`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 determining the number of alpha (α) and beta (β) particles emitted during the decay of Actinium-231 (Ac-231) to Lead-207 (Pb-207), we can follow these steps:

### Step 1: Write down the nuclear reaction
The nuclear reaction can be represented as:
\[ \text{Ac}_{89}^{231} \rightarrow \text{Pb}_{82}^{207} + n \, \text{α} + y \, \text{β} \]
where \( n \) is the number of alpha particles emitted and \( y \) is the number of beta particles emitted.

### Step 2: Balance the mass numbers
The mass number on the left side is 231 (from Ac-231), and on the right side, it is 207 (from Pb-207) plus the contributions from the alpha particles. Each alpha particle has a mass number of 4. Therefore, we can set up the equation:
\[ 231 = 207 + 4n \]
Rearranging this gives:
\[ 4n = 231 - 207 \]
\[ 4n = 24 \]
Dividing both sides by 4:
\[ n = 6 \]

### Step 3: Balance the atomic numbers
Next, we balance the atomic numbers. The atomic number of Ac-231 is 89, and for Pb-207 it is 82. The contribution from the alpha particles (each with an atomic number of 2) and beta particles (each with an atomic number of -1) can be expressed as:
\[ 89 = 82 + 2n - y \]
Substituting \( n = 6 \) into the equation:
\[ 89 = 82 + 2(6) - y \]
This simplifies to:
\[ 89 = 82 + 12 - y \]
\[ 89 = 94 - y \]
Rearranging gives:
\[ y = 94 - 89 \]
\[ y = 5 \]

### Conclusion
Thus, the number of alpha particles emitted is \( n = 6 \) and the number of beta particles emitted is \( y = 5 \).

### Final Answer
The number of alpha and beta particles emitted are respectively: **6 and 5**.

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