Answer
a. Xe-141.
b. See answers.
c. 175.8 MeV.
Work Step by Step
a. On the left side of the reaction, the total number of nucleons is 236. This must also be true on the right hand side. Sr-92 has 92 nucleons, and there are 3 neutrons, so the mystery nucleus has 141 nucleons.
On the left side of the reaction, the total number of protons is 92. This must also be true on the right hand side. Sr-92 has 38 protons, so the mystery nucleus has 54 protons.
The mystery nucleus must be Xe-141.
b. The reactor is running at “barely critical.” Each nucleus that fissions leads to just one more nucleus that fissions. Of the 3 produced neutrons, 1 causes another fission. The other two neutrons either escape the reactor or are absorbed by a nucleus that does not fission.
c. Calculate the Q-value, using data from Appendix B and the given website (data retrieved 2019 August 13). We use atomic masses.
$$Q=\left( m_n+m(^{235}_{92}U)-m(^{92}_{38}Sr)-m(^{141}_{54}Xe)-3m_n \right)c^2$$
$$ =\left(1.008665u+235.043930u-91.911038u-140.926787u-3(1.008665u)\right)c^2\left( \frac{931.49MeV/c^2}{u}\right)$$
$$=175.8MeV$$
The Q-value is positive. The reaction is exothermic, and releases 175.8 MeV of energy.
