On decay of `""_(92)U^(238)` to a stable nucleus `""_(82)Pb^(206)` , the number of `alpha` and `beta` particles emitted ar

Options

A. 9,6
B. 8,6
C. 7,9
D. 10,6

Explanation

To solve the problem of determining the number of alpha and beta particles emitted during the decay of Uranium-238 to Lead-206, we can follow these steps:

### Step 1: Write the decay equation
The decay process can be represented as:
\[ _{92}^{238}U \rightarrow _{82}^{206}Pb + x \, _{2}^{4}He + y \, _{-1}^{0}e \]
where \(x\) is the number of alpha particles emitted and \(y\) is the number of beta particles emitted.

### Step 2: Balance the atomic numbers
The total atomic number on the left side (Uranium) is 92, and on the right side (Lead) it is 82. Each alpha particle contributes 2 to the atomic number, and each beta particle contributes -1. Therefore, we can set up the equation:
\[ 92 = 82 + 2x - y \]
This is our first equation.

### Step 3: Balance the mass numbers
The total mass number on the left side is 238, and on the right side it is 206. Each alpha particle contributes 4 to the mass number, and beta particles do not contribute to the mass number. Thus, we can set up the second equation:
\[ 238 = 206 + 4x + 0y \]
This simplifies to:
\[ 238 = 206 + 4x \]
Rearranging gives us:
\[ 4x = 238 - 206 \]
\[ 4x = 32 \]
\[ x = \frac{32}{4} = 8 \]

### Step 4: Substitute \(x\) back to find \(y\)
Now that we have \(x = 8\), we can substitute this value back into the first equation to find \(y\):
\[ 92 = 82 + 2(8) - y \]
\[ 92 = 82 + 16 - y \]
\[ 92 = 98 - y \]
Rearranging gives us:
\[ y = 98 - 92 \]
\[ y = 6 \]

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

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

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