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)