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 MeV
C. 23.8 MeV
D. 931 MeV

Explanation

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: Calculate the total binding energy of the reactants (deuterons)
The binding energy per nucleon for deuteron \( _1^2H \) is given as \( 1.112 \, \text{MeV} \). Since there are two deuterons, we first calculate the total binding energy for the two deuterons.

\[ \text{Total binding energy of 2 deuterons} = 2 \times \text{Binding energy per nucleon} \times \text{Number of nucleons} \]

Each deuteron has 2 nucleons, so:

\[ \text{Total binding energy of 2 deuterons} = 2 \times 1.112 \, \text{MeV} \times 2 = 4.448 \, \text{MeV} \]

### Step 2: Calculate the total binding energy of the product (alpha particle)
The binding energy per nucleon for the alpha particle \( _2^4He \) is given as \( 7.047 \, \text{MeV} \). The alpha particle has 4 nucleons, so:

\[ \text{Total binding energy of alpha particle} = \text{Binding energy per nucleon} \times \text{Number of nucleons} \]

\[ \text{Total binding energy of alpha particle} = 7.047 \, \text{MeV} \times 4 = 28.188 \, \text{MeV} \]

### Step 3: Calculate the energy released \( Q \)
The energy released in the reaction can be calculated as the difference between the total binding energy of the products and the total binding energy of the reactants:

\[ Q = \text{Total binding energy of products} - \text{Total binding energy of reactants} \]

Substituting the values we found:

\[ Q = 28.188 \, \text{MeV} - 4.448 \, \text{MeV} = 23.74 \, \text{MeV} \]

### Final Answer
The energy \( Q \) released in the fusion reaction is approximately \( 23.74 \, \text{MeV} \).

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