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(D) Let the number of $\alpha$-particles emitted be $n_1$ and the number of $\beta$-particles emitted be $n_2$.The decay equation is: ${}^{238}U_{92}

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$8 \alpha$ and $6 \beta$

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(D) Let the number of $\alpha$-particles emitted be $n_1$ and the number of $\beta$-particles emitted be $n_2$.The decay equation is: ${}^{238}U_{92} \rightarrow {}^{206}Pb_{82} + n_1({}^{4}He_{2}) + n_2({}^{0}e_{-1})$.Equating the mass numbers:$238 = 206 + 4n_1$$32 = 4n_1 \Rightarrow n_1 = 8$.Equating the atomic numbers:$92 = 82 + 2n_1 - n_2$$92 = 82 + 2(8) - n_2$$92 = 82 + 16 - n_2$$92 = 98 - n_2$$n_2 = 98 - 92 = 6$.Thus,$8 \alpha$ and $6 \beta$ particles are emitted.

The particles emitted during the sequential radioactive decay of ${}^{238}U_{92}$ to ${}^{206}Pb_{82}$ are

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