The number of alpha and beta^- particles emit | vedclass

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(C) Let the number of $\alpha$ particles emitted be $x$ and the number of $\beta^-$ particles emitted be $y$.The decay reaction is: ${}_{88}^{226}Ra \

Vedclass · Accepted Answer

$5\alpha$ and $4\beta^-$

Vedclass · Answer

(C) Let the number of $\alpha$ particles emitted be $x$ and the number of $\beta^-$ particles emitted be $y$.The decay reaction is: ${}_{88}^{226}Ra \rightarrow {}_{82}^{206}Pb + x({}_{2}^{4}\alpha) + y({}_{-1}^{0}\beta^-)$.Balancing the mass number:$226 = 206 + 4x + 0$$4x = 226 - 206$$4x = 20$$x = 5$.Balancing the atomic number:$88 = 82 + 2x - y$$88 = 82 + 2(5) - y$$88 = 82 + 10 - y$$88 = 92 - y$$y = 92 - 88$$y = 4$.Therefore,$5$ $\alpha$ particles and $4$ $\beta^-$ particles are emitted.

The number of $\alpha$ and $\beta^-$ particles emitted during the radioactive decay chain starting from ${}_{88}^{226}Ra$ and ending at ${}_{82}^{206}Pb$ is:

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