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  • VITEEE- When U(92,235) is bombarded with a neutron, it undergoes as

_{0}n^{1} + _{92}U^{235} \rightarrow _{92}U^{236} \rightarrow _{56}Ba^{144}+ _{36}Kr^{89} + neutrons. In this process, number of neutrons released is

  1. 6

  2. 3

  3. 5

  4. 2

Answers (1)

best_answer

Total number of neutrons and protons initiaaly = 235 + 1 = 236

Total number of neutrons and protons present in atoms of Ba and Kr = 144 + 89 = 233

Number of neutrons released = 236 - 233 = 3

Posted by

Sayak

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