ExamBro
ExamBro
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.
Verified Solution

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

Same subject
Explore more questions on app
From JEE Mains
Explore more questions on app