Question

Complete the equation for the following fission process ${}_{92}{U}^{235}{+}_0{n}^1{\to }_{38}S{r}^{90}+...$

• A. ${}_{54}X{e}^{143}+3_0{n}^1$
• B. ${}_{54}X{e}^{145}$
• C. ${}_{57}X{e}^{142}$
• D. ${}_{54}X{e}^{142}+3_{0}n$

Answer 1

The correct option is A ${}_{54}X{e}^{143}+3_0{n}^1$

Sum of mass number in $reactant$ side is $235+1=236$

and Sum of atomic number in $reactant$ side is $92+0=92$

so the product side sum of mass number should also be $236$ and that of atomic number should be $92$

so ${}_{54}X{e}^{143}+3_0{n}^1$ is

suitable because $143+3\times 1+90=236$ and $38+54=92$