RankLab
Back to Questions
PhysicsHardNumerical2024 · 09 Apr Shift 1

Q30.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 = 4.0026u, mass of 12C = 12u ]

What This Question Tests

This question involves calculating the energy released per nuclear reaction (Q-value) using mass defect, and then using the star's energy generation rate to find the rate at which helium nuclei are converted to carbon.

Concepts Tested

Mass defectEnergy released (Q-value)Rate of energy generationNumber of particles reacted per second

Formulas Used

Q = (Σm_reactants - Σm_products)c²

Power = Q × (rate of reactions)

📚 NCERT Sections This Tests

13.5The Q Value Of A Nuclear Reaction A + B ® C + D Is Defined By

Physics Class 12 · Chapter 13

82% match

13.5 The Q value of a nuclear reaction A + b ® C + d is defined by Q = [ mA + mb – mC – md]c2 where the masses refer to the respective nuclei. Determine from the given data the Q-value of the following reactions and state whether the reactions are exothermic or endothermic. (i) 11 H+13 H →12 H+12 H (ii) 126 C+126 C →1020 Ne+ 24 He Atomic masses are given to be m ( 12 H ) = 2.014102 u m ( 13 H) = 3.016049 u m ( 126 C ) = 12.000000 u m ( 1020 Ne ) = 19.992439 u

13.3A Given Coin Has A Mass Of 3.0 G. Calculate The Nuclear Energy That

Physics Class 12 · Chapter 13

79% match

13.3 A given coin has a mass of 3.0 g. Calculate the nuclear energy that would be required to separate all the neutrons and protons from each other. For simplicity assume that the coin is entirely made of 6329Cu atoms (of mass 62.92960 u).

13.2Obtain The Binding Energy Of The Nuclei 5626Fe And 20983 Bi In Units Of

Physics Class 12 · Chapter 13

78% match

13.2 Obtain the binding energy of the nuclei 5626Fe and 20983 Bi in units of MeV from the following data: m ( 5626Fe ) = 55.934939 u m ( 20983 Bi ) = 208.980388 u