Question

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

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

Answer 1

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

By balancing the equations with mass number and atomic number, we get the correct answer.

For option $\left(C\right)$,

${}_{92}{\text{U}}^{235}+{}_0{n}^1\to {}_{38}{\text{Sr}}^{90}+{}_{54}{X}^{143}+3_0{n}^1$

Considering the Mass number for reactants and products,

$235+1=90+143+3$
$\Rightarrow 236=236$

Considering the Atomic number for reactants and products,

$92=38+54$
$\Rightarrow 92=92$

Hence, option $\left(C\right)$ is correct.