A star has 100 % helium composition. It starts to convert three ⁴ He into one ¹² C via triple alpha process as ⁴ He+ ⁴ He+ ⁴ He → ¹² C+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 ⁴ He to ¹² C is n × 10⁴² s⁻¹, where n is _________. [ Take, mass of ⁴ He=4.0026 u, mass of ¹² C=12 u]

Solution

Physics sir · 3 steps

Find the energy released by one triple-alpha reaction, then divide the star's power by it.

1. Mass defect. Three helium nuclei have mass 3 × 4.0026 = 12.0078 u and the carbon nucleus has 12 u. So Δ m = 0.0078 u is converted to energy in each reaction.
2. Energy per reaction. With 1 u = 931.5 MeV, Q = 0.0078 × 931.5 = 7.27 MeV. In joules, Q = 7.27 × 1.6 × 10⁻¹³ ≈ 1.164 × 10⁻¹² J.

   Reactions per second = P/Q = (5.808 × 10³⁰ J s⁻¹)/(1.164 × 10⁻¹² J) ≈ 4.99 × 10⁴² s⁻¹

3. So. The conversion rate is about 5 × 10⁴² s⁻¹, so n = 5. The star's mass is not needed for this rate.

The common slip: 15. Counting the helium nuclei used instead of the reactions multiplies the answer by 3 and gives 15. The question asks for the rate of the three-to-one conversion.

The idea to keep. For a star or reactor, the number of reactions per second is the power divided by the energy one reaction releases. Get Q from the mass defect first.

Answer from the NTA final answer key. Solution by PrepWiser.