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 determining the number of alpha and beta particles emitted during the disintegration of the nucleus of \( ^{238}U_{92} \) to \( ^{206}U_{82} \), we can follow these steps:

### Step 1: Write down the nuclear reaction
The disintegration can be represented as:
\[ ^{238}U_{92} \rightarrow ^{206}U_{82} + x \, ^{4}He + y \, \beta^- \]
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 atomic number must be balanced on both sides of the equation. The atomic number of uranium is 92, and after disintegration, it becomes 82. The equation for the atomic numbers is:
\[ 92 = 82 + 2x - y \tag{1} \]

### Step 3: Balance the mass numbers
The mass number must also be balanced. The mass number of uranium is 238, and after disintegration, it becomes 206. The equation for the mass numbers is:
\[ 238 = 206 + 4x \tag{2} \]

### Step 4: Solve the mass number equation
From equation (2):
\[ 238 - 206 = 4x \]
\[ 32 = 4x \]
\[ x = \frac{32}{4} = 8 \]
So, \( x = 8 \), meaning 8 alpha particles are emitted.

### Step 5: Substitute \( x \) into the atomic number equation
Now substitute \( x = 8 \) into equation (1):
\[ 92 = 82 + 2(8) - y \]
\[ 92 = 82 + 16 - y \]
\[ 92 = 98 - y \]
\[ y = 98 - 92 = 6 \]
So, \( y = 6 \), meaning 6 beta particles are emitted.

### Conclusion
The final answer is:
- Number of alpha particles emitted: \( 8 \)
- Number of beta particles emitted: \( 6 \)