Question

${}_{92}{U}^{235}$, 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. $1-neutrons$
• D. $4-neutrons$

Answer 1

Answer: (A)

$3-neutrons$

${}_{92}{U}^{235}$ + ${}_0{n}^1\to {}_{54}X{e}^{139}+{}_{38}S{r}^{94}+x$
92 + 0 = 54 + 38 +a
a = 0
235 + 1 =139+94 +b
b=3
So x = ${}_0{n}^1$