Three `alpha-`particle and one `beta-`particle decaying takes place in series from an isotope `._88 Ra^238`. Finally the isotope obtained will be.

Options

A. `._84 X^220`
B. `._86 X^222`
C. `._83 X^224`
D. `._83 X^215`

Explanation

To determine the final isotope obtained after the decay of an isotope \( _{88}^{238}Ra \) through the emission of three alpha particles and one beta particle, we can follow these steps:

### Step 1: Understand the decay process
- An alpha particle (\( \alpha \)) consists of 2 protons and 2 neutrons. Therefore, each alpha decay decreases the atomic mass number (A) by 4 and the atomic number (Z) by 2.
- A beta particle (\( \beta \)) is an electron emitted from a nucleus when a neutron is converted into a proton. This process increases the atomic number (Z) by 1 while keeping the mass number (A) unchanged.

### Step 2: Calculate the change in mass number (A)
- The initial isotope is \( _{88}^{238}Ra \).
- After emitting 3 alpha particles:
\[ A' = 238 - 3 \times 4 = 238 - 12 = 226 \]

### Step 3: Calculate the change in atomic number (Z)
- The initial atomic number of radium (Ra) is 88.
- After emitting 3 alpha particles:
\[ Z' = 88 - 3 \times 2 = 88 - 6 = 82 \]
- After emitting 1 beta particle:
\[ Z'' = 82 + 1 = 83 \]

### Step 4: Write the final isotope
- After the decay process, we have:
- Mass number (A) = 226
- Atomic number (Z) = 83
- The final isotope can be represented as \( _{83}^{226}Bi \) (Bismuth).

### Final Answer
The isotope obtained after the decay of \( _{88}^{238}Ra \) through the emission of three alpha particles and one beta particle is \( _{83}^{226}Bi \).

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