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`

Answer

To solve the problem of determining the respective number of alpha and beta particles emitted in the radioactive decay reaction:

\[ _{90}^{200}X \rightarrow _{80}^{168}Y \]

we will follow these steps:

### Step 1: Determine the number of alpha particles emitted
The mass number of the parent nucleus \(X\) is 200, and the mass number of the daughter nucleus \(Y\) is 168. The difference in mass number gives us the total number of nucleons lost during the decay.

\[ \text{Number of alpha particles} = \frac{\text{Mass number of } X - \text{Mass number of } Y}{4} \]

Calculating this:

\[ \text{Number of alpha particles} = \frac{200 - 168}{4} = \frac{32}{4} = 8 \]

### Step 2: Determine the change in atomic number
Next, we find the change in atomic number. The atomic number of \(X\) is 90, and the atomic number of \(Y\) is 80. The difference in atomic number is:

\[ \text{Change in atomic number} = 90 - 80 = 10 \]

### Step 3: Relate alpha and beta particles to the change in atomic number
Each alpha particle emitted decreases the atomic number by 2. If \(n\) is the number of alpha particles emitted, the equation relating the number of beta particles \(b\) can be expressed as:

\[ \text{Atomic number of } Y = \text{Atomic number of } X - 2n + b \]

Substituting the known values:

\[ 80 = 90 - 2(8) + b \]

### Step 4: Solve for the number of beta particles
Now we can simplify and solve for \(b\):

\[ 80 = 90 - 16 + b \]
\[ 80 = 74 + b \]
\[ b = 80 - 74 = 6 \]

### Conclusion
Thus, the respective number of alpha and beta particles emitted in the decay is:

- Number of alpha particles = 8
- Number of beta particles = 6

### Final Answer
The answer is \(8\) alpha particles and \(6\) beta particles.

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