Question

The number of neutrons released when ${}_{92}^{235}U$ undergoes fission by absorbing ${}_0^{1}n$ and $\left({}_{56}^{144}Ba{+}_{36}^{89}Kr\right)$ are formed

• A. $0$
• B. $1$
• C. $2$
• D. $3$

Answer 1

The correct option is D $3$

Fission reaction cen be written as,
${}_{92}^{235}U{+}_0^{1}U{\to }_{56}^{144}Ba{+}_{36}^{89}Kr{+}_0^{x}n+Q$

Due to mass conservation,
$235+1=144+89+x$
$\therefore x=3$