The binding energy of deuteron `._1^2 H` is `1.112 MeV` per nucleon and an `alpha-`particle `._2^4 He` has a binding energy of `7.047 MeV` per nucleon. Then in the fusion reaction `._1^2H + ._1^2h rarr ._2^4 He + Q`, the energy `Q` released is.

Options

A. `1 MeV`
B. `11.9 Me V`
C. `23 ,8 Me V`
D. `931 Me V`

Answer

To find the energy \( Q \) released in the fusion reaction \( _1^2H + _1^2H \rightarrow _2^4He + Q \), we can follow these steps:

### Step 1: Identify the Binding Energies
- The binding energy of deuteron \( _1^2H \) is given as \( 1.112 \, \text{MeV} \) per nucleon.
- The binding energy of the alpha particle \( _2^4He \) is given as \( 7.047 \, \text{MeV} \) per nucleon.

### Step 2: Calculate the Total Binding Energy of the Products
- The alpha particle \( _2^4He \) has 4 nucleons.
- Therefore, the total binding energy for the alpha particle is:
\[ \text{Total Binding Energy of } _2^4He = 4 \times 7.047 \, \text{MeV} = 28.188 \, \text{MeV} \]

### Step 3: Calculate the Total Binding Energy of the Reactants
- The reactants consist of two deuterons, each having 2 nucleons, so the total number of nucleons is:
\[ \text{Total Nucleons in Reactants} = 2 + 2 = 4 \]
- Therefore, the total binding energy for the two deuterons is:
\[ \text{Total Binding Energy of Reactants} = 4 \times 1.112 \, \text{MeV} = 4.448 \, \text{MeV} \]

### Step 4: Calculate the Energy Released \( Q \)
- The energy released \( Q \) in the reaction can be calculated by taking the difference between the total binding energy of the products and the total binding energy of the reactants:
\[ Q = \text{Total Binding Energy of Products} - \text{Total Binding Energy of Reactants} \]
\[ Q = 28.188 \, \text{MeV} - 4.448 \, \text{MeV} = 23.740 \, \text{MeV} \]

### Final Answer
Thus, the energy \( Q \) released in the fusion reaction is approximately:
\[ Q \approx 23.740 \, \text{MeV} \]

---