What is the respective number of `alpha` and `beta` particles emitted in the following radioactive decay? `._(90)X^(200) rarr ._(80)Y^(168)`

Options

A. 6 and 8
B. 6 and 6
C. 8 and 8
D. 8 and 6

Explanation

To solve the problem of determining the respective number of alpha and beta particles emitted in the radioactive decay process from \(_{90}X^{200}\) to \(_{80}Y^{168}\), we can follow these steps:

### Step 1: Understand the decay process
The decay process involves a parent nucleus \(_{90}X^{200}\) transforming into a daughter nucleus \(_{80}Y^{168}\). In this process, alpha and beta particles are emitted.

### Step 2: Define the particles
- An **alpha particle** (\(_{2}^{4}He\)) consists of 2 protons and 2 neutrons, contributing 4 to the mass number and 2 to the atomic number.
- A **beta particle** is an electron (\(e^-\)) emitted from the nucleus, which has negligible mass and a charge of -1, contributing 0 to the mass number and -1 to the atomic number.

### Step 3: Set up equations
Let \(X\) be the number of alpha particles emitted and \(Y\) be the number of beta particles emitted.

1. **Mass number equation**:
\[ 200 = 168 + 4X + 0Y \]
Simplifying this gives:
\[ 200 - 168 = 4X \implies 32 = 4X \implies X = 8 \]

2. **Atomic number equation**:
\[ 90 = 80 + 2X - Y \]
Substituting \(X = 8\) into the equation:
\[ 90 = 80 + 2(8) - Y \]
Simplifying this gives:
\[ 90 = 80 + 16 - Y \implies 90 = 96 - Y \implies Y = 96 - 90 \implies Y = 6 \]

### Step 4: Conclusion
The respective number of alpha and beta particles emitted during the decay is:
- Alpha particles (\(X\)) = 8
- Beta particles (\(Y\)) = 6

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

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