Question

The answer to this question is a single digit integer, ranging from 0 to 9.

The number of neutrons accompanying the formation of ${}_{54}^{139}Xe$ and ${}_{39}^{94}Sr$ from the absorption of slow neutrons by ${}_{92}^{235}U$ followed by fission is :

Answer 1

$U_{92}^{235}+n_0^3+P_1^{\to }X{e}_{54}^{135}+S{r}_{39}^{94}$
The total mass number difference is three between products and reactants
So one proton is added and remaining two are neutrons.
Hence the number of protons accompanying the above reaction is 2 neutrons.