The binding energy of the `""^(14)N` nucleus is 16.76 pJ and `""^(14)C` nucleus is 16.86 pJ. Which nucleus is the decay product of the other?

Answer

To determine which nucleus is the decay product of the other between \(^{14}N\) (nitrogen) and \(^{14}C\) (carbon), we can analyze their binding energies.

### Step-by-Step Solution:

1. **Understand Binding Energy**:
The binding energy of a nucleus is the energy required to disassemble it into its constituent protons and neutrons. A higher binding energy indicates a more stable nucleus.

2. **Given Data**:
- Binding energy of \(^{14}N\) = 16.76 pJ
- Binding energy of \(^{14}C\) = 16.86 pJ

3. **Compare Binding Energies**:
- Compare the binding energies of both nuclei:
- \(^{14}N\) has a binding energy of 16.76 pJ.
- \(^{14}C\) has a binding energy of 16.86 pJ.
- Since 16.86 pJ (for \(^{14}C\)) is greater than 16.76 pJ (for \(^{14}N\)), this indicates that \(^{14}C\) is more stable than \(^{14}N\).

4. **Determine Decay Relationship**:
- In nuclear decay processes, a less stable nucleus (with lower binding energy) will decay into a more stable nucleus (with higher binding energy).
- Since \(^{14}N\) has a lower binding energy than \(^{14}C\), it suggests that \(^{14}N\) can decay into \(^{14}C\).

5. **Conclusion**:
- Therefore, \(^{14}C\) is the decay product of \(^{14}N\).

### Final Answer:
The nucleus \(^{14}C\) is the decay product of the nucleus \(^{14}N\).

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