In the nuclear reaction `U_92^238 to Pb_82^206 + xHe_2^4 + y beta_(-1)^0` the value of x and y are respectively ………

Options

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

Answer

To solve the nuclear reaction \( U_{92}^{238} \to Pb_{82}^{206} + xHe_{2}^{4} + y\beta_{-1}^{0} \) and find the values of \( x \) and \( y \), we will use the conservation laws for atomic mass and atomic number.

### Step-by-Step Solution:

1. **Identify the components of the reaction:**
- Reactant: Uranium \( U_{92}^{238} \)
- Products: Lead \( Pb_{82}^{206} \), \( x \) alpha particles \( He_{2}^{4} \), and \( y \) beta particles \( \beta_{-1}^{0} \)

2. **Apply the conservation of atomic mass:**
- The total atomic mass on the left side (reactants) must equal the total atomic mass on the right side (products).
- The mass of uranium is 238, and the mass of lead is 206. Each alpha particle has a mass of 4.
- Therefore, the equation for mass conservation is:
\[ 238 = 206 + 4x + 0y \]
- Simplifying this gives:
\[ 4x = 238 - 206 = 32 \]
- Solving for \( x \):
\[ x = \frac{32}{4} = 8 \]

3. **Apply the conservation of atomic number:**
- The total atomic number on the left side must equal the total atomic number on the right side.
- The atomic number of uranium is 92, and the atomic number of lead is 82. Each alpha particle contributes 2 to the atomic number, and each beta particle contributes -1.
- Therefore, the equation for atomic number conservation is:
\[ 92 = 82 + 2x - y \]
- Substituting the value of \( x \) we found:
\[ 92 = 82 + 2(8) - y \]
- Simplifying this gives:
\[ 92 = 82 + 16 - y \]
\[ 92 = 98 - y \]
- Rearranging to solve for \( y \):
\[ y = 98 - 92 = 6 \]

4. **Final values:**
- We have found \( x = 8 \) and \( y = 6 \).

### Conclusion:
The values of \( x \) and \( y \) are respectively \( 8 \) and \( 6 \).