Modern Physics 5 Question 15

15. A star initially has 1040 deuterons. It produces energy via the processes 1H2+1H21H3+p and 1H2+1H32He4+n. If the average power radiated by the star is 1016W, the deuteron supply of the star is exhausted in a time of the order of

(1993, 2M)

(a) 106s

(c) 1012s

(b) 108s

(d) 1016s

The mass of the nuclei are as follows

M(H2)=2.014amu;M(n)=1.008amu

M(p)=1.007amu;M(He4)=4.001amu

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Answer:

Correct Answer: 15. (c)

Solution:

  1. The given reactions are :

1H2+1H21H3+p1H2+1H32He4+n31H22He4+n+p

Mass defect

Δm=(3×2.0144.0011.0071.008)amu=0.026amu

Energy released =0.026×931MeV

=0.026×931×1.6×1013J=3.87×1012J

This is the energy produced by the consumption of three deuteron atoms.

Total energy released by 1040 deuterons

=10403×3.87×1012J=1.29×1028J

The average power radiated is P=1016W or 1016J/s.

Therefore, total time to exhaust all deuterons of the star will be t=1.29×10281016=1.29×1012s1012s



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