A radioactive disintegration of `""_(90)Th^(232)` yields `""_(82)Pb^(208)` in the end . The number of `alpha` and `beta` -particle emitted will be

Options

A. `6 alpha` and `6 beta`
B. `5 alpha` and `5 beta`
C. `6 alpha` and `4 beta`
D. `4 alpha` and `6 beta`

Explanation

To determine the number of alpha and beta particles emitted during the radioactive disintegration of \( _{90}^{232}\text{Th} \) to \( _{82}^{208}\text{Pb} \), we can follow these steps:

### Step 1: Write the initial and final nuclear equations
The initial nucleus is \( _{90}^{232}\text{Th} \) and the final nucleus is \( _{82}^{208}\text{Pb} \). We need to find the number of alpha (\( \alpha \)) and beta (\( \beta \)) particles emitted.

### Step 2: Define variables for alpha and beta particles
Let:
- \( X \) = number of alpha particles emitted
- \( Y \) = number of beta particles emitted

### Step 3: Set up the equations based on conservation of mass and charge
1. **Mass number conservation**:
\[ 232 - 4X + Y = 208 \]
(Each alpha particle reduces the mass number by 4, and each beta particle does not change the mass number.)

2. **Atomic number conservation**:
\[ 90 - 2X + Y = 82 \]
(Each alpha particle reduces the atomic number by 2, and each beta particle increases it by 1.)

### Step 4: Solve the mass number equation
From the mass number equation:
\[ 232 - 4X + Y = 208 \]
Rearranging gives:
\[ Y = 208 - 232 + 4X = 4X - 24 \quad \text{(Equation 1)} \]

### Step 5: Solve the atomic number equation
From the atomic number equation:
\[ 90 - 2X + Y = 82 \]
Rearranging gives:
\[ Y = 82 - 90 + 2X = 2X - 8 \quad \text{(Equation 2)} \]

### Step 6: Set the two equations for \( Y \) equal to each other
From Equation 1 and Equation 2:
\[ 4X - 24 = 2X - 8 \]
Solving for \( X \):
\[ 4X - 2X = 24 - 8 \]
\[ 2X = 16 \]
\[ X = 8 \]

### Step 7: Substitute \( X \) back to find \( Y \)
Using \( X = 8 \) in Equation 2:
\[ Y = 2(8) - 8 = 16 - 8 = 8 \]

### Conclusion
The number of alpha particles emitted \( (X) \) is 8 and the number of beta particles emitted \( (Y) \) is 8.

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