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. 23.74 Mev
B. 32.82 MeV
C. 11.9 MeV
D. 4.94 MeV

Answer

To find the energy \( Q \) released in the fusion reaction \( \, _1^2H + \, _1^2H \rightarrow \, _2^4He + Q \), we will use the concept of binding energy.

### Step 1: Calculate the total binding energy of the reactants

The reactants are two deuterium nuclei \( \, _1^2H \). The binding energy per nucleon for deuterium is given as \( 1.112 \, \text{MeV} \).

Since each deuterium nucleus has 2 nucleons, the total binding energy for one deuterium nucleus is:
\[ \text{Binding Energy of one } \, _1^2H = 2 \times 1.112 \, \text{MeV} = 2.224 \, \text{MeV} \]

For two deuterium nuclei, the total binding energy is:
\[ \text{Total Binding Energy of Reactants} = 2 \times 2.224 \, \text{MeV} = 4.448 \, \text{MeV} \]

### Step 2: Calculate the total binding energy of the products

The product of the reaction is one alpha particle \( \, _2^4He \). The binding energy per nucleon for the alpha particle is given as \( 7.047 \, \text{MeV} \).

Since the alpha particle has 4 nucleons, the total binding energy for the alpha particle is:
\[ \text{Binding Energy of } \, _2^4He = 4 \times 7.047 \, \text{MeV} = 28.188 \, \text{MeV} \]

### Step 3: Calculate the energy released \( Q \)

The energy released \( Q \) in the reaction can be calculated by finding the difference between the total binding energy of the products and the total binding energy of the reactants:
\[ Q = \text{Total Binding Energy of Products} - \text{Total Binding Energy of Reactants} \]
\[ Q = 28.188 \, \text{MeV} - 4.448 \, \text{MeV} = 23.74 \, \text{MeV} \]

Thus, the energy \( Q \) released in the fusion reaction is approximately:
\[ Q \approx 23.74 \, \text{MeV} \]

### Final Answer
The energy \( Q \) released is \( \approx 23.74 \, \text{MeV} \).

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