Answer
a. 5 neutrons.
b. 171.1 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. The two daughter nuclei have 133+98 = 231 nucleons all together, so there must be 5 neutrons produced.
b. Calculate the Q-value, using data from Appendix B. We use atomic masses.
$$Q=\left(m_{n}+m(^{235}_{92}U)-m(^{133}_{51}Sb)-m(^{98}_{41}Nb)-5m_{n} \right)c^2$$
$$ =\left(1.008665u+235.043930u-132.915250u-97.910328u-5(1.008665u)\right)c^2\left( \frac{931.49MeV/c^2}{u}\right)$$
$$=171.1MeV$$
The Q-value is positive. The fission reaction is exothermic, and releases 171.1 MeV of energy.
