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JEE Mains · Physics · STD 12 - Nuclei
- JEE Mains 2013
- Medium
- MCQ
When Uranium is bombarded with neutrons, it undergoes fission. The fission reaction can be written as \({}_{92}{U^{235}} + {}_0{n^1} \to {}_{56}B{a^{141}} + {}_{36}K{r^{92}} + 3x + Q{\rm{( energy)}}\) where three particles named \(x\) are produced and energy \(Q\) is released. What is the name of the particle \(x\) ?
Pick an option to check your answer.
Answer & Solution
Correct Answer
(C) neutron
Step-by-step Solution
Detailed explanation
Nuclear fission equation
\({}_{92}{U^{235}} + {}_0{n^1} \to {}_{56}B{a^{141}} + {}_{36}K{r^{92}} + 3\,{}_0{n^1} + Q{\rm{( energy)}}\)
Hence particle \(x\) is neutron
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