Question

In the nuclear fusion reaction \( 4^1H + 2e^- \to 2v + 6\gamma + \), what is the missing product?

• A. \( 3_2 He, 0.42 \, \text{MeV} \)
• B. \( 4_2 He, 22.86 \, \text{MeV} \)
• C. \( 4_2 He, 26.7 \, \text{MeV} \)
• D. \( 3_2 He, 5.49 \, \text{MeV} \)

Hint:

In fusion reactions, the energy released depends on the mass difference between the reactants and products. The mass defect is converted into energy, following \( E = mc^2 \).

Answer 1

The Correct Option is C

Step 1: Understanding the nuclear fusion reaction.
In nuclear fusion reactions, when hydrogen nuclei fuse, the product is typically a helium nucleus and energy. The missing product is the helium nucleus (\( 4_2 He \)) with a specific energy value.
Step 2: Applying the known energy values.
The energy released during the fusion of hydrogen atoms to form helium in this reaction is known to be approximately \( 26.7 \, \text{MeV} \).
Step 3: Conclusion.
The correct answer is (C) \( 4_2 He, 26.7 \, \text{MeV} \).