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JEE Main 2024 (9 Apr, Shift 1) · Nuclear Fission & Fusion

Nuclear Fission & Fusion question from JEE Main 2024 (9 Apr, Shift 1)

Q15JEE Main 2024 · 9 APR · SHIFT 1Difficult

Nuclear Fission & Fusion

A star has 100%100\% helium composition. It starts to convert three 4He{}^{4}\text{He} into one 12C{}^{12}\text{C} via triple alpha process as 4He+4He+4He→12C+Q{}^{4}\text{He} + {}^{4}\text{He} + {}^{4}\text{He} \rightarrow {}^{12}\text{C} + \text{Q}. The mass of the star is 2.0×1032 kg2.0 \times 10^{32}\text{ kg} and it generates energy at the rate of 5.808×1030 W5.808 \times 10^{30}\text{ W}. The rate of converting these 4He{}^{4}\text{He} to 12C{}^{12}\text{C} is n×1042 s−1\text{n} \times 10^{42}\text{ s}^{-1}, where n\text{n} is _________. [ Take, mass of 4He=4.0026u{}^{4}\text{He} = 4.0026\text{u}, mass of 12C=12u{}^{12}\text{C} = 12\text{u}]

Numerical answer type. Work it out, then open the answer.

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Answer: 5

  1. Δm=3×4.0026−12=0.0078 u\Delta m = 3\times4.0026 - 12 = 0.0078\text{ u}, so Q=0.0078×931.5≈7.27 MeV≈1.16×10−12 JQ = 0.0078\times931.5 \approx 7.27\text{ MeV} \approx 1.16\times10^{-12}\text{ J}.
  2. Reactions per second =PQ=5.808×10301.16×10−12≈5×1042 s−1= \dfrac{P}{Q} = \dfrac{5.808\times10^{30}}{1.16\times10^{-12}} \approx 5\times10^{42}\text{ s}^{-1}.
  3. So n=5n = 5.
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