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(C) The initial nucleus is ${}_{90}^{232}Th$.Each $\alpha$-particle emission reduces the mass number $A$ by $4$ and the atomic number $Z$ by $2$.Each

Vedclass · Accepted Answer

$A = 208; Z = 82$

Vedclass · Answer

(C) The initial nucleus is ${}_{90}^{232}Th$.Each $\alpha$-particle emission reduces the mass number $A$ by $4$ and the atomic number $Z$ by $2$.Each $\beta$-particle emission does not change the mass number $A$ and increases the atomic number $Z$ by $1$.After $6$ $\alpha$-decays:$A' = 232 - (6 	imes 4) = 232 - 24 = 208$$Z' = 90 - (6 	imes 2) = 90 - 12 = 78$After $4$ $\beta$-decays:$A = A' = 208$$Z = Z' + (4 	imes 1) = 78 + 4 = 82$Thus,the final nucleus is ${}_{82}^{208}X$,where $A = 208$ and $Z = 82$.

In a radioactive decay chain,the initial nucleus is ${}_{90}^{232}Th$. At the end,$6$ $\alpha$-particles and $4$ $\beta$-particles are emitted. If the final nucleus is ${}_{Z}^{A}X$,then $A$ and $Z$ are given by:

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