The number of `alpha and beta`-particles emitted during the transformation of `._(90)Th^(232) " to " ._(82)Pb^(208)` are respectively

Options

A. 4, 2
B. 2, 2
C. 8, 6
D. 6, 4

Explanation

To solve the problem of determining the number of alpha and beta particles emitted during the transformation of \( _{90}^{232}\text{Th} \) to \( _{82}^{208}\text{Pb} \), we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Initial and Final Nuclei**:
- Initial nucleus: \( _{90}^{232}\text{Th} \)
- Final nucleus: \( _{82}^{208}\text{Pb} \)

2. **Calculate the Mass Change**:
- Mass change = Initial mass - Final mass
- Mass change = \( 232 - 208 = 24 \)

3. **Determine the Number of Alpha Particles Emitted**:
- Each alpha particle has a mass of 4 (2 protons and 2 neutrons).
- Number of alpha particles emitted = Mass change / Mass of one alpha particle
- Number of alpha particles = \( 24 / 4 = 6 \)

4. **Calculate the Change in Atomic Number**:
- Each alpha particle emitted decreases the atomic number by 2.
- If \( n \) is the number of alpha particles emitted, the change in atomic number due to alpha decay is \( 2n \).
- Therefore, the new atomic number after emitting 6 alpha particles = \( 90 - 2 \times 6 = 90 - 12 = 78 \).

5. **Determine the Final Atomic Number**:
- We need to reach \( 82 \) (the atomic number of lead).
- The difference in atomic number = Final atomic number - New atomic number after alpha decay
- Difference = \( 82 - 78 = 4 \)

6. **Calculate the Number of Beta Particles Emitted**:
- Each beta particle increases the atomic number by 1.
- Therefore, to increase the atomic number from 78 to 82, we need to emit 4 beta particles.

7. **Final Count of Particles Emitted**:
- Number of alpha particles emitted = 6
- Number of beta particles emitted = 4

### Final Answer:
The number of alpha and beta particles emitted during the transformation of \( _{90}^{232}\text{Th} \) to \( _{82}^{208}\text{Pb} \) are respectively **6 alpha particles and 4 beta particles**.