When the nucleus of `.^(238)U_(92)` disintegrates to give one nuclues of `.^(206)U_(82)`, the number of `alpha-` particles emitted and the number of `beta-` particles emitted is

Options

A. 4 and 8 respectively
B. 6 and 8 respectively
C. 8 and 6 respectively
D. 8 and 10 resepctively

Answer

To solve the problem of how many alpha and beta particles are emitted when the nucleus of \( ^{238}_{92}U \) disintegrates to give one nucleus of \( ^{206}_{82}U \), we can follow these steps:

### Step 1: Identify the initial and final nuclei
- The initial nucleus is \( ^{238}_{92}U \) (Uranium-238).
- The final nucleus is \( ^{206}_{82}U \) (Uranium-206).

### Step 2: Write the equations for conservation of atomic number and mass number
- Let \( x \) be the number of alpha particles emitted.
- Let \( y \) be the number of beta particles emitted.

#### Conservation of Atomic Number:
The atomic number on the left side (initial) must equal the atomic number on the right side (final):
\[ 92 = 82 + 2x - y \quad \text{(Equation 1)} \]

#### Conservation of Mass Number:
The mass number on the left side (initial) must equal the mass number on the right side (final):
\[ 238 = 206 + 4x + 0y \quad \text{(Equation 2)} \]

### Step 3: Simplify the equations
From Equation 2:
\[ 238 - 206 = 4x \implies 32 = 4x \implies x = 8 \]

### Step 4: 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 5: Conclusion
Thus, the number of alpha particles emitted \( x \) is 8, and the number of beta particles emitted \( y \) is 6.

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

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