The binding energy of deuteron is `2.2` MeV and that of `._(2)^(4)He` is `28` MeV. If two deuterons are fused to form one `._(2)^(4)He`, th `n` the energy released is

Options

A. `25.8` MeV
B. `23.6` MeV
C. `19.2` MeV
D. `30.2` MeV

Answer

To solve the problem of calculating the energy released when two deuterons fuse to form one helium nucleus, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Binding Energies**:
- The binding energy of a deuteron (D) is given as \(2.2 \, \text{MeV}\).
- The binding energy of helium-4 (\(^{4}_{2}\text{He}\)) is given as \(28 \, \text{MeV}\).

2. **Calculate the Total Binding Energy of Reactants**:
- When two deuterons are fused, the total binding energy of the reactants (two deuterons) is calculated as:
\[ \text{Total Binding Energy of Reactants} = 2 \times \text{Binding Energy of Deuteron} = 2 \times 2.2 \, \text{MeV} = 4.4 \, \text{MeV} \]

3. **Calculate the Energy Released**:
- The energy released during the fusion process can be found by subtracting the total binding energy of the reactants from the binding energy of the product (helium nucleus):
\[ \text{Energy Released} = \text{Binding Energy of Helium} - \text{Total Binding Energy of Reactants} \]
\[ \text{Energy Released} = 28 \, \text{MeV} - 4.4 \, \text{MeV} = 23.6 \, \text{MeV} \]

4. **Final Answer**:
- The energy released when two deuterons fuse to form one helium nucleus is \(23.6 \, \text{MeV}\).

### Summary:
The energy released during the fusion of two deuterons to form helium-4 is \(23.6 \, \text{MeV}\).