When Uranium is bombarded with neutrons, it undergoes fission. The fission reaction can be written as :
\({}_{92}\mathrm{U}^{235} + {}_{0}n^{1} \rightarrow {}_{56}\mathrm{Ba}^{141} + {}_{36}\mathrm{Kr}^{92} + 3x + Q\) (energy) where three particles named \(x\) are produced and energy Q is released. What is the name of the particle \(x\) ?
\({}_{92}\mathrm{U}^{235} + {}_{0}n^{1} \rightarrow {}_{56}\mathrm{Ba}^{141} + {}_{36}\mathrm{Kr}^{92} + 3x + Q\) (energy) where three particles named \(x\) are produced and energy Q is released. What is the name of the particle \(x\) ?
Show Hint
- electron
- \(\alpha\)-particle
- neutron
- neutrino
The Correct Option is C
Solution and Explanation
Atomic number: LHS \(= 92 + 0 = 92\). RHS \(= 56 + 36 + 3Z = 92 + 3Z\). So \(Z = 0\).
A particle with A = 1 and Z = 0 is a neutron, \({}_0n^1\).
Answer: (3) neutron
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