_{90}^{232}Th,an isotope of thorium,decays in | vedclass

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(C) The initial nucleus is $_{90}^{232}Th$. Each $\alpha$-particle emission reduces the mass number by $4$ and the atomic number by $2$. Each $\beta$-

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$_{82}^{208}Pb$

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(C) The initial nucleus is $_{90}^{232}Th$. Each $\alpha$-particle emission reduces the mass number by $4$ and the atomic number by $2$. Each $\beta$-particle emission increases the atomic number by $1$ and leaves the mass number unchanged. Let the final nucleus be $_{Z'}^{A'}X$. The new mass number $A' = A - 4n_{\alpha} = 232 - 4 	imes 6 = 232 - 24 = 208$. The new atomic number $Z' = Z - 2n_{\alpha} + 1n_{\beta} = 90 - 2(6) + 4 = 90 - 12 + 4 = 82$. The element with atomic number $82$ is Lead $(Pb)$. Therefore,the end product is $_{82}^{208}Pb$.

$_{90}^{232}Th$,an isotope of thorium,decays in ten stages,emitting six $\alpha$-particles and four $\beta$-particles in total. The end product of the decay is:

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