The number of neutrons accompanying the formation of `._(54)Xe^(139)` and `._(38)Sr^(94)` from the absorption of a slow neutron by `._(92)U^(235)`, followed by nuclear fission is

Answer

3

Explanation

since the charge in mass is only due to the emmission of `alpha`-particle, we have
Number of `alpha`-particle emitted `=(234-206)/(4)=7`
Now the associated decrease in atomic number would be `14 (=2xx7)` and thus the atomic number of the daughter atom would be `76(=90-14)`. But the actual atomic number of lead is `82 e`. the atomic number is six more than expected. This is because of the emission `beta`-particle. Since there is an increase of one in atomic number due to the emission of one `beta`-particle, we have
Number of `beta`-particles emitted `=(82-76)/(1)=6`
Hence, number of `alpha`-particles emitted `=7` ltbr,. Number of `beta`-particles emittes `=6`
Answer is `6+7=13`