$_{92} U^{235}$ undergoes successive disintegrations with the end product of $_{82} {P b}^{203}$ . The number of $α$ and $β$ particles emitted are

α = 6

β = 4

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α = 6

β = 0

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α = 8

β = 6

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α = 3

β = 3

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Solution

The correct option is C $α = 8$ , $β = 6$
During $1 α$ decay, mass number and atomic number of parent nucleus decrease by $4$ and $2$ , respectively.
During $1 β$ decay, atomic number of parent nucleus increases by $1$ whereas mass number remains the same.
Hence number of alpha particles emitted $n_{α} = \frac{235 - 203}{4} = 8$
Due to $α$ decay, atomic number of new element so formed $Z_{n e w} = 92 - 2 (8) = 76$
Thus number of $β$ particles emitted $n_{β} = \frac{82 - 76}{1} = 6$
Thus the nuclear reaction : $_{92} U^{235} \to_{82} P b^{203} + 8 α + 6 β$

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92U235 undergoes successive disintegrations with the end product of 82Pb203. The number of α and β particles emitted are

$_{92} U^{235}$ undergoes successive disintegrations with the end product of $_{82} {P b}^{203}$ . The number of $α$ and $β$ particles emitted are