An artificial radioactive decay series begins | vedclass

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(A) The initial nucleus is $_{94}^{241}Pu$.Each $\alpha$-decay reduces the mass number by $4$ and the atomic number by $2$.Each $\beta$-decay does not

Vedclass · Accepted Answer

$_{83}^{209}Bi$

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(A) The initial nucleus is $_{94}^{241}Pu$.Each $\alpha$-decay reduces the mass number by $4$ and the atomic number by $2$.Each $\beta$-decay does not change the mass number but increases the atomic number by $1$.After $8$ $\alpha$-decays,the mass number becomes $241 - (8 	imes 4) = 241 - 32 = 209$.The atomic number after $8$ $\alpha$-decays becomes $94 - (8 	imes 2) = 94 - 16 = 78$.After $5$ $\beta$-decays,the mass number remains $209$.The atomic number becomes $78 + 5 = 83$.Thus,the final stable nuclide is $_{83}^{209}Bi$.

An artificial radioactive decay series begins with unstable $_{94}^{241}Pu$. The stable nuclide obtained after eight $\alpha$-decays and five $\beta$-decays is

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