`Ac_89^231 gives Pb_82^207` after emission of some `alpha and beta`-particles. The number of such  and -particles are respectively:

Options

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

Answer

To solve the problem of how many alpha (α) and beta (β) particles are emitted when Actinium-231 (Ac-89^231) decays to Lead-207 (Pb-82^207), we need to balance both the mass number and the atomic number of the reactants and products.

### Step 1: Write down the known values
- **Initial nucleus:** Actinium-231 (Ac-89^231)
- Atomic number (Z) = 89
- Mass number (A) = 231
- **Final nucleus:** Lead-207 (Pb-82^207)
- Atomic number (Z) = 82
- Mass number (A) = 207

### Step 2: Set up the equations for mass and atomic numbers
1. **Mass number equation:**
\[ A_{\text{initial}} = A_{\text{final}} + 4n + 0m \]
where \( n \) is the number of alpha particles and \( m \) is the number of beta particles. The mass number of an alpha particle is 4, and that of a beta particle is 0.

Substituting the known values:
\[ 231 = 207 + 4n \]

2. **Atomic number equation:**
\[ Z_{\text{initial}} = Z_{\text{final}} + 2n - m \]
where the atomic number of an alpha particle is 2, and that of a beta particle is -1.

Substituting the known values:
\[ 89 = 82 + 2n - m \]

### Step 3: Solve the mass number equation
From the mass number equation:
\[ 231 - 207 = 4n \]
\[ 24 = 4n \]
\[ n = \frac{24}{4} = 6 \]

### Step 4: Substitute \( n \) into the atomic number equation
Now substitute \( n = 6 \) into the atomic number equation:
\[ 89 = 82 + 2(6) - m \]
\[ 89 = 82 + 12 - m \]
\[ 89 = 94 - m \]
Rearranging gives:
\[ m = 94 - 89 = 5 \]

### Step 5: Conclusion
Thus, the number of alpha (α) particles emitted is \( n = 6 \) and the number of beta (β) particles emitted is \( m = 5 \).

### Final Answer
The number of alpha and beta particles emitted are respectively:
- **Alpha particles (α): 6**
- **Beta particles (β): 5**

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