$$_{92}{U^{235}}$$ nucleus absorb a neutron and disintegrate in $$_{54}X{e^{139}},{\,_{38}}S{r^{94}}$$ and $$X,$$ so what will be the product $$X ?$$
A.
3-neutrons
B.
2-neutrons
C.
$$\alpha $$ -particle
D.
$$\beta $$ -particle
Answer :
3-neutrons
Solution :
$$_{92}{U^{235}}$$ nucleus absorbs a neutron and then disintegrate in $$_{54}X{e^{139}},{\,_{38}}S{r^{94}}$$ and $$X.$$
Thus,
$$_{92}{U^{235}}{ + _0}{n^1}{ \to _{54}}X{e^{139}}{ + _{38}}S{r^{94}} + {3_0}{n^1}$$
Hence, the product is 3-neutrons.
Releted MCQ Question on Physical Chemistry >> Nuclear Chemistry
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