The atomic weight of thorium is A = 232 and t | vedclass

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(B) The nuclear decay reaction can be represented as: $_{90}Th^{232} \rightarrow _{82}Pb^{208} + n_{1} \alpha + n_{2} \beta^{-}$.First,we perform a ma

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$6$ and $4$

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(B) The nuclear decay reaction can be represented as: $_{90}Th^{232} \rightarrow _{82}Pb^{208} + n_{1} \alpha + n_{2} \beta^{-}$.First,we perform a mass balance (where the mass of an $\alpha$ particle is $4$ and a $\beta$ particle is $0$):$232 = 208 + 4n_{1}$$4n_{1} = 232 - 208 = 24$$n_{1} = 6$ (number of $\alpha$ particles).Next,we perform a charge (atomic number) balance (where the charge of an $\alpha$ particle is $+2$ and a $\beta^{-}$ particle is $-1$):$90 = 82 + 2n_{1} - n_{2}$$90 = 82 + 2(6) - n_{2}$$90 = 82 + 12 - n_{2}$$90 = 94 - n_{2}$$n_{2} = 94 - 90 = 4$ (number of $\beta$ particles).Thus,the number of emitted $\alpha$ and $\beta$ particles are $6$ and $4$ respectively.

The atomic weight of thorium is $A = 232$ and the atomic number is $Z = 90$. After disintegration,the final element obtained is $Pb$,which has $A = 208$ and $Z = 82$. In this process,the number of emitted $\alpha$ and $\beta$ particles are:

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