 c^{2} \cdot u \right)\left(\frac{931.5 MeV }{c^{2} \cdot u }\right)=185 MeV\end{aligned}

Even if the fission is induced by a thermal neutron of negligible kinetic energy on the nuclear scale, a tremendous amount of energy is released.