In the radioactive disintegration series {}_{ | vedclass

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(B) An $\alpha$-particle corresponds to a helium nucleus $({}_{2}^{4}He)$. It reduces the mass number by $4$ and the atomic number by $2$. In $\beta$-

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$6 \alpha$ and $4 \beta$

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(B) An $\alpha$-particle corresponds to a helium nucleus $({}_{2}^{4}He)$. It reduces the mass number by $4$ and the atomic number by $2$. In $\beta$-decay,the atomic number increases by $1$ while the mass number remains unchanged. For the reaction: ${}_{90}^{232}Th \rightarrow {}_{82}^{208}Pb + x({}_{2}^{4}He) + y({}_{-1}^{0}e)$. Equating the mass numbers: $232 = 208 + 4x$,which gives $4x = 24$,so $x = 6$. Equating the atomic numbers: $90 = 82 + 2x - y$. Substituting $x = 6$: $90 = 82 + 12 - y$,which gives $90 = 94 - y$,so $y = 4$. Thus,$6 \alpha$ and $4 \beta$ particles are emitted.

In the radioactive disintegration series ${}_{90}^{232}Th \rightarrow {}_{82}^{208}Pb$ involving $\alpha$ and $\beta$ decay,the total number of $\alpha$ and $\beta$ particles emitted are:

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