In a radioactive decay chain, the initial nucleus is `overset(232)(90).^(Th)`. At the end there are 6 `alpha`-particles and 4 `beta`-particles which are emitted. If the end nucleus is `overset(A)(Z).^(X),A` and Z are given by :

Options

A. `A=208 , Z=80`
B. `A=202 , Z=80`
C. `A=208 , Z=82`
D. `A=200 , Z=81`

Explanation

To solve the problem, we need to analyze the decay chain of the initial nucleus, which is thorium-232 (represented as \( \overset{232}{90}\text{Th} \)). We will track the changes in the mass number (A) and atomic number (Z) as alpha and beta particles are emitted.

### Step-by-Step Solution:

1. **Identify the Initial Values:**
- The initial nucleus is thorium-232, which has:
- Mass number (A) = 232
- Atomic number (Z) = 90

2. **Effect of Alpha Particles:**
- Each alpha particle emission decreases the atomic number (Z) by 2 and the mass number (A) by 4.
- Since 6 alpha particles are emitted:
- Change in Z due to alpha particles = \( 6 \times 2 = 12 \)
- Change in A due to alpha particles = \( 6 \times 4 = 24 \)

3. **Calculate New Values After Alpha Emission:**
- New atomic number (Z'):
\[ Z' = Z - 12 = 90 - 12 = 78 \]
- New mass number (A'):
\[ A' = A - 24 = 232 - 24 = 208 \]

4. **Effect of Beta Particles:**
- Each beta particle emission increases the atomic number (Z) by 1, while the mass number (A) remains unchanged.
- Since 4 beta particles are emitted:
- Change in Z due to beta particles = \( 4 \times 1 = 4 \)

5. **Calculate Final Values After Beta Emission:**
- Final atomic number (Z_final):
\[ Z_{\text{final}} = Z' + 4 = 78 + 4 = 82 \]
- Final mass number (A_final):
\[ A_{\text{final}} = A' = 208 \]

6. **Final Result:**
- The end nucleus is represented as \( \overset{A_{\text{final}}}{Z_{\text{final}}}X \).
- Therefore, we have:
- \( A = 208 \)
- \( Z = 82 \)

### Conclusion:
The end nucleus after the decay chain is \( \overset{208}{82}X \).