Question

The number of neutrons emitted when ${}_{92}^{235}U$ undergoes controlled nuclear fission to ${}_{54}^{142}Xe$ and ${}_{38}^{90}Sr$ is:

• A. $0$
• B. $1$
• C. $2$
• D. $3$

Answer 1

Answer: (D)

$3$

${}_{92}{U}^{235}{\to }_{54}X{e}^{142}{+}_{38}S{r}^{90}+x^{}{n}^1$

Now, by balancing weight on both sides of the reaction, we get:

$235=90+142+x$

$x=3$

Hence, option D is correct.