In a radioactive decay chain, `._(90)^(232)Th` nucleus decays to `._(82)^(212)Pb` nucleus. Let `N_(alpha) and N_(beta)` be the number of `alpha and beta`- particles, respectively, emitted in this decay process. Which of the following statements is (are) true ?

Options

A. `N_(alpha) = 5`
B. `N_(alpha) = 6`
C. `N_(beta) = 2`
D. `N_(beta) = 4`

Answer

To solve the problem, we need to analyze the decay process of the thorium-232 (Th-232) nucleus to lead-212 (Pb-212) and determine the number of alpha (α) and beta (β) particles emitted during this decay chain.

### Step-by-Step Solution:

1. **Identify the Initial and Final Nuclei:**
- Initial nucleus: Thorium-232, represented as \( _{90}^{232}Th \)
- Final nucleus: Lead-212, represented as \( _{82}^{212}Pb \)

2. **Calculate the Change in Mass Number:**
- The mass number of Th-232 is 232, and that of Pb-212 is 212.
- Change in mass number = \( 232 - 212 = 20 \)

3. **Determine the Number of Alpha Particles Emitted:**
- Each alpha particle (α) emitted decreases the mass number by 4.
- Therefore, the number of alpha particles emitted, \( N_{\alpha} \), can be calculated as:
\[ N_{\alpha} = \frac{\text{Change in mass number}}{4} = \frac{20}{4} = 5 \]

4. **Calculate the Change in Atomic Number:**
- The atomic number of Th-232 is 90, and that of Pb-212 is 82.
- Change in atomic number = \( 90 - 82 = 8 \)

5. **Relate Alpha and Beta Particles:**
- Each alpha particle decreases the atomic number by 2.
- Let \( N_{\beta} \) be the number of beta particles emitted. The relationship can be expressed as:
\[ 90 - 2N_{\alpha} + N_{\beta} = 82 \]
- Substituting \( N_{\alpha} = 5 \):
\[ 90 - 2(5) + N_{\beta} = 82 \]
- Simplifying this gives:
\[ 90 - 10 + N_{\beta} = 82 \implies N_{\beta} = 82 - 80 = 2 \]

6. **Conclusion:**
- The number of alpha particles emitted, \( N_{\alpha} = 5 \)
- The number of beta particles emitted, \( N_{\beta} = 2 \)

### Final Answer:
- The correct statements are:
- \( N_{\alpha} = 5 \)
- \( N_{\beta} = 2 \)