{ }_{92} U^{235} undergoes successive disinte | vedclass

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(C) Let the number of $\alpha$ particles emitted be $x$ and the number of $\beta$ particles emitted be $y$.The nuclear decay reaction is represented a

Vedclass · Accepted Answer

$\alpha=8, \beta=6$

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(C) Let the number of $\alpha$ particles emitted be $x$ and the number of $\beta$ particles emitted be $y$.The nuclear decay reaction is represented as:${ }_{92} U^{235} \longrightarrow x({ }_{2} \alpha^{4}) + y({ }_{-1} \beta^{0}) + { }_{82} Pb^{203}$Equating the mass numbers on both sides:$235 = 4x + 203$$4x = 235 - 203 = 32$$x = 8$Equating the atomic numbers on both sides:$92 = 2x - y + 82$Substituting $x = 8$ into the equation:$92 = 2(8) - y + 82$$92 = 16 - y + 82$$92 = 98 - y$$y = 98 - 92 = 6$Therefore,$8$ $\alpha$ particles and $6$ $\beta$ particles are emitted.

${ }_{92} U^{235}$ undergoes successive disintegrations with the end product of ${ }_{82} Pb^{203}$. The number of $\alpha$ and $\beta$ particles emitted are

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