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(A) Let $n_{\alpha}$ be the number of $\alpha$-particles and $n_{\beta}$ be the number of $\beta$-particles emitted.For the mass number change: $238 =

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$8 \alpha, 6 \beta$

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(A) Let $n_{\alpha}$ be the number of $\alpha$-particles and $n_{\beta}$ be the number of $\beta$-particles emitted.For the mass number change: $238 = 206 + 4n_{\alpha} + 0n_{\beta}$.$4n_{\alpha} = 238 - 206 = 32 \implies n_{\alpha} = 8$.For the atomic number change: $92 = 82 + 2n_{\alpha} - 1n_{\beta}$.Substituting $n_{\alpha} = 8$: $92 = 82 + 2(8) - n_{\beta}$.$92 = 82 + 16 - n_{\beta} \implies 92 = 98 - n_{\beta}$.$n_{\beta} = 98 - 92 = 6$.Thus,$8$ $\alpha$-particles and $6$ $\beta$-particles are emitted.

In the uranium radioactive series,the initial nucleus is ${ }_{92}^{238} U$ and the final nucleus is ${ }_{82}^{206} Pb$. When the uranium nucleus decays into lead,the number of $\alpha$-particles and $\beta$-particles emitted are

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