Question

Energy released when two deuterons $ \text{(H}_2\text{)} $ fuse to form a helium nucleus $ \text{(He}_4\text{)} $ is:

• A. 8.1 MeV
• B. 5.9 MeV
• C. 23.6 MeV
• D. 26.8 MeV

Hint:

The energy released in a nuclear fusion reaction can be calculated using the binding energy per nucleon of the reactants and products.

Answer 1

The Correct Option is C

Given: - Binding energy per nucleon of \( \text{H}_2^1 \) = 1.1 MeV 
- Binding energy per nucleon of \( \text{He}_4^2 \) = 7.0 MeV 
The energy released \( Q \) is the difference between the binding energy of the reactants and products: \[ E_B = \text{BE}_{\text{reactant}} - \text{BE}_{\text{product}} \] \[ E_B = 1.1 \times 2 + 1.1 \times 2 - 7 \times 4 = 23.6 \, \text{MeV} \] Thus, the energy released is: \[ Q = 23.6 \, \text{MeV} \]

Answer 2

We are asked to find the energy released when two deuterons \( (\text{H}_2) \) fuse to form one helium nucleus \( (\text{He}_4) \).

Concept Used:

The energy released in a nuclear fusion reaction is due to the mass defect and is given by Einstein’s relation:

\[ E = \Delta m\,c^2 \]

where \( \Delta m \) is the decrease in mass (mass defect) between reactants and products.

Step-by-Step Solution:

Step 1: Write the nuclear fusion reaction.

\[ {}^2_1\text{H} + {}^2_1\text{H} \longrightarrow {}^4_2\text{He} \]

Step 2: Write the approximate atomic masses involved (in atomic mass units, u):

\[ m({}^2_1\text{H}) = 2.0141\,\text{u}, \quad m({}^4_2\text{He}) = 4.0026\,\text{u} \]

Step 3: Compute the mass defect:

\[ \Delta m = \text{mass of reactants} - \text{mass of product} \] \[ \Delta m = [2 \times 2.0141] - 4.0026 = 4.0282 - 4.0026 = 0.0256\,\text{u} \]

Step 4: Convert this mass defect into energy using the conversion factor \( 1\,\text{u} = 931.5\,\text{MeV}/c^2 \):

\[ E = \Delta m \times 931.5 = 0.0256 \times 931.5 = 23.85\,\text{MeV} \]

Final Computation & Result:

The energy released when two deuterons fuse to form a helium nucleus is approximately:

\[ \boxed{E \approx 23.8\,\text{MeV}} \]

Hence, about 23.8 MeV of energy is released in the fusion of two deuterons to form one helium nucleus.