Find the energy required for separation of `a_(10)Ne^(20)` nucleus into two `alpha`-particles and `a_(6)C^(12)` nucleus if it is known that the binding energies per nucleon in `._(10)Ne^(20), ._(2)He^(4)` and `._(6)C^(12)` nuclei are equal to `8.03, 7.07` and `7.68 MeV` respectively.

Answer

To find the energy required for the separation of a \( _{10}^{20}\text{Ne} \) nucleus into two alpha particles and a \( _{6}^{12}\text{C} \) nucleus, we will use the concept of binding energy. The binding energy per nucleon for each nucleus is given, and we can calculate the total binding energy for each nucleus involved in the reaction.

### Step-by-Step Solution:

1. **Identify the Nuclei and Their Binding Energies**:
- For \( _{10}^{20}\text{Ne} \): Binding energy per nucleon = 8.03 MeV
- For \( _{2}^{4}\text{He} \) (alpha particle): Binding energy per nucleon = 7.07 MeV
- For \( _{6}^{12}\text{C} \): Binding energy per nucleon = 7.68 MeV

2. **Calculate the Total Binding Energy of Each Nucleus**:
- Total binding energy of \( _{10}^{20}\text{Ne} \):
\[ BE_{\text{Ne}} = \text{(Number of nucleons)} \times \text{(Binding energy per nucleon)} = 20 \times 8.03 \text{ MeV} = 160.6 \text{ MeV} \]
- Total binding energy of two alpha particles:
\[ BE_{\text{He}} = 2 \times \text{(Number of nucleons in He)} \times \text{(Binding energy per nucleon)} = 2 \times 4 \times 7.07 \text{ MeV} = 56.56 \text{ MeV} \]
- Total binding energy of \( _{6}^{12}\text{C} \):
\[ BE_{\text{C}} = \text{(Number of nucleons)} \times \text{(Binding energy per nucleon)} = 12 \times 7.68 \text{ MeV} = 92.16 \text{ MeV} \]

3. **Calculate the Total Binding Energy After Separation**:
- The total binding energy after separation (for two alpha particles and one carbon nucleus):
\[ BE_{\text{after}} = BE_{\text{He}} + BE_{\text{C}} = 56.56 \text{ MeV} + 92.16 \text{ MeV} = 148.72 \text{ MeV} \]

4. **Calculate the Energy Required for Separation**:
- The energy required for separation is the difference between the binding energy of the original nucleus and the total binding energy after separation:
\[ E_{\text{required}} = BE_{\text{Ne}} - BE_{\text{after}} = 160.6 \text{ MeV} - 148.72 \text{ MeV} = 11.88 \text{ MeV} \]

5. **Final Result**:
- The energy required for the separation of the \( _{10}^{20}\text{Ne} \) nucleus into two alpha particles and one \( _{6}^{12}\text{C} \) nucleus is approximately **11.88 MeV**.