The nuclear reaction `.^2H+.^2H rarr .^4 He` (mass of deuteron `= 2.0141 a.m.u` and mass of `He = 4.0024 a.m.u`) is

Options

A. fusion reaction releasing `24 MeV` energy
B. fusion reaction absorbing `24 MeV` energy
C. fission reaction releasing `0.0258 MeV` energy
D. fission reaction absorbing`0.0258 MeV` energy

Answer

To solve the problem regarding the nuclear reaction \( ^2H + ^2H \rightarrow ^4He \), we will follow these steps:

### Step 1: Identify the masses involved in the reaction
We know the following:
- Mass of deuteron (\( ^2H \)) = 2.0141 a.m.u
- Mass of helium (\( ^4He \)) = 4.0024 a.m.u

### Step 2: Calculate the total mass of the reactants
Since there are two deuterons in the reaction, we calculate the total mass of the reactants:
\[ \text{Total mass of reactants} = 2 \times \text{mass of deuteron} = 2 \times 2.0141 \, \text{a.m.u} = 4.0282 \, \text{a.m.u} \]

### Step 3: Calculate the mass defect
The mass defect (\( \Delta m \)) is the difference between the total mass of the reactants and the mass of the products:
\[ \Delta m = \text{Total mass of reactants} - \text{mass of products} \]
\[ \Delta m = 4.0282 \, \text{a.m.u} - 4.0024 \, \text{a.m.u} = 0.0258 \, \text{a.m.u} \]

### Step 4: Calculate the energy released using Einstein's equation
Using the mass-energy equivalence relation \( E = \Delta m c^2 \), where \( c^2 \) in terms of energy is given as 931.5 MeV/a.m.u:
\[ E = \Delta m \times c^2 = 0.0258 \, \text{a.m.u} \times 931.5 \, \text{MeV/a.m.u} \]
\[ E = 24.05 \, \text{MeV} \]

### Step 5: Conclusion
The nuclear reaction \( ^2H + ^2H \rightarrow ^4He \) is a fusion reaction that releases approximately 24 MeV of energy.