Question:
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 _____
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 _____
Updated On: May 7, 2025
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Correct Answer: 2
Solution and Explanation
The radioactive decay process involves alpha and beta decay sequences. We can break down the steps as follows:
The original \(^{230}_{90}Th\) nucleus undergoes alpha decay, emitting an alpha particle (\(\alpha\)). After this decay, the resulting nucleus is \(^{226}_{88}Ra\).
The \(^{226}_{88}Ra\) nucleus undergoes alpha decay again, resulting in \(^{222}_{86}Rn\), emitting another alpha particle (\(\alpha\)).
The \(^{222}_{86}Rn\) nucleus undergoes alpha decay to form \(^{218}_{84}Po\), emitting another alpha particle (\(\alpha\)).
The \(^{218}_{84}Po\) nucleus undergoes alpha decay to form \(^{214}_{82}Pb\), emitting yet another alpha particle (\(\alpha\)).
The \(^{214}_{82}Pb\) nucleus undergoes beta decay, emitting a beta particle (\(\beta\)), and transforming into \(^{214}_{83}Bi\).
The \(^{214}_{83}Bi\) nucleus undergoes beta decay again, emitting a second beta particle (\(\beta\)), and transforming into \(^{214}_{84}Po\), which is the final product.
Thus, we see that there are 4 alpha decays and 2 beta decays. Therefore, the ratio of the number of alpha particles to beta particles is:
\[ \text{Ratio of } \alpha \text{ to } \beta = \frac{4}{2} = 2 \]
Final Answer:
The ratio of the number of alpha particles to the number of beta particles emitted in this process is \(2\).
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