The binding energy per nucleon of deuterium and helium nuclei are 1.1 MeV and 7.0 MeV respectively. When two deuterium nuclei fuse to form a helium nucleus the energy released in the fusion is

Options

A. 23.6 MeV
B. 2.2 MeV
C. 28.0 MeV
D. 30.2 MeV

Explanation

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

### Step 1: Identify the binding energies
- The binding energy per nucleon for deuterium (²H) is given as 1.1 MeV.
- The binding energy per nucleon for helium (⁴He) is given as 7.0 MeV.

### Step 2: Calculate the total binding energy for the reactants (deuterium)
- Each deuterium nucleus has 2 nucleons (1 proton and 1 neutron).
- Therefore, the total binding energy for one deuterium nucleus is:
\[ \text{Binding Energy (deuterium)} = \text{Binding Energy per nucleon} \times \text{Number of nucleons} = 1.1 \, \text{MeV} \times 2 = 2.2 \, \text{MeV} \]
- Since we have two deuterium nuclei, the total binding energy for both is:
\[ \text{Total Binding Energy (2 deuterium)} = 2.2 \, \text{MeV} \times 2 = 4.4 \, \text{MeV} \]

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

### Step 4: Calculate the energy released in the fusion process
- The energy released during the fusion is the difference between the total binding energy of the products and the total binding energy of the reactants:
\[ \text{Energy Released} = \text{Total Binding Energy (helium)} - \text{Total Binding Energy (2 deuterium)} \]
\[ \text{Energy Released} = 28.0 \, \text{MeV} - 4.4 \, \text{MeV} = 23.6 \, \text{MeV} \]

### Final Answer
The energy released in the fusion of two deuterium nuclei to form a helium nucleus is **23.6 MeV**.

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