If _{92}U^{238} undergoes successively 8 alph | vedclass

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(B) The initial nucleus is $_{92}U^{238}$.An $\alpha$-decay reduces the mass number $A$ by $4$ and the atomic number $Z$ by $2$.$A$ $\beta$-decay does

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$_{82}Pb^{206}$

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(B) The initial nucleus is $_{92}U^{238}$.An $\alpha$-decay reduces the mass number $A$ by $4$ and the atomic number $Z$ by $2$.$A$ $\beta$-decay does not change the mass number $A$ and increases the atomic number $Z$ by $1$.After $8 \alpha$-decays,the new mass number $A' = 238 - (8 	imes 4) = 238 - 32 = 206$.The new atomic number $Z' = 92 - (8 	imes 2) + (6 	imes 1) = 92 - 16 + 6 = 82$.The element with atomic number $82$ is Lead $(Pb)$.Therefore,the resulting nucleus is $_{82}Pb^{206}$.

If $_{92}U^{238}$ undergoes successively $8 \alpha$-decays and $6 \beta$-decays,then the resulting nucleus is

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