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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]

Asked in JEE Main 9th April 1st Shift 2024 · Energy released in fission and fusion

Answer: 5

Step-by-step solution

Δ m=3×4.0026-12=0.0078 u; Q=0.0078×931.5=7.27 MeV=1.16×10⁻¹² J

Reactions per second =(5.808×10³⁰)/(1.16×10⁻¹²)≈5×10⁴² s⁻¹

→ 5

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