Actinium 231,{}^{231}Ac_{89},emits in success | vedclass

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Question

(D) The initial nucleus is ${}^{231}Ac_{89}$.Total number of $\alpha$ particles emitted = $4 + 1 = 5$.Total number of $\beta^-$ particles emitted = $2

Vedclass · Accepted Answer

$^{211}Pb_{82}$

Vedclass · Answer

(D) The initial nucleus is ${}^{231}Ac_{89}$.Total number of $\alpha$ particles emitted = $4 + 1 = 5$.Total number of $\beta^-$ particles emitted = $2 + 1 = 3$.Each $\alpha$ decay reduces the mass number $A$ by $4$ and the atomic number $Z$ by $2$.Each $\beta^-$ decay leaves the mass number $A$ unchanged and increases the atomic number $Z$ by $1$.Final mass number $A' = 231 - (5 	imes 4) = 231 - 20 = 211$.Final atomic number $Z' = 89 - (5 	imes 2) + (3 	imes 1) = 89 - 10 + 3 = 82$.The element with atomic number $82$ is Lead $(Pb)$.Therefore,the resultant isotope is ${}^{211}Pb_{82}$.

Actinium $231$,${}^{231}Ac_{89}$,emits in succession two $\beta^-$ particles,four $\alpha$ particles,one $\beta^-$ particle,and one $\alpha$ particle,plus several $\gamma$ rays. What is the resultant isotope?

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