The end product of the decay of { }_{90} Th^ | vedclass

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(D) Let the number of $ \alpha $-particles emitted be $ x $ and the number of $ \beta $-particles emitted be $ y $. The decay reaction is: $ { }_{90}

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$ 6, 4 $

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(D) Let the number of $ \alpha $-particles emitted be $ x $ and the number of $ \beta $-particles emitted be $ y $. The decay reaction is: $ { }_{90} Th^{232} \rightarrow { }_{82} Pb^{208} + x({ }_{2} He^{4}) + y({ }_{-1} e^{0}) $. Equating the mass numbers: $ 232 = 208 + 4x \implies 4x = 24 \implies x = 6 $. Equating the atomic numbers: $ 90 = 82 + 2x - y $. Substituting $ x = 6 $: $ 90 = 82 + 2(6) - y \implies 90 = 82 + 12 - y \implies 90 = 94 - y \implies y = 4 $. Thus,$ 6 $ $ \alpha $-particles and $ 4 $ $ \beta $-particles are emitted.

The end product of the decay of $ { }_{90} Th^{232} $ is $ { }_{82} Pb^{208} $. The number of $ \alpha $ and $ \beta $ particles emitted are respectively:

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