{}_{90}Th^{232} rightarrow {}_{82}Pb^{208}. T | vedclass

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Question

(A) Let the reaction be ${}_{90}Th^{232} \rightarrow {}_{82}Pb^{208} + n_{\alpha} ({}_{2}He^{4}) + n_{\beta} ({}_{-1}e^{0})$.Mass balance: $232 = 208

Vedclass · Accepted Answer

$6 \alpha$ and $4 \beta$

Vedclass · Answer

(A) Let the reaction be ${}_{90}Th^{232} \rightarrow {}_{82}Pb^{208} + n_{\alpha} ({}_{2}He^{4}) + n_{\beta} ({}_{-1}e^{0})$.Mass balance: $232 = 208 + 4n_{\alpha}$ $\Rightarrow 4n_{\alpha} = 24$ $\Rightarrow n_{\alpha} = 6$.Atomic number balance: $90 = 82 + 2n_{\alpha} - n_{\beta}$ $\Rightarrow 90 = 82 + 2(6) - n_{\beta}$ $\Rightarrow 90 = 94 - n_{\beta}$ $\Rightarrow n_{\beta} = 4$.Therefore,$6 \alpha$ and $4 \beta$ particles are emitted.

${}_{90}Th^{232} \rightarrow {}_{82}Pb^{208}$. The number of $\alpha$ and $\beta$-particles emitted during the above reaction is:

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