The number of `alpha and beta` particles emitted in the nucleur disintegration series `_90^228 Th to _83^212 Bi` are:

Options

A. `8alpha,1beta`
B. `4alpha,7beta`
C. `3alpha,7beta`
D. `4alpha,1beta`

Answer

To solve the problem of determining the number of alpha and beta particles emitted in the nuclear disintegration series from \(_{90}^{228}Th\) to \(_{83}^{212}Bi\), we can follow these steps:

### Step 1: Understand the Conservation Laws
In nuclear reactions, both mass number (A) and atomic number (Z) are conserved. This means that the total mass number before the disintegration must equal the total mass number after the disintegration, and the same applies to the atomic number.

### Step 2: Set Up the Equations
Let:
- \(X\) = number of alpha particles emitted
- \(Y\) = number of beta particles emitted

The mass number before disintegration is 228 (for Th), and after disintegration, it is 212 (for Bi). Each alpha particle has a mass number of 4, and beta particles have a mass number of 0.

**Mass Number Conservation:**
\[ 228 = 212 + 4X \]

**Atomic Number Conservation:**
The atomic number of thorium (Th) is 90, and that of bismuth (Bi) is 83. Each alpha particle decreases the atomic number by 2, and each beta particle increases it by 1.

\[ 90 = 83 + 2X - Y \]

### Step 3: Solve the Mass Number Equation
From the mass number equation:
\[ 228 = 212 + 4X \]
Subtract 212 from both sides:
\[ 16 = 4X \]
Divide by 4:
\[ X = 4 \]

### Step 4: Solve the Atomic Number Equation
Now substitute \(X = 4\) into the atomic number equation:
\[ 90 = 83 + 2(4) - Y \]
This simplifies to:
\[ 90 = 83 + 8 - Y \]
Combine like terms:
\[ 90 = 91 - Y \]
Rearranging gives:
\[ Y = 91 - 90 = 1 \]

### Step 5: Conclusion
Thus, the number of alpha particles emitted is \(4\) and the number of beta particles emitted is \(1\).

### Final Answer
The number of alpha and beta particles emitted in the nuclear disintegration series from \(_{90}^{228}Th\) to \(_{83}^{212}Bi\) is:
- **Alpha particles (X)**: 4
- **Beta particles (Y)**: 1