An atomic nucleus _{90}^{232}Th emits several | vedclass

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(C) Let the number of $\alpha$-particles emitted be $n_{\alpha}$ and the number of $\beta$-particles emitted be $n_{\beta}$.The change in mass number

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$6 \alpha$ and $4 \beta$

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(C) Let the number of $\alpha$-particles emitted be $n_{\alpha}$ and the number of $\beta$-particles emitted be $n_{\beta}$.The change in mass number is due only to $\alpha$-decay:$n_{\alpha} = \frac{A - A'}{4} = \frac{232 - 208}{4} = \frac{24}{4} = 6$.The change in atomic number is given by $Z' = Z - 2n_{\alpha} + n_{\beta}$.Substituting the values: $82 = 90 - 2(6) + n_{\beta}$.$82 = 90 - 12 + n_{\beta}$.$82 = 78 + n_{\beta}$.$n_{\beta} = 82 - 78 = 4$.Therefore,the nucleus emitted $6 \alpha$ and $4 \beta$ particles.

An atomic nucleus $_{90}^{232}Th$ emits several $\alpha$ and $\beta$ radiations and finally reduces to $_{82}^{208}Pb$. It must have emitted:

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