Question

For nuclear reaction: ${}_{92}^{235}U{+}_0^{1}n{\to }_{56}^{144}Ba+.....+3_0^{1}n$

• A. ${}_{26}K{r}^{89}$
• B. ${}_{36}K{r}^{89}$
• C. ${}_{26}S{r}^{90}$
• D. ${}_{38}K{r}^{89}$

Answer 1

Answer: (B)

${}_{36}K{r}^{89}$

$\text{given a nuclear reaction}$

${}_{92}^{235}U+\frac{1}{0}n{\to }_{144}^{56}Ba+{}_a^{b}X+3_0^{1}n$

$\begin{array}{c}\text{by balancing atomic no.}\\ \begin{array}{c}92+0=56+a+0\\ a=36\end{array}\end{array}$

$\begin{array}{c}\text{by balancing mass no.}\\ 235+1=144+b+3\\ b=89\\ \text{So,}{}_{36}K{r}^{89}\text{option (b) is correct}\end{array}$