The nucleus { }_{88}^{226} Ra is converted in | vedclass

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Question

(A) Let $n_{\alpha}$ be the number of alpha particles and $n_{\beta}$ be the number of beta particles emitted.For the mass number $(A)$: $226 = 206 +

Vedclass · Accepted Answer

$5$,$4$

Vedclass · Answer

(A) Let $n_{\alpha}$ be the number of alpha particles and $n_{\beta}$ be the number of beta particles emitted.For the mass number $(A)$: $226 = 206 + 4n_{\alpha} + 0n_{\beta}$$20 = 4n_{\alpha} \implies n_{\alpha} = 5$.For the atomic number $(Z)$: $88 = 82 + 2n_{\alpha} - 1n_{\beta}$$88 = 82 + 2(5) - n_{\beta}$$88 = 82 + 10 - n_{\beta}$$88 = 92 - n_{\beta}$$n_{\beta} = 92 - 88 = 4$.Thus,the number of alpha particles is $5$ and the number of beta particles is $4$.

The nucleus ${ }_{88}^{226} Ra$ is converted into ${ }_{82}^{206} Pb$ by the emission of alpha $(\alpha)$ and beta $(\beta)$ particles. The number of alpha and beta particles emitted are respectively:

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