Question 667: Atoms & Nuclei

Question

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

Accepted Answer

To solve this problem, we need to analyze the changes in the atomic number and mass number during the decay chain reaction from ${ }_{90}^{230}\mathrm{Th}$ (Thorium) to ${ }_{84}^{214}\mathrm{Po}$ (Polonium). Let's denote the number of alpha particles emitted by $n_{\alpha}$ and the number of beta-minus particles emitted by $n_{\beta^{-}}$.
First, consider the changes in the mass number (A) and atomic number (Z). For alpha decay, which emits an $\alpha$ particle (${ }_{2}^{4}\mathrm{He}$), the mass number decreases by 4, and the atomic number decreases by 2. For beta-minus decay, which emits a $\beta^{-}$ particle (an electron, $e^{-}$), the atomic number increases by 1, but the mass number remains unchanged.
The initial and final nuclei are given as:
Initial: ${ }_{90}^{230}\mathrm{Th}$
Final: ${ }_{84}^{214}\mathrm{Po}$
The changes in the mass number (A) and atomic number (Z) are:
Change in mass number (A): $230 - 214 = 16$
Change in atomic number (Z): $90 - 84 = 6$
Since each alpha particle decreases the mass number by 4 and the atomic number by 2, if $n_{\alpha}$ is the number of alpha particles emitted, we have:
$$\Delta A = 4n_{\alpha}$$
$$\Delta Z_{\alpha} = 2n_{\alpha}$$
From the given changes, we can write:
$$4n_{\alpha} = 16 \Rightarrow n_{\alpha} = \frac{16}{4} = 4$$
The contribution to the change in atomic number due to the alpha particles is:
$$\Delta Z_{\alpha} = 2 	imes 4 = 8$$
However, since the change in atomic number is 6, the number of beta-minus particles emitted must account for the difference:
$$\Delta Z - \Delta Z_{\alpha} = n_{\beta^{-}} \Rightarrow 6 - 8 = n_{\beta^{-}} \Rightarrow n_{\beta^{-}} = 2 	ext{ (increase)}$$
Note that an increase of 2 units due to $\beta^-$ (beta-minus decay) particles means:
$$\Delta Z_{\beta^{-}} = n_{\beta^{-}} = 2$$
Thus the number of $\beta^{-}$ particles emitted is 2.
Therefore, the ratio of the number of $\alpha$ particles to the number of $\beta^{-}$ particles emitted is:
$$\frac{n_{\alpha}}{n_{\beta^{-}}} = \frac{4}{2} = 2$$
So, the ratio is
2
.

Atoms & Nuclei

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

Integer / Numeric Answer Type