`""_(90)^(232)Th` disintegrates to `""_(y)^(x)Pb` by emitting six `alpha`-and four `beta`-particles. Find x and y

Answer

To solve the problem of the disintegration of Thorium-232 \((^{232}_{90}Th)\) into Lead \((^{x}_{y}Pb)\) by emitting six alpha particles and four beta particles, we will follow these steps:

### Step 1: Understand the effect of alpha and beta emissions

1. **Alpha Particle Emission**: Each alpha particle emitted decreases the atomic mass by 4 and the atomic number by 2.
2. **Beta Particle Emission**: Each beta particle emitted does not change the atomic mass but increases the atomic number by 1.

### Step 2: Calculate the changes due to alpha emissions

- **Initial Atomic Mass and Atomic Number**:
- Atomic Mass of Thorium (Th) = 232
- Atomic Number of Thorium (Th) = 90

- **Alpha Emissions**:
- Total mass decrease from 6 alpha particles = \(6 \times 4 = 24\)
- Total atomic number decrease from 6 alpha particles = \(6 \times 2 = 12\)

- **New Atomic Mass and Atomic Number after Alpha Emissions**:
- New Atomic Mass = \(232 - 24 = 208\)
- New Atomic Number = \(90 - 12 = 78\)

### Step 3: Calculate the changes due to beta emissions

- **Beta Emissions**:
- Total atomic number increase from 4 beta particles = \(4 \times 1 = 4\)

- **Final Atomic Number after Beta Emissions**:
- Final Atomic Number = \(78 + 4 = 82\)
- The atomic mass remains the same at 208.

### Step 4: Write the final result

Thus, after the disintegration process, we find:
- The atomic mass \(x = 208\)
- The atomic number \(y = 82\)

Therefore, the final answer is:
\[ x = 208, \quad y = 82 \]