The nucleus _{90}Th^{228} transforms into _{8 | vedclass

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(A) 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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$4\,\alpha, 1\,\beta$

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(A) 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 given by $A - A' = 4n_{\alpha}$.$228 - 212 = 4n_{\alpha} \implies 16 = 4n_{\alpha} \implies n_{\alpha} = 4$.The change in atomic number is given by $Z - Z' = 2n_{\alpha} - n_{\beta}$.$90 - 83 = 2(4) - n_{\beta}$.$7 = 8 - n_{\beta} \implies n_{\beta} = 1$.Thus,$4$ $\alpha$-particles and $1$ $\beta$-particle are emitted.

The nucleus $_{90}Th^{228}$ transforms into $_{83}Bi^{212}$. How many $\alpha$-particles and $\beta$-particles are emitted during this process?

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