The binding energies per nucleon for deuteron (`._1H^2`) and helium (`._2He^4)` are `1.1 MeV` and `7.0 MeV` respectively. The energy released when two deutrons fuse to form a helium nucleus (`._2He^4`) is……..

Options

A. `13.9 MeV`
B. `26.9 MeV`
C. `23.6 MeV`
D. `19.2 MeV`

Explanation

To find the energy released when two deuterons fuse to form a helium nucleus, we can follow these steps:

### Step 1: Understand the Binding Energies
- The binding energy per nucleon for deuteron (\( _1H^2 \)) is given as \( 1.1 \, \text{MeV} \).
- The binding energy per nucleon for helium (\( _2He^4 \)) is given as \( 7.0 \, \text{MeV} \).

### Step 2: Calculate Total Binding Energy of Deuterons
- Each deuteron has 2 nucleons (1 proton and 1 neutron).
- Therefore, the total binding energy of two deuterons (which has \( 2 \times 2 = 4 \) nucleons) is:
\[ \text{Total binding energy of deuterons} = 4 \times 1.1 \, \text{MeV} = 4.4 \, \text{MeV} \]

### Step 3: Calculate Total Binding Energy of Helium
- Helium has 4 nucleons (2 protons and 2 neutrons).
- Therefore, the total binding energy of helium is:
\[ \text{Total binding energy of helium} = 4 \times 7.0 \, \text{MeV} = 28.0 \, \text{MeV} \]

### Step 4: Calculate the Energy Released
- The energy released during the fusion process is equal to the change in binding energy, which can be calculated as:
\[ \text{Energy released} = \text{Binding energy of helium} - \text{Binding energy of deuterons} \]
\[ \text{Energy released} = 28.0 \, \text{MeV} - 4.4 \, \text{MeV} = 23.6 \, \text{MeV} \]

### Step 5: Conclusion
- The energy released when two deuterons fuse to form a helium nucleus is \( 23.6 \, \text{MeV} \).

### Final Answer
The energy released is \( 23.6 \, \text{MeV} \).

---