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(D) The nuclear reaction is represented as: $_{90}Th^{232} 	o {}_{82}Pb^{208} + x({}_{2}He^{4}) + y({}_{-1}\beta^{0})$Equating the mass number on bot

Vedclass · Accepted Answer

$6, 4$

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(D) The nuclear reaction is represented as: $_{90}Th^{232} 	o {}_{82}Pb^{208} + x({}_{2}He^{4}) + y({}_{-1}\beta^{0})$Equating the mass number on both sides:$232 = 208 + 4x + 0y$$4x = 232 - 208 = 24$$x = 6$Equating the atomic number on both sides:$90 = 82 + 2x - y$$90 = 82 + 2(6) - y$$90 = 82 + 12 - y$$90 = 94 - y$$y = 4$Therefore,$6$ $\alpha$-particles and $4$ $\beta$-particles are emitted.

The number of $\alpha$ and $\beta$-particles emitted during the transformation of $_{90}Th^{232}$ to $_{82}Pb^{208}$ are respectively

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