{ }_{90}^{232} text{Th} emits 6 alpha and 4 b | vedclass

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(A) In a radioactive decay process,the total mass number and total atomic number are conserved.Let the final product be ${ }_{Z}^{A} 	ext{Pb}$.The de

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$208, 82$

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(A) In a radioactive decay process,the total mass number and total atomic number are conserved.Let the final product be ${ }_{Z}^{A} 	ext{Pb}$.The decay reaction is: ${ }_{90}^{232} 	ext{Th} ightarrow { }_{Z}^{A} 	ext{Pb} + 6({ }_{2}^{4} 	ext{He}) + 4({ }_{-1}^{0} 	ext{e})$.For the mass number $(A)$:$232 = A + 6(4) + 4(0)$$232 = A + 24$$A = 232 - 24 = 208$.For the atomic number $(Z)$:$90 = Z + 6(2) + 4(-1)$$90 = Z + 12 - 4$$90 = Z + 8$$Z = 90 - 8 = 82$.Thus,the mass number is $208$ and the atomic number is $82$.

${ }_{90}^{232} \text{Th}$ emits $6 \alpha$ and $4 \beta$ particles and gets converted into lead. The mass number and atomic number of lead is $......$

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