Question

The species $A$ in the reaction is:

${}_{92}{U}^{236}{\to }_{54}X{e}^{144}{+}_{38}S{r}^{90}+A$

• A. ${}_1{H}^1$
• B. ${}_{2}H{e}^4$
• C. ${}_{-1}{\beta }^0$
• D. ${}_0{n}^1$

Answer 1

The correct option is D ${}_0{n}^1$

The species A in the reaction is ${}_0{n}^1$

${}_{92}{U}^{236}\to {}_{54}X{e}^{144}{+}_{38}S{r}^{90}+2_0{n}^1$

Atomic number balance:

$92=54+38+0$

Mass number balance:

$236=144+90+2$