Write the initial and final nuclei

${}_{90}^{232}\text{Th} \rightarrow {}_{82}^{212}\text{Pb}$

Let the number of emitted $\alpha$ particles be $N_\alpha$ and the number of emitted $\beta^-$ particles be $N_\beta$.

Use change in mass number to find $N_\alpha$

• Each $\alpha$ decay reduces mass number by $4$.
• $\beta^-$ decay does not change mass number.

So, $232 - 4N_\alpha = 212$ $4N_\alpha = 20$ $N_\alpha = 5$

Hence, Option A is true and Option B is false.

Use change in atomic number to find $N_\beta$

• Each $\alpha$ decay reduces atomic number by $2$.
• Each $\beta^-$ decay increases atomic number by $1$.

Starting from $Z=90$: after $5$ alpha decays, $90 - 2\times 5 = 80$

Final atomic number is $82$, so we need increase of $2$: $80 + N_\beta = 82$ $N_\beta = 2$

Hence, Option C is true and Option D is false.

Final evaluation of options

• A: $N_\alpha = 5$ ✅
• B: $N_\alpha = 6$ ❌
• C: $N_\beta = 2$ ✅
• D: $N_\beta = 4$ ❌

Conclusion

The correct options are: $\boxed{A,\ C}$

Comparison with stored correct answer

Stored correct answer: A, C

This matches the derived answer exactly.