How many alpha and beta particles are emitted when uranium `._(92)^(238)U` decays to lead `._(82)^(206)Pb` ?

Answer

To determine how many alpha and beta particles are emitted when uranium-238 decays to lead-206, we can follow these steps:

### Step 1: Write the decay equation
The decay can be represented as:
\[ _{92}^{238}U \rightarrow _{82}^{206}Pb + \text{alpha particles} + \text{beta particles} \]

### Step 2: Identify the mass and atomic numbers
- The atomic number of uranium (U) is 92 and its mass number is 238.
- The atomic number of lead (Pb) is 82 and its mass number is 206.

### Step 3: Set up the equations for mass and charge conservation
Let:
- \( n \) = number of alpha particles emitted
- \( y \) = number of beta particles emitted

The decay can be expressed in terms of mass and charge conservation:
1. For mass numbers:
\[ 238 = 206 + 4n + 0y \]
2. For atomic numbers:
\[ 92 = 82 + n - y \]

### Step 4: Solve the mass number equation
From the mass number equation:
\[ 238 = 206 + 4n \]
Subtract 206 from both sides:
\[ 32 = 4n \]
Divide by 4:
\[ n = 8 \]
This means 8 alpha particles are emitted.

### Step 5: Solve the atomic number equation
Now, substitute \( n = 8 \) into the atomic number equation:
\[ 92 = 82 + 8 - y \]
Simplifying gives:
\[ 92 = 90 - y \]
Rearranging gives:
\[ y = 90 - 92 \]
Thus:
\[ y = 2 \]
This means 2 beta particles are emitted.

### Final Answer
Therefore, when uranium-238 decays to lead-206, it emits:
- **8 alpha particles**
- **2 beta particles**

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