The number of neutrons accompanyng the formation of `_(54)^(139)Xe` and `_(38)^(139)Xe` from the absorption of a slow neutron by `_(92)^(235)U` followed by nuclear fision is:
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Correct Answer - 3
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`
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