In a radioactive decay chain reaction,{ }_{90 | vedclass

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(B) Let the number of $\alpha$-particles emitted be $n$ and the number of $\beta^{-}$-particles emitted be $m$.The decay reaction is: ${ }_{90}^{230}

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$2$

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(B) Let the number of $\alpha$-particles emitted be $n$ and the number of $\beta^{-}$-particles emitted be $m$.The decay reaction is: ${ }_{90}^{230} Th \rightarrow { }_{84}^{214} Po + n({ }_{2}^{4} He) + m({ }_{-1}^{0} e)$.Equating the mass numbers: $230 = 214 + 4n \Rightarrow 4n = 16 \Rightarrow n = 4$.Equating the atomic numbers: $90 = 84 + 2n - m$.Substituting $n = 4$: $90 = 84 + 2(4) - m \Rightarrow 90 = 84 + 8 - m \Rightarrow 90 = 92 - m$.Thus,$m = 92 - 90 = 2$.The ratio of the number of $\alpha$-particles to $\beta^{-}$-particles is $\frac{n}{m} = \frac{4}{2} = 2$.

In a radioactive decay chain reaction,${ }_{90}^{230} Th$ nucleus decays into ${ }_{84}^{214} Po$ nucleus. The ratio of the number of $\alpha$ to number of $\beta^{-}$ particles emitted in this process is. . . . .

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