What are the respective number of `alpha`- and `beta`- particle emitted in the following radioactive decay ?
`""_(90)^(200)X to ""_(80)^(168)Y`

Options

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

Answer

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

### Step 1: Identify Initial and Final Atomic Numbers and Mass Numbers
- The initial nucleus is \( _{90}^{200}X \) with:
- Atomic number (Z) = 90
- Mass number (A) = 200

- The final nucleus is \( _{80}^{168}Y \) with:
- Atomic number (Z) = 80
- Mass number (A) = 168

### Step 2: Calculate Change in Mass Number
- Change in mass number (A) = Initial mass number - Final mass number
\[ \Delta A = 200 - 168 = 32 \]

### Step 3: Calculate Change in Atomic Number
- Change in atomic number (Z) = Initial atomic number - Final atomic number
\[ \Delta Z = 90 - 80 = 10 \]

### Step 4: Determine Number of Alpha Particles Emitted
- Each alpha particle (α) has:
- Mass number = 4
- Atomic number = 2

Let \( n \) be the number of alpha particles emitted. The mass number decreases by \( 4n \) and the atomic number decreases by \( 2n \).

From the mass number:
\[ 4n = 32 \implies n = \frac{32}{4} = 8 \]

### Step 5: Determine Number of Beta Particles Emitted
- The change in atomic number due to alpha particles is:
\[ \Delta Z_{\text{alpha}} = 2n = 2 \times 8 = 16 \]

Since the total change in atomic number is 10, we can find the number of beta particles (β) emitted. Each beta particle does not change the mass number but increases the atomic number by 1.

Let \( m \) be the number of beta particles emitted:
\[ \Delta Z_{\text{total}} = \Delta Z_{\text{alpha}} + m \]
\[ 10 = 16 + m \implies m = 10 - 16 = -6 \]

Since we cannot have a negative number of beta particles, we realize that the total change in atomic number must account for the alpha particles reducing the atomic number. Thus:
\[ m = 10 - 16 = -6 \text{ (not possible)} \]
This indicates that we have accounted for all changes through alpha decay.

### Conclusion
- Therefore, the respective number of alpha and beta particles emitted in the decay of \( _{90}^{200}X \) to \( _{80}^{168}Y \) is:
- Number of alpha particles emitted = 8
- Number of beta particles emitted = 0