_{86}A^{222} to _{84}B^{210}. In this reactio | vedclass

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(B) Let the number of $\alpha$ particles emitted be $n_{\alpha}$ and the number of $\beta$ particles emitted be $n_{\beta}$.For the mass number: $222

Vedclass · Accepted Answer

$3\alpha, 4\beta$

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(B) Let the number of $\alpha$ particles emitted be $n_{\alpha}$ and the number of $\beta$ particles emitted be $n_{\beta}$.For the mass number: $222 = 210 + 4n_{\alpha} \implies 4n_{\alpha} = 12 \implies n_{\alpha} = 3$.For the atomic number: $86 = 84 + 2n_{\alpha} - 1n_{\beta}$.Substituting $n_{\alpha} = 3$: $86 = 84 + 2(3) - n_{\beta} \implies 86 = 84 + 6 - n_{\beta} \implies 86 = 90 - n_{\beta} \implies n_{\beta} = 4$.Thus,$3$ $\alpha$ particles and $4$ $\beta$ particles are emitted.

$_{86}A^{222} \to _{84}B^{210}$. In this reaction,how many $\alpha$ and $\beta$ particles are emitted?

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