Question

In the neutron-induced fission of uranium-235 that produces \(_{51}^{133}Sb\) and \(_{41}^{99}Nb\) as fragments, the number of neutrons released is:

विकल्प

• 2 neutrons

• 4 neutrons

• 3 neutrons

• 5 neutrons

Answer 1

Conserving the mass number: 236 = 133 + 99 + x, which gives x = 4 neutrons. The atomic numbers also balance since 51 + 41 = 92, confirming the reaction \[_{0}^{1}n + _{92}^{235}U \to _{92}^{236}U \to _{51}^{133}Sb + _{41}^{99}Nb + 4\,_{0}^{1}n\].