4He+4He+4He→12C+Q power generated =tNQ where, N→ No. of reaction /sec. Q=(3 mHe−mC)C2Q=(3×4.0026−12)(3×108)2Q=7.266MeV tN=Q power =7.266×106×1.6×10−195.808×1030tN=5×1042 rate of conversion of 4He into 12C=15×1042 Hence, n=15
JEE Main 2024 — Physics Modern Physics
A star has 100% helium composition. It starts to convert three 4He into one 12C via triple alpha process as 4He+4He+4He→12C+Q. The mass of the star is 2.0×1032 kg and it generates energy at the rate of 5.808×1030 W. The rate of converting these 4He to 12C is n×1042 s−1, where n is _________ [ Take, mass of 4He=4.0026u, mass of 12C=12u ]
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