An artifical radioactive decay series begins with unstable `._94 ^241 Pu`. The stable nuclide obtained after eight `alpha-`decays and five `beta^+ -` decays is.

Options

A. `._83^209 Bi`
B. `._82^209 Pb`
C. `._82^205 Ti`
D. `._82^201 Hg`

Explanation

To solve the problem of determining the stable nuclide obtained after eight alpha decays and five beta-plus decays starting from the unstable nuclide \( _{94}^{241}Pu \), we will follow these steps:

### Step 1: Calculate the effect of alpha decays
Each alpha decay reduces the mass number (A) by 4 and the atomic number (Z) by 2.

- **Initial values:**
- Mass number (A) = 241
- Atomic number (Z) = 94

- **After 8 alpha decays:**
- Mass number decreases: \( A' = 241 - 8 \times 4 = 241 - 32 = 209 \)
- Atomic number decreases: \( Z' = 94 - 8 \times 2 = 94 - 16 = 78 \)

### Step 2: Calculate the effect of beta-plus decays
Each beta-plus decay increases the atomic number (Z) by 1 but does not change the mass number (A).

- **After 5 beta-plus decays:**
- Mass number remains the same: \( A'' = 209 \)
- Atomic number increases: \( Z'' = 78 + 5 = 83 \)

### Step 3: Identify the stable nuclide
Now we have the final values after all the decays:
- Mass number (A) = 209
- Atomic number (Z) = 83

We need to identify the nuclide with these values.

- The nuclide with atomic number 83 is Bismuth (Bi).

### Final Result
The stable nuclide obtained after eight alpha decays and five beta-plus decays from \( _{94}^{241}Pu \) is \( _{83}^{209}Bi \).

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