Question

In the nuclear reaction
$n+{}_{92}^{235}U\to {}_{54}^{a}Xe+{}_b^{94}Sr+2n$ determine the values of a and b.

Answer 1

The neutron will be written as ${}_0{n}^1$

so conserving the atomic number in both sides of the equation we get $0+92=54+b+2\times 0$ so $b=48$

and on conserving the atomic mass in both sides we get $1+235=a+94+2\times 1$ or $a=140$

Finally we get $a=140$ and $b=48$