A nucleus of mass `M + Deltam` is at rest and decays into two daughter nuclei of equal mass M/2 each. Speed of light is C.
The binding energy per nucleon for the parent nucleus is `E_1` and that for the daughter nuclei is `E_2`. Then

Options

A. `E_2 = 2E_1`
B. `E_1 gt E_2`
C. `E_2 gt E_1`
D. `E_1 = 2E_2`

Answer

To solve the problem, we need to analyze the decay of the parent nucleus into two daughter nuclei and understand the relationship between their binding energies per nucleon.

### Step-by-Step Solution:

1. **Understanding the Decay Process**:
- We have a parent nucleus with mass \( M + \Delta m \) that is at rest.
- It decays into two daughter nuclei, each with mass \( \frac{M}{2} \).

2. **Conservation of Energy**:
- The total energy before decay is the rest energy of the parent nucleus, which can be expressed as \( E_{\text{parent}} = (M + \Delta m)c^2 \).
- After the decay, the total energy of the two daughter nuclei will be the sum of their rest energies, which is \( E_{\text{daughters}} = 2 \left( \frac{M}{2} \right)c^2 = Mc^2 \).

3. **Binding Energy Considerations**:
- The binding energy per nucleon for the parent nucleus is given as \( E_1 \).
- The binding energy per nucleon for the daughter nuclei is given as \( E_2 \).
- The binding energy of a nucleus is a measure of how stable it is; a higher binding energy per nucleon indicates a more stable nucleus.

4. **Stability and Binding Energy**:
- When a nucleus decays, it typically forms more stable products. This means that the daughter nuclei will have a higher binding energy per nucleon compared to the parent nucleus.
- Therefore, we can conclude that \( E_2 > E_1 \).

5. **Final Conclusion**:
- Since the daughter nuclei are more stable than the parent nucleus, the binding energy per nucleon of the daughter nuclei \( E_2 \) is greater than that of the parent nucleus \( E_1 \).
- Thus, we can write:
\[ E_2 > E_1 \]