How may `alpha-` and `beta-` particles will be emitted when `._(90)Th^(232)` changes into `._(82)Pb^(208)`?

Answer

To determine how many alpha and beta particles will be emitted when \( _{90}^{232}\text{Th} \) (Thorium-232) transforms into \( _{82}^{208}\text{Pb} \) (Lead-208), we can follow these steps:

### Step 1: Understand the decay process
Thorium-232 decays into Lead-208 through a series of alpha and beta emissions. An alpha particle is represented as \( _{2}^{4}\text{He} \) (mass number 4, atomic number 2), and a beta particle is represented as \( _{-1}^{0}\text{e} \) (mass number 0, atomic number -1).

### Step 2: Set up the equations
We need to balance both the mass numbers and atomic numbers. Let \( x \) be the number of alpha particles emitted and \( y \) be the number of beta particles emitted.

1. **Mass Number Equation:**
\[ 232 = 208 + 4x \]

2. **Atomic Number Equation:**
\[ 90 = 82 + 2x - y \]

### Step 3: Solve the mass number equation
From the mass number equation:
\[ 232 = 208 + 4x \]
Subtract 208 from both sides:
\[ 232 - 208 = 4x \]
\[ 24 = 4x \]
Now, divide by 4:
\[ x = 6 \]

### Step 4: Solve the atomic number equation
Now substitute \( x = 6 \) into the atomic number equation:
\[ 90 = 82 + 2(6) - y \]
Calculate \( 2(6) \):
\[ 90 = 82 + 12 - y \]
Combine the terms:
\[ 90 = 94 - y \]
Now, isolate \( y \):
\[ y = 94 - 90 \]
\[ y = 4 \]

### Conclusion
The number of alpha particles emitted is \( x = 6 \) and the number of beta particles emitted is \( y = 4 \).

### Final Answer
- Alpha particles emitted: 6
- Beta particles emitted: 4

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