A radioactive element X emits six alpha-parti | vedclass

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(C) Let the radioactive element $X$ be represented as ${ }_{Z}^{A} X$.When an $\alpha$-particle $({ }_{2}^{4} He)$ is emitted,the mass number decrease

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${ }_{90}^{232} Th$

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(C) Let the radioactive element $X$ be represented as ${ }_{Z}^{A} X$.When an $\alpha$-particle $({ }_{2}^{4} He)$ is emitted,the mass number decreases by $4$ and the atomic number decreases by $2$.When a $\beta$-particle $({ }_{-1}^{0} e)$ is emitted,the mass number remains unchanged and the atomic number increases by $1$.Given that $6$ $\alpha$-particles and $4$ $\beta$-particles are emitted to form ${ }_{82}^{208} Pb$,we can write the conservation equations:For the mass number $A$:$A - (6 	imes 4) - (4 	imes 0) = 208$$A - 24 = 208$$A = 208 + 24 = 232$For the atomic number $Z$:$Z - (6 	imes 2) + (4 	imes 1) = 82$$Z - 12 + 4 = 82$$Z - 8 = 82$$Z = 82 + 8 = 90$Thus,the element $X$ is ${ }_{90}^{232} Th$.

$A$ radioactive element $X$ emits six $\alpha$-particles and four $\beta$-particles,leading to the final product ${ }_{82}^{208} Pb$. The element $X$ is:

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