What are the respective number of `alpha` and `beta`-particles emitted in the following radioactive decay?

Options

A. (a) 6 and 8
B. (b) 6 and 6
C. (c) 8 and 8
D. (d) 8 and 6

Explanation

To solve the problem of determining the respective number of alpha and beta particles emitted in the given radioactive decay, we will follow these steps:

### Step 1: Understand the decay equation
The decay is represented as:
\[ \text{ }^{200}_{90}X \rightarrow \text{ }^{168}_{80}Y \]
Here, \(X\) is the parent nucleus with a mass number of 200 and an atomic number of 90, and \(Y\) is the daughter nucleus with a mass number of 168 and an atomic number of 80.

### Step 2: Define the number of alpha and beta particles
Let:
- \(n\) = number of alpha particles emitted
- \(m\) = number of beta particles emitted

### Step 3: Write the equations for atomic number and mass number
1. **For Atomic Number**:
The atomic number decreases by 2 for each alpha particle emitted and increases by 1 for each beta particle emitted. Thus, we can write:
\[ 90 - 2n + m = 80 \quad \text{(Equation 1)} \]

2. **For Mass Number**:
The mass number decreases by 4 for each alpha particle emitted and remains unchanged for beta particles. Thus, we can write:
\[ 200 - 4n = 168 \quad \text{(Equation 2)} \]

### Step 4: Solve Equation 2 for \(n\)
From Equation 2:
\[ 200 - 4n = 168 \]
Rearranging gives:
\[ 4n = 200 - 168 \]
\[ 4n = 32 \]
\[ n = \frac{32}{4} = 8 \]

### Step 5: Substitute \(n\) into Equation 1 to find \(m\)
Now, substitute \(n = 8\) into Equation 1:
\[ 90 - 2(8) + m = 80 \]
This simplifies to:
\[ 90 - 16 + m = 80 \]
\[ 74 + m = 80 \]
\[ m = 80 - 74 = 6 \]

### Step 6: Conclusion
Thus, the respective number of alpha and beta particles emitted are:
- Number of alpha particles (\(n\)) = 8
- Number of beta particles (\(m\)) = 6

### Final Answer:
The respective number of alpha and beta particles emitted are 8 and 6.

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