If three helium nuclei combine to form a carbon nucleus then the energy released in this reaction is ________ × 10⁻² MeV. (Given 1 u=931 MeV / c², atomic mass of helium =4.002603 u)

Solution

Physics sir · 3 steps

Carbon-12 weighs exactly 12 u by definition, so the mass lost is just 3 helium masses minus 12.

1. Write the reaction. Three alpha particles fuse: 3 ⁴₂He → ¹²₆C. The mass of ¹²C is not given because the atomic mass unit is defined from it: its atomic mass is exactly 12 u.
2. Find mass defect. Mass before is 3 × 4.002603 = 12.007809 u. Mass after is 12.000000 u, so Δ m = 0.007809 u. Atomic masses work here because both sides carry 6 electrons, so the electron masses cancel.

   E = Δ m × 931 MeV = 0.007809 × 931 = 7.270 MeV

3. So. E ≈ 7.27 MeV = 727 × 10⁻² MeV, so the blank is 727.

The common slip: 7.27. Writing the energy in MeV and forgetting the 10⁻² in the blank gives 7.27, not 727.

The idea to keep. Energy released is the mass that goes missing, times 931 MeV per u. Remember ¹²C is exactly 12 u; questions often leave it out on purpose.

Answer from the NTA final answer key. Solution by PrepWiser.