_{89}Ac^{231} gives _{82}Pb^{207} after the e | vedclass

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Question

(B) Let the number of $\alpha$-particles emitted be $x$ and the number of $\beta$-particles emitted be $y$.The change in mass number is given by $231

Vedclass · Accepted Answer

$6, 5$

Vedclass · Answer

(B) Let the number of $\alpha$-particles emitted be $x$ and the number of $\beta$-particles emitted be $y$.The change in mass number is given by $231 - 207 = 4x$.$24 = 4x \implies x = 6$.The change in atomic number is given by $89 - 82 = 2x - y$.$7 = 2(6) - y$.$7 = 12 - y \implies y = 5$.Thus,the number of $\alpha$-particles is $6$ and the number of $\beta$-particles is $5$.

$_{89}Ac^{231}$ gives $_{82}Pb^{207}$ after the emission of some $\alpha$ and $\beta$ particles. The number of such $\alpha$ and $\beta$ particles are respectively:

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