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

Options

A. 12,6
B. 10, 4
C. 8, 6
D. 8,8

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: Identify the initial and final elements
The decay process can be represented as:
\[ _{92}^{238}U \rightarrow _{82}^{206}Pb + \text{(alpha and beta particles)} \]

### Step 2: Calculate the change in mass number (A)
The initial mass number (A) of uranium-238 is 238, and the final mass number (A) of lead-206 is 206.
To find the change in mass number:
\[ \Delta A = A_{\text{initial}} - A_{\text{final}} = 238 - 206 = 32 \]

### Step 3: Determine the number of alpha particles emitted
Each alpha particle has a mass number of 4 (since it is represented as \( _{2}^{4}He \)). To find the number of alpha particles emitted:
\[ \text{Number of alpha particles} = \frac{\Delta A}{4} = \frac{32}{4} = 8 \]

### Step 4: Calculate the change in atomic number (Z)
The initial atomic number (Z) of uranium-238 is 92, and the final atomic number (Z) of lead-206 is 82.
To find the change in atomic number:
\[ \Delta Z = Z_{\text{initial}} - Z_{\text{final}} = 92 - 82 = 10 \]

### Step 5: Determine the effect of alpha particles on atomic number
Each alpha particle decreases the atomic number by 2. Therefore, for 8 alpha particles:
\[ \text{Change in atomic number due to alpha particles} = 8 \times 2 = 16 \]

### Step 6: Calculate the remaining change in atomic number
Now, we need to find out how many beta particles are emitted to account for the remaining change in atomic number.
The total change in atomic number from uranium to lead is 10, while the change due to alpha particles is 16. Thus, we need to find how many beta particles (which increase the atomic number by 1) are needed to compensate for this difference:
\[ \text{Remaining change} = 16 - 10 = 6 \]
This means that 6 beta particles must be emitted.

### Conclusion
In summary, when uranium-238 decays to lead-206, the number of particles emitted is:
- **Alpha particles**: 8
- **Beta particles**: 6

### Final Answer
8 alpha particles and 6 beta particles are emitted.

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