JEE Main20249 Apr 2024Morning ShiftPhysicsNuclear PhysicsActual
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 ]
Correct answer
0
Step-by-step solution
^4 He + ^4 He + ^4 He ¹² C + Q power generated = N t Q where, N No. of reaction / sec . aligned & Q = (3 ~m _ He - m _ C ) C ^2 & Q =(3 4.0026-12) (3 10^8 )^2 & Q =7.266 MeV aligned aligned & N t = power Q = 5.808 10³⁰ 7.266 10^6 1.6 10⁻¹⁹ & N t =5 10⁴² aligned rate of conversion of ^4 He into ¹² C =15 10⁴² Hence, n =15