A radioactive reaction is _{92}U^{238} to _{8 | vedclass

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(A) Let the number of $\alpha$ particles emitted be $n_{\alpha}$ and the number of $\beta$ particles emitted be $n_{\beta}$.The change in mass number

Vedclass · Accepted Answer

$8\alpha, 6\beta$

Vedclass · Answer

(A) Let the number of $\alpha$ particles emitted be $n_{\alpha}$ and the number of $\beta$ particles emitted be $n_{\beta}$.The change in mass number is given by $A - A' = 4n_{\alpha}$.$238 - 206 = 4n_{\alpha} \implies 32 = 4n_{\alpha} \implies n_{\alpha} = 8$.The change in atomic number is given by $Z - Z' = 2n_{\alpha} - n_{\beta}$.$92 - 82 = 2(8) - n_{\beta}$.$10 = 16 - n_{\beta} \implies n_{\beta} = 6$.Thus,$8$ $\alpha$ particles and $6$ $\beta$ particles are emitted.

$A$ radioactive reaction is $_{92}U^{238} \to _{82}Pb^{206}$. How many $\alpha$ and $\beta$ particles are emitted?

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