When _{90}Th^{238} transforms into _{83}Bi^{2 | 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 decay equation is: $_{

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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 decay equation is: $_{90}Th^{238} \rightarrow _{83}Bi^{222} + n_{\alpha}(_{2}He^{4}) + n_{\beta}(-_{1}e^{0})$.Equating the mass numbers: $238 = 222 + 4n_{\alpha} \Rightarrow 4n_{\alpha} = 16 \Rightarrow n_{\alpha} = 4$.Equating the atomic numbers: $90 = 83 + 2n_{\alpha} - n_{\beta}$.Substituting $n_{\alpha} = 4$: $90 = 83 + 2(4) - n_{\beta} \Rightarrow 90 = 83 + 8 - n_{\beta} \Rightarrow 90 = 91 - n_{\beta} \Rightarrow n_{\beta} = 1$.Thus,$4 \alpha$ particles and $1 \beta$ particle are emitted.

When $_{90}Th^{238}$ transforms into $_{83}Bi^{222}$,the number of $\alpha$ and $\beta$ particles emitted are:

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