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^(226)`
D. `._(83)X^(215)`

Answer

To solve the problem of determining the isotope obtained after three alpha decays and one beta decay from the isotope \( _{88}^{238} \text{Ra} \), we can follow these steps:

### Step 1: Understand the decay process
- An alpha particle (\( \alpha \)) consists of 2 protons and 2 neutrons, which means it decreases the atomic mass by 4 and the atomic number by 2.
- A beta particle (\( \beta \)) is an electron emitted from a nucleus, which effectively increases the atomic number by 1 (since a neutron is converted into a proton).

### Step 2: Calculate the change in atomic mass (A)
- Starting with \( A = 238 \) (the mass number of radium).
- Since three alpha particles are emitted, the total decrease in mass due to alpha decay is:
\[ \text{Decrease in mass} = 3 \times 4 = 12 \]
- Therefore, the new mass number \( A' \) after the alpha decays will be:
\[ A' = 238 - 12 = 226 \]

### Step 3: Calculate the change in atomic number (Z)
- Starting with \( Z = 88 \) (the atomic number of radium).
- Each alpha decay decreases the atomic number by 2, so for three alpha particles:
\[ \text{Decrease in atomic number} = 3 \times 2 = 6 \]
- After three alpha decays, the atomic number becomes:
\[ Z' = 88 - 6 = 82 \]
- Now, we have one beta decay, which increases the atomic number by 1:
\[ Z'' = 82 + 1 = 83 \]

### Step 4: Write the final isotope
- After all the decays, the final isotope will have:
- Mass number \( A = 226 \)
- Atomic number \( Z = 83 \)
- The resulting isotope can be represented as:
\[ _{83}^{226} \text{Bi} \]
- Therefore, the final isotope obtained after the decay process is \( \text{Bismuth} \) (Bi).

### Final Answer
The isotope obtained after the decay process is \( _{83}^{226} \text{Bi} \).

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