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 \( U_{92}^{238} \rightarrow Pb_{82}^{206} \) and determine the number of alpha and beta particles that have decayed, we can follow these steps:

### Step 1: Understand the Reaction
We start with Uranium-238 (U) and end with Lead-206 (Pb). We need to find how many alpha (\( \alpha \)) and beta (\( \beta \)) particles are emitted during this decay process.

### Step 2: Define Variables
Let:
- \( x \) = number of alpha particles emitted
- \( y \) = number of beta particles emitted

### Step 3: Write the Mass Number Equation
The mass number (A) must be conserved in the reaction. The mass number of Uranium-238 is 238, and the mass number of Lead-206 is 206. Each alpha particle has a mass number of 4, and beta particles have a mass number of 0. Therefore, we can write the equation:

\[ 238 = 206 + 4x \]

### Step 4: Solve for \( x \)
Rearranging the equation gives:

\[ 238 - 206 = 4x \]
\[ 32 = 4x \]
\[ x = \frac{32}{4} = 8 \]

So, \( x = 8 \). This means 8 alpha particles are emitted.

### Step 5: Write the Atomic Number Equation
Next, we need to balance the atomic numbers (Z). The atomic number of Uranium is 92, and that of Lead is 82. Each alpha particle decreases the atomic number by 2, and each beta particle increases it by 1. Therefore, we can write the equation:

\[ 92 = 82 + 2x - y \]

### Step 6: Substitute \( x \) and Solve for \( y \)
Substituting \( x = 8 \) into the equation gives:

\[ 92 = 82 + 2(8) - y \]
\[ 92 = 82 + 16 - y \]
\[ 92 = 98 - y \]
\[ y = 98 - 92 = 6 \]

So, \( y = 6 \). This means 6 beta particles are emitted.

### Final Answer
The number of alpha particles decayed is 8, and the number of beta particles decayed is 6.

### Summary
- Number of alpha particles (\( x \)) = 8
- Number of beta particles (\( y \)) = 6

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