Nuclear Chemistry Ques 11
11. ${ } _{92} \mathrm{U}^{235}$, nucleus absorbs a neutron and disintegrates into ${ } _{54} \mathrm{Xe}^{139},{ } _{38} \mathrm{Sr}^{94}$ and $x$. So what will be the product $x$ ?
$[2002]$
(a) $3$-neutrons
(b) $2$-neutrons
(c) $\alpha$-particle
(d) $\beta$-particle
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Answer:
Correct Answer: 11.(a)
Solution: (a)
${ } _{92} U^{235}+{ } _0 n^1 \rightarrow{ } _{54} X^{139}+{ } _{38} \mathrm{Sr}^{94}+X $
$ 92+0=54+38+\mathrm{a} \Rightarrow \mathrm{a}=0 $
$ 235+1=139+94+\mathrm{b} \Rightarrow \mathrm{b}=3 \mathrm{So}, \mathrm{X}=3{ }_0 \mathrm{n}^1 $
i.e 3 neutrons.
BETA
