The number of `alpha and beta`- particles emitted in the nuclear reaction `Th_90^228 to Th_90^228` are respectively

Options

A. 4, 1
B. 3, 7
C. 8, 1
D. 4, 7

Answer

To determine the number of alpha and beta particles emitted in the nuclear reaction from Thorium-228 to Bismuth-212, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Reaction**:
The nuclear reaction is given as:
\[ \text{Th}_{90}^{228} \rightarrow \text{Bi}_{83}^{212} \]

2. **Define Variables**:
Let \( x \) be the number of alpha particles emitted, and \( y \) be the number of beta particles emitted.
- An alpha particle has a mass number of 4 and an atomic number of 2.
- A beta particle has a mass number of 0 and an atomic number of -1.

3. **Balance Mass Numbers**:
The total mass number on the left side must equal the total mass number on the right side:
\[ 228 = 212 + 4x \]
Rearranging gives:
\[ 4x = 228 - 212 = 16 \]
Dividing both sides by 4:
\[ x = 4 \]

4. **Balance Atomic Numbers**:
The total atomic number on the left side must equal the total atomic number on the right side:
\[ 90 = 83 + 2x - y \]
Substituting \( x = 4 \):
\[ 90 = 83 + 2(4) - y \]
Simplifying gives:
\[ 90 = 83 + 8 - y \]
\[ 90 = 91 - y \]
Rearranging gives:
\[ y = 1 \]

5. **Conclusion**:
The number of alpha particles emitted is \( x = 4 \) and the number of beta particles emitted is \( y = 1 \).

### Final Answer:
The number of alpha and beta particles emitted in the reaction is:
- Alpha particles: 4
- Beta particles: 1

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