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(B) Let the number of $\alpha$-particles emitted be $x$ and the number of $\beta$-particles emitted be $y$.The change in mass number is due only to $\

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$8, 6$

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(B) Let the number of $\alpha$-particles emitted be $x$ and the number of $\beta$-particles emitted be $y$.The change in mass number is due only to $\alpha$-particles,as each $\alpha$-particle reduces the mass number by $4$.$4x = 238 - 206 = 32$$x = 8$The change in atomic number (charge) is given by $2x - y = 92 - 82$,where $2x$ is the decrease due to $\alpha$-decay and $y$ is the increase due to $\beta$-decay.$2(8) - y = 10$$16 - y = 10$$y = 6$Thus,the number of $\alpha$-particles is $8$ and the number of $\beta$-particles is $6$.

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

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