Question

Complete the fission reaction: \[_{0}^{1}n + _{92}^{235}U \to _{92}^{236}U \to _{56}^{144}Ba + _{36}^{89}Kr + \,?\]

Options

• \[4\,_{0}^{1}n\]

• \[1\,_{0}^{1}n\]

• \[2\,_{0}^{1}n\]

• \[3\,_{0}^{1}n\]

Answer 1

By conservation of nucleon number, the total mass number must remain 236 on both sides. Since the fragments have A = 144 and A = 89, the number of released neutrons is 236 - 144 - 89 = 3. Charge is also conserved: 56 + 36 = 92.