A nuclear of mass `M +deltam ` is at rest and decay into two daughter nuclei of equal mass `(M)/(2)` each speed is `c`
The binding energy per nucleon for the nucleus is `E_(1)` and that for the daugther nuclei is `E_(2)` Then

Options

A. `E_(2)=2E_(1)`
B. `E_(1)gtE_(2)`
C. `E_(2)gtE_(1)`
D. `E_(1)=2E_(2)`

Answer

To solve the problem, we need to analyze the decay of a nucleus and the relationship between the binding energies of the parent and daughter nuclei.

### Step-by-Step Solution:

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

2. **Conservation of Mass-Energy**:
- According to the conservation of mass-energy, the total mass-energy before and after the decay must remain constant.
- The extra mass \( \Delta m \) is converted into energy during the decay process.

3. **Binding Energy Definitions**:
- The binding energy per nucleon for the parent nucleus is denoted as \( E_1 \).
- The binding energy per nucleon for each daughter nucleus is denoted as \( E_2 \).

4. **Stability of Nuclei**:
- Generally, smaller nuclei (daughter nuclei) are more stable than larger nuclei (parent nucleus).
- This is because the binding energy per nucleon tends to increase as the size of the nucleus decreases, up to a certain point.

5. **Reasoning for Binding Energy Comparison**:
- When the parent nucleus decays into two smaller nuclei, the proton-to-neutron ratio typically increases, leading to a higher binding energy per nucleon.
- Thus, the binding energy per nucleon for the daughter nuclei \( E_2 \) will be greater than that of the parent nucleus \( E_1 \).

6. **Conclusion**:
- Therefore, we conclude that \( E_2 > E_1 \).

### Final Result:
The relationship between the binding energies is:
\[ E_2 > E_1 \]