When _{90}^{228}Th transforms to _{83}^{212}B | vedclass

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(D) Let the number of emitted $\alpha$-particles be $n_{\alpha}$ and the number of emitted $\beta$-particles be $n_{\beta}$.For the mass number: $228

Vedclass · Accepted Answer

$4\,\alpha, 1\beta$

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(D) Let the number of emitted $\alpha$-particles be $n_{\alpha}$ and the number of emitted $\beta$-particles be $n_{\beta}$.For the mass number: $228 = 212 + 4n_{\alpha} + 0n_{\beta}$.$4n_{\alpha} = 228 - 212 = 16$,so $n_{\alpha} = 4$.For the atomic number: $90 = 83 + 2n_{\alpha} - 1n_{\beta}$.Substituting $n_{\alpha} = 4$: $90 = 83 + 2(4) - n_{\beta}$.$90 = 83 + 8 - n_{\beta} = 91 - n_{\beta}$.$n_{\beta} = 91 - 90 = 1$.Thus,the number of $\alpha$-particles is $4$ and the number of $\beta$-particles is $1$.

When $_{90}^{228}Th$ transforms to $_{83}^{212}Bi$,the number of emitted $\alpha$- and $\beta$- particles is,respectively:

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