A) 3-neutrons
B) 2-neutrons
C) or-particle
D) \[\beta -\]particle
Correct Answer: A
Solution :
\[{{\,}_{92}}{{U}^{235}}+{{\,}_{0}}{{n}^{1}}\to {{\,}_{54}}X{{e}^{139}}+{{\,}_{38}}S{{r}^{94}}+X\] \[92+0=54+38+a\Rightarrow a=0\]\[235+1=139+94+b\Rightarrow b=3\] So, \[X=3\,{{\,}_{0}}{{n}^{1}}\] That is, 3 neutrons.You need to login to perform this action.
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