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 deuterons fuse to form a helium nucleus (`_2He^4`) is

Explanation

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

### Step 1: Identify the binding energies
- The binding energy per nucleon for deuteron (\(_1H^2\)) is given as 1.1 MeV.
- The binding energy per nucleon for helium (\(_2He^4\)) is given as 7.0 MeV.

### Step 2: Calculate the total binding energy of the reactants (deuterons)
- Each deuteron has 2 nucleons (1 proton and 1 neutron).
- Therefore, the total binding energy for two deuterons is:
\[ E_1 = \text{Binding energy per nucleon} \times \text{Number of nucleons in two deuterons} \]
\[ E_1 = 1.1 \, \text{MeV} \times 2 = 2.2 \, \text{MeV} \]

### Step 3: Calculate the total binding energy of the products (helium nucleus)
- The helium nucleus has 4 nucleons (2 protons and 2 neutrons).
- Therefore, the total binding energy for helium is:
\[ E_2 = \text{Binding energy per nucleon} \times \text{Number of nucleons in helium} \]
\[ E_2 = 7.0 \, \text{MeV} \times 4 = 28.0 \, \text{MeV} \]

### Step 4: Calculate the energy released in the fusion reaction
- The energy released (Q) during the fusion reaction can be calculated by finding the difference between the binding energies of the products and the reactants:
\[ Q = E_2 - E_1 \]
\[ Q = 28.0 \, \text{MeV} - 2.2 \, \text{MeV} = 25.8 \, \text{MeV} \]

### Final Answer:
The energy released when two deuterons fuse to form a helium nucleus is **25.8 MeV**.

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