The binding energy of deuteron is 2.2 MeV and that of H`""_(2)^(4)He` is 28MeV. If two deuterons are fused to form one `""_(2)^(4)He` then the energy released is:-

Options

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

Explanation

To find the energy released when two deuterons are fused to form one helium atom, we will use the binding energy values given in the problem.

### Step-by-Step Solution:

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

2. **Write the Fusion Reaction:**
- The fusion of two deuterons can be represented as:
\[ 2D \rightarrow ^4He + \text{Energy} \]

3. **Calculate the Total Binding Energy of the Reactants:**
- Since we have two deuterons, the total binding energy for the reactants is:
\[ \text{Total Binding Energy of Reactants} = 2 \times B_D = 2 \times 2.2 \, \text{MeV} = 4.4 \, \text{MeV} \]

4. **Calculate the Binding Energy of the Product:**
- The binding energy of the product (helium-4) is:
\[ \text{Binding Energy of Product} = B_{He} = 28 \, \text{MeV} \]

5. **Calculate the Energy Released:**
- The energy released during the fusion process can be calculated using the formula:
\[ \text{Energy Released} = \text{Binding Energy of Product} - \text{Total Binding Energy of Reactants} \]
- Substituting the values:
\[ \text{Energy Released} = 28 \, \text{MeV} - 4.4 \, \text{MeV} = 23.6 \, \text{MeV} \]

6. **Final Answer:**
- The energy released when two deuterons are fused to form one helium atom is \( \boxed{23.6 \, \text{MeV}} \).