JEE Main 2024 Physics question
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 = 4.0026 u, mass of ¹²C = 12 u]
Answer: 5
Solution
Q = (3 × 4.0026 − 12)c² = 0.0078 u × 931.5 MeV/u = 7.266 MeV = 1.163 × 10⁻¹² J. The rate of reactions = power/Q = 5.808 ×
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