Question

A neutron strikes a ${}_{92}{U}^{235}$ nucleus and as a result ${}_{36}K{r}^{93}$ and ${}_{56}B{a}^{140}$ are produced with

• A. $\alpha$- particle
• B. $1-neutron$
• C. $3-neutron$
• D. $2-\beta -$ particle

Answer 1

The correct option is C $3-neutron$

$\begin{array}{c}\text{We have to conserve the number of}\\ \text{neutrons of reaction.}\end{array}$

$\text{1 Neutron}+\underset{92}{U^{235}}⟶{K_r^{93}}_{36}+B{a}_{56}^{140}+nneutron$

$\begin{array}{c}n+\left(84\right)+\left(57\right)=1+143\\ n+141=144\\ n=3-\text{neutron}\end{array}$