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(D) Let the number of $\alpha$-particles emitted be $n_{\alpha}$ and the number of $\beta$-particles emitted be $n_{\beta}$.For the decay ${}^{238}_{9

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$8$ and $6$

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(D) Let the number of $\alpha$-particles emitted be $n_{\alpha}$ and the number of $\beta$-particles emitted be $n_{\beta}$.For the decay ${}^{238}_{92}U \longrightarrow {}^{206}_{82}Pb + n_{\alpha}({}^{4}_{2}He) + n_{\beta}({}^{0}_{-1}e)$:Equating the mass number: $238 = 206 + 4n_{\alpha} \implies 4n_{\alpha} = 32 \implies n_{\alpha} = 8$.Equating the atomic number: $92 = 82 + 2n_{\alpha} - n_{\beta}$.Substituting $n_{\alpha} = 8$: $92 = 82 + 2(8) - n_{\beta} \implies 92 = 82 + 16 - n_{\beta} \implies 92 = 98 - n_{\beta} \implies n_{\beta} = 6$.Thus, $8$ $\alpha$-particles and $6$ $\beta$-particles are emitted.

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

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