The end product of `4n` series is

Options

A. `._(82)Pb^(208)`
B. `._(82)Pb^(207)`
C. `._(82)Pb^(209)`
D. `._(82)Pb^(204)`

Answer

To solve the question regarding the end product of the `4n` series, we can follow these steps:

### Step-by-Step Solution:

1. **Understanding the `4n` Series**:
- The `4n` series refers to a series of radioactive decay processes where the mass number of the parent nuclide is divisible by 4. This means that when the mass number is divided by 4, there is no remainder.

2. **Identifying the End Product**:
- The end product of the `4n` series is a stable nuclide. In the case of the `4n` series, which is also known as the thorium series, the final stable product is lead (Pb).

3. **Determining the Specific Isotope**:
- The specific isotope of lead that is the end product of the `4n` series is lead-208 (Pb-208). This isotope has a mass number of 208, which is indeed divisible by 4 (208 ÷ 4 = 52, remainder 0).

4. **Conclusion**:
- Therefore, the end product of the `4n` series is lead-208 (Pb-208).

### Final Answer:
The end product of the `4n` series is lead-208 (Pb-208).

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