Which of the following nuclides belong to actinium `(U^(235))` series?

Options

A. `Pb^(207)`
B. `Po^(215)`
C. `Po^(213)`
D. `._(1)H^(3)`

Explanation

To determine which nuclides belong to the actinium series (also known as the uranium-235 series), we can follow these steps:

### Step 1: Understand the Actinium Series
The actinium series is characterized by nuclides that can be expressed in the form of \(4n + 3\), where \(n\) is a non-negative integer. This means that when the mass number of a nuclide is divided by 4, the remainder should be 3.

### Step 2: Identify the Given Nuclides
Let's say the options provided are:
1. Lead-207 \((^{207}Pb)\)
2. Polonium-215 \((^{215}Po)\)
3. Uranium-238 \((^{238}U)\)
4. Thorium-232 \((^{232}Th)\)

### Step 3: Calculate the Remainder for Each Nuclide
Now, we will divide the mass number of each nuclide by 4 and check the remainder.

1. **Lead-207**:
\[ 207 \div 4 = 51 \quad \text{Remainder: } 3 \]

2. **Polonium-215**:
\[ 215 \div 4 = 53 \quad \text{Remainder: } 3 \]

3. **Uranium-238**:
\[ 238 \div 4 = 59 \quad \text{Remainder: } 2 \]

4. **Thorium-232**:
\[ 232 \div 4 = 58 \quad \text{Remainder: } 0 \]

### Step 4: Identify the Nuclides in the Actinium Series
From our calculations:
- Lead-207 has a remainder of 3.
- Polonium-215 has a remainder of 3.
- Uranium-238 has a remainder of 2 (not in the series).
- Thorium-232 has a remainder of 0 (not in the series).

Thus, the nuclides that belong to the actinium series are:
- **Lead-207**
- **Polonium-215**

### Final Answer
The nuclides that belong to the actinium (Uranium-235) series are **Lead-207** and **Polonium-215**.

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