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 (α) and beta (β) particles emitted during a radioactive decay, we can follow these steps:

### Step 1: Define Variables
Let:
- \( n \) = number of alpha particles emitted
- \( m \) = number of beta particles emitted

### Step 2: Understand the Effects of Emissions
1. **Alpha particles**: Each alpha particle emitted decreases the atomic number (Z) by 2 and the mass number (A) by 4.
2. **Beta particles**: Each beta particle emitted increases the atomic number (Z) by 1 but does not affect the mass number (A).

### Step 3: Set Up Equations
From the problem, we can derive two equations based on the changes in atomic and mass numbers.

1. **For Atomic Number (Z)**:
\[ Z_{\text{final}} = Z_{\text{initial}} - 2n + m \]
Given \( Z_{\text{final}} = 80 \) and \( Z_{\text{initial}} = 90 \):
\[ 80 = 90 - 2n + m \quad \text{(Equation 1)} \]

2. **For Mass Number (A)**:
\[ A_{\text{final}} = A_{\text{initial}} - 4n \]
Given \( A_{\text{final}} = 168 \) and \( A_{\text{initial}} = 200 \):
\[ 168 = 200 - 4n \quad \text{(Equation 2)} \]

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

### Step 5: Substitute n into Equation 1 to Solve for m
Substituting \( n = 8 \) into Equation 1:
\[ 80 = 90 - 2(8) + m \]
\[ 80 = 90 - 16 + m \]
\[ 80 = 74 + m \]
\[ m = 80 - 74 = 6 \]

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

### Final Answer:
The answer is 8 alpha particles and 6 beta particles.

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