QJEEAdv22Ph7992: JEE Advanced 2022 Physics - Class 12 JEEAdv22Ph7992 Solution: In a radioactive decay chain reaction, nucleus decays into…

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:

$\Delta Z_{\alpha} = 2 \times 4 = 8$

However, since the change in atomic number is 6, the number of beta-minus particles emitted must account for the difference:

$\Delta Z - \Delta Z_{\alpha} = n_{\beta^{-}} \Rightarrow 6 - 8 = n_{\beta^{-}} \Rightarrow n_{\beta^{-}} = 2 \text{ (increase)}$

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.