When the radioactive isotope `._(88)Ra^(228)` decays in series by the emission of `3 alpha` and `1 beta` particle, the isotope finally formed is

Options

A. `._84X^(228)`
B. `._86X^(222)`
C. `._83X^(216)`
D. `._86X^(215)`

Answer

To solve the problem of determining the final isotope formed when the radioactive isotope \( _{88}^{228}\text{Ra} \) decays by the emission of 3 alpha particles and 1 beta particle, we can follow these steps:

### Step 1: Understand the decay process
- **Alpha decay**: When an isotope emits an alpha particle (\( \alpha \)), it loses 2 protons and 4 nucleons (2 protons + 2 neutrons).
- **Beta decay**: When an isotope emits a beta particle (\( \beta \)), it loses 1 neutron and gains 1 proton, so the mass number remains the same, but the atomic number increases by 1.

### Step 2: Calculate the changes due to alpha decay
- The initial isotope is \( _{88}^{228}\text{Ra} \).
- After 3 alpha decays:
- Mass number decreases by \( 3 \times 4 = 12 \).
- Atomic number decreases by \( 3 \times 2 = 6 \).

Calculating the new mass number (A) and atomic number (Z):
- New mass number \( A = 228 - 12 = 216 \)
- New atomic number \( Z = 88 - 6 = 82 \)

### Step 3: Calculate the changes due to beta decay
- After 1 beta decay:
- Mass number remains the same (216).
- Atomic number increases by 1.

Calculating the final atomic number (Z):
- New atomic number \( Z = 82 + 1 = 83 \)

### Step 4: Identify the final isotope
- The final isotope can be represented as \( _{Z}^{A}X \).
- Therefore, the final isotope is \( _{83}^{216}\text{Bi} \) (Bismuth).

### Final Answer
The isotope finally formed is \( _{83}^{216}\text{Bi} \).

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