Nuclear fusion A + B → C takes place with kinetic energies 5 MeV (A) and 3 MeV (B). C forms in an excited state of excitation energy 9.23 MeV. Given nuclear masses 25.001 u, 10.010 u and 35.000 u, respectively, and 1 u = 930 MeV/c², find the kinetic energy of C in MeV.
Physics · Mixed Chapter
JEE Mains

Question

Nuclear fusion A + B → C takes place with kinetic energies 5 MeV (A) and 3 MeV (B). C forms in an excited state of excitation energy 9.23 MeV. Given nuclear masses 25.001 u, 10.010 u and 35.000 u, respectively, and 1 u = 930 MeV/c², find the kinetic energy of C in MeV.

Physics · Mixed Chapter

G1 Physics · Accepted Answer

Step 1 — Mass defect of the reaction: Δm = (25.001+10.010−35.000) u = 0.011 u. Step 2 — Nuclear Q-value: Q = Δmc² = 0.011×930 = 10.23 MeV. Step 3 — Initial kinetic energy: K₁ = 5+3 = 8 MeV. Step 4 — Energy conservation including product excitation: K₁+Q = K₂+Ex = K₂+9.23. Hence K₂ = 8+10.23−9.23 = 9 MeV. Answer: 9.