In the radioactive decay _{92}X^{232} to _{89 | vedclass

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(A) The radioactive decay reaction is given by: $_{92}X^{232} 	o _{89}Y^{220} + x(_{2}\alpha^{4}) + y(_{-1}\beta^{0})$.First,calculate the number of

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$3\alpha$ and $3\beta$

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(A) The radioactive decay reaction is given by: $_{92}X^{232} 	o _{89}Y^{220} + x(_{2}\alpha^{4}) + y(_{-1}\beta^{0})$.First,calculate the number of $\alpha$-particles $(x)$ by comparing the mass numbers: $232 = 220 + 4x \implies 4x = 12 \implies x = 3$.Next,calculate the number of $\beta$-particles $(y)$ by comparing the atomic numbers: $92 = 89 + 2x - y \implies 92 = 89 + 2(3) - y \implies 92 = 89 + 6 - y \implies 92 = 95 - y \implies y = 3$.Therefore,$3$ $\alpha$-particles and $3$ $\beta$-particles are ejected.

In the radioactive decay $_{92}X^{232} \to _{89}Y^{220}$,how many $\alpha$ and $\beta$-particles are ejected from $X$ to form $Y$?

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