Which of the following pair is isodiaphers?

Options

A. (a) `C_6^14 and Na_11^23`
B. (b) `Mg_12^24 and Na_11^23`
C. (c) `He_2^4 and O_8^16`
D. (d) `C_6^12 and N_7^15`

Answer

To determine which of the following pairs are isodiaphers, we need to understand the definition of isodiaphers. Isodiaphers are pairs of atoms of different elements that have the same difference between the number of neutrons (n) and the number of protons (p). This difference is represented as \( n - p \).

Let's analyze the options step by step:

### Step 1: Understand the terms
- **Mass Number (A)**: Total number of protons and neutrons in an atom.
- **Atomic Number (Z)**: Number of protons in an atom.
- **Number of Neutrons (n)**: Can be calculated as \( n = A - Z \).

### Step 2: Calculate \( n - p \) for each option

#### Option 1: Carbon-14 (C-14) and Sodium-23 (Na-23)
- For Carbon-14:
- \( A = 14 \), \( Z = 6 \)
- Number of Neutrons \( n = A - Z = 14 - 6 = 8 \)
- Difference \( n - p = 8 - 6 = 2 \)

- For Sodium-23:
- \( A = 23 \), \( Z = 11 \)
- Number of Neutrons \( n = A - Z = 23 - 11 = 12 \)
- Difference \( n - p = 12 - 11 = 1 \)

**Conclusion for Option 1**: \( n - p \) values are different (2 and 1), so they are not isodiaphers.

#### Option 2: Magnesium-24 (Mg-24) and Neon-23 (Ne-23)
- For Magnesium-24:
- \( A = 24 \), \( Z = 12 \)
- Number of Neutrons \( n = A - Z = 24 - 12 = 12 \)
- Difference \( n - p = 12 - 12 = 0 \)

- For Neon-23:
- \( A = 23 \), \( Z = 10 \)
- Number of Neutrons \( n = A - Z = 23 - 10 = 13 \)
- Difference \( n - p = 13 - 10 = 3 \)

**Conclusion for Option 2**: \( n - p \) values are different (0 and 3), so they are not isodiaphers.

#### Option 3: Helium-4 (He-4) and Oxygen-16 (O-16)
- For Helium-4:
- \( A = 4 \), \( Z = 2 \)
- Number of Neutrons \( n = A - Z = 4 - 2 = 2 \)
- Difference \( n - p = 2 - 2 = 0 \)

- For Oxygen-16:
- \( A = 16 \), \( Z = 8 \)
- Number of Neutrons \( n = A - Z = 16 - 8 = 8 \)
- Difference \( n - p = 8 - 8 = 0 \)

**Conclusion for Option 3**: Both have the same \( n - p \) value (0), so they are isodiaphers.

#### Option 4: Carbon-12 (C-12) and Nitrogen-15 (N-15)
- For Carbon-12:
- \( A = 12 \), \( Z = 6 \)
- Number of Neutrons \( n = A - Z = 12 - 6 = 6 \)
- Difference \( n - p = 6 - 6 = 0 \)

- For Nitrogen-15:
- \( A = 15 \), \( Z = 7 \)
- Number of Neutrons \( n = A - Z = 15 - 7 = 8 \)
- Difference \( n - p = 8 - 7 = 1 \)

**Conclusion for Option 4**: \( n - p \) values are different (0 and 1), so they are not isodiaphers.

### Final Answer
The correct pair that are isodiaphers is **Option 3: Helium-4 (He-4) and Oxygen-16 (O-16)**.