In the nuclear reaction `U_92^238 to Pb_82^206`, the number of alpha and beta particles decayed are:

Options

A. `4 alpha, 3 beta`
B. `8alpha, 6beta`
C. `6alpha,4beta`
D. `7alpha,5beta`

Answer

To solve the nuclear reaction \( U_{92}^{238} \to Pb_{82}^{206} \) and find the number of alpha and beta particles that have decayed, we will follow these steps:

### Step 1: Identify the initial and final atomic and mass numbers
- The initial nucleus is Uranium (U) with atomic number \( Z = 92 \) and mass number \( A = 238 \).
- The final nucleus is Lead (Pb) with atomic number \( Z = 82 \) and mass number \( A = 206 \).

### Step 2: Set up equations for alpha and beta decay
Let:
- \( x \) = number of alpha particles emitted
- \( y \) = number of beta particles emitted

**Alpha particles** have:
- Atomic number = 2
- Mass number = 4

**Beta particles** have:
- Atomic number = -1
- Mass number = 0

### Step 3: Write the equations based on conservation laws
1. **Conservation of Atomic Number**:
\[ 92 = 82 + 2x - y \quad \text{(Equation 1)} \]

2. **Conservation of Mass Number**:
\[ 238 = 206 + 4x + 0y \quad \text{(Equation 2)} \]

### Step 4: Simplify the equations
From Equation 2:
\[ 238 - 206 = 4x \implies 32 = 4x \implies x = \frac{32}{4} = 8 \]

### Step 5: Substitute \( x \) into Equation 1
Now substitute \( x = 8 \) into Equation 1:
\[ 92 = 82 + 2(8) - y \]
\[ 92 = 82 + 16 - y \]
\[ 92 = 98 - y \]
\[ y = 98 - 92 = 6 \]

### Step 6: Conclusion
Thus, the number of alpha particles \( x \) is 8 and the number of beta particles \( y \) is 6.

### Final Answer:
- Number of alpha particles = 8
- Number of beta particles = 6

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