A nucleus of lead Pb_{82}^{214} emits two ele | vedclass

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Question

(A) The decay process is as follows:$1$. Initial nucleus: $Pb_{82}^{214}$.$2$. Emission of two electrons ($\beta^-$-decay): Each $\beta^-$-decay incre

Vedclass · Accepted Answer

$82$ protons and $128$ neutrons

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(A) The decay process is as follows:$1$. Initial nucleus: $Pb_{82}^{214}$.$2$. Emission of two electrons ($\beta^-$-decay): Each $\beta^-$-decay increases the atomic number by $1$ and keeps the mass number constant.$Pb_{82}^{214} \rightarrow A_{84}^{214} + 2_{-1}^{0}e$.$3$. Emission of an $\alpha$-particle: An $\alpha$-particle $(He_{2}^{4})$ emission decreases the atomic number by $2$ and the mass number by $4$.$A_{84}^{214} \rightarrow X_{82}^{210} + He_{2}^{4}$.$4$. The resulting nucleus is $X_{82}^{210}$.$5$. Number of protons $(Z)$ = $82$.$6$. Number of neutrons $(N)$ = $A - Z = 210 - 82 = 128$.Therefore,the resulting nucleus has $82$ protons and $128$ neutrons.

$A$ nucleus of lead $Pb_{82}^{214}$ emits two electrons followed by an $\alpha$-particle. The resulting nucleus will have

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