Which of the following pair is isodiaphers?

Options

A. `C_6^14 and Na_11^23`
B. `Mg_12^24 and Na_11^23`
C. `He_2^4 and O_8^16`
D. `C_6^12 and N_7^15`

Explanation

To determine which of the given pairs is isodiaphers, we need to understand the definition of isodiaphers. Isodiaphers are pairs of nuclides that have the same difference between the number of neutrons (N) and protons (Z). This means that for two species to be considered isodiaphers, the difference (N - Z) must be equal for both.

Let's analyze the options step by step:

### Step 1: Identify the pairs and their atomic numbers
Assuming we have the following pairs (for example):
1. Carbon (C) and Sodium (Na)
2. Helium (He) and Oxygen (O)

### Step 2: Calculate N - Z for each pair

#### Pair 1: Carbon (C) and Sodium (Na)
- **Carbon (C)**:
- Atomic number (Z) = 6 (number of protons)
- Atomic mass (A) = 12 (for the most common isotope)
- Number of neutrons (N) = A - Z = 12 - 6 = 6
- Difference (N - Z) = 6 - 6 = 0

- **Sodium (Na)**:
- Atomic number (Z) = 11 (number of protons)
- Atomic mass (A) = 23 (for the most common isotope)
- Number of neutrons (N) = A - Z = 23 - 11 = 12
- Difference (N - Z) = 12 - 11 = 1

**Conclusion for Pair 1**: The differences are not equal (0 ≠ 1), so Carbon and Sodium are not isodiaphers.

#### Pair 2: Helium (He) and Oxygen (O)
- **Helium (He)**:
- Atomic number (Z) = 2 (number of protons)
- Atomic mass (A) = 4 (for the most common isotope)
- Number of neutrons (N) = A - Z = 4 - 2 = 2
- Difference (N - Z) = 2 - 2 = 0

- **Oxygen (O)**:
- Atomic number (Z) = 8 (number of protons)
- Atomic mass (A) = 16 (for the most common isotope)
- Number of neutrons (N) = A - Z = 16 - 8 = 8
- Difference (N - Z) = 8 - 8 = 0

**Conclusion for Pair 2**: The differences are equal (0 = 0), so Helium and Oxygen are isodiaphers.

### Final Answer:
The pair that is isodiaphers is **Helium (He) and Oxygen (O)**.