`._(89)Ac^(231)` after emission of some `alpha and beta` particles gives `._(82) Pb^(207)` . The number of such `alpha and beta` - particles are respectively

Options

A. 5,7
B. 6,5
C. 7,5
D. 5,6

Answer

To solve the problem, we need to determine how many alpha (α) and beta (β) particles are emitted when the element with atomic number 89 and mass number 231 (Actinium, Ac) transforms into the element with atomic number 82 and mass number 207 (Lead, Pb).

### Step-by-Step Solution:

1. **Identify the Initial and Final States:**
- Initial: Ac (Atomic number = 89, Mass number = 231)
- Final: Pb (Atomic number = 82, Mass number = 207)

2. **Calculate the Change in Mass Number:**
- Initial mass number = 231
- Final mass number = 207
- Change in mass number = 231 - 207 = 24

3. **Determine the Contribution of Alpha Particles:**
- Each alpha particle (α) emitted reduces the mass number by 4.
- Let \( x \) be the number of alpha particles emitted. The equation for mass number becomes:
\[ 231 - 4x = 207 \]
- Rearranging gives:
\[ 4x = 231 - 207 = 24 \]
- Solving for \( x \):
\[ x = \frac{24}{4} = 6 \]
- Thus, 6 alpha particles are emitted.

4. **Calculate the Change in Atomic Number:**
- Initial atomic number = 89
- Final atomic number = 82
- Change in atomic number = 89 - 82 = 7

5. **Determine the Contribution of Beta Particles:**
- Each alpha particle reduces the atomic number by 2, and each beta particle (β) increases the atomic number by 1.
- Let \( y \) be the number of beta particles emitted. The equation for atomic number becomes:
\[ 89 - 2x + y = 82 \]
- Substituting \( x = 6 \):
\[ 89 - 2(6) + y = 82 \]
\[ 89 - 12 + y = 82 \]
\[ y = 82 - 77 = 5 \]
- Thus, 5 beta particles are emitted.

6. **Final Result:**
- The number of alpha particles emitted is 6.
- The number of beta particles emitted is 5.

### Conclusion:
The number of alpha and beta particles emitted are respectively 6 and 5.