Atomic StructureHard
Question
A star initially has 104 deutetron. It produces energy via the processes
1H2 + 1H2 → 1H3 + p and 1H2 + 1H3 → 2He4 + n.
If the average power radiated by the star is 1016 W, the deuteron supply of the star is exhausted in a time of the order of
The mass of the nuclei are follows
M(H2) = 2.014 amu; M(n) = 1.008amu;
M(p) = 1.007 amu; M(He4) = 4.001 amu.
1H2 + 1H2 → 1H3 + p and 1H2 + 1H3 → 2He4 + n.
If the average power radiated by the star is 1016 W, the deuteron supply of the star is exhausted in a time of the order of
The mass of the nuclei are follows
M(H2) = 2.014 amu; M(n) = 1.008amu;
M(p) = 1.007 amu; M(He4) = 4.001 amu.
Options
A.106s
B.108s
C.1012s
D.1016s
Solution
The given reactions are :
1H2 + 1H2 → 1H3 + p
1H2 + 1H3 → 2He4 + n
3 1H2 → 2He4 + n + p
Mass defect
ᐃm = (3 × 2.014 - 4.001 - 1.007 - 1.008) amu
= 0.026 amu
Energy released = 0.026 × 931 MeV
= 0.026 × 931 × 1.6 × 10-13 J
= 3.87 × 10-12 J
This is the energy produced by the consumption of three deutron atoms.
∴ Total energy released by 1040 deutrons
× 3.87 × 1012 J = 1.29 × 1028 J
The average power radiated is P = 1016 W or 1016 J/s.
Therefore, total time to exhaust all deutrons of the star will be
1.29 × 1012s 
1012s
1H2 + 1H2 → 1H3 + p
1H2 + 1H3 → 2He4 + n
3 1H2 → 2He4 + n + p
Mass defect
ᐃm = (3 × 2.014 - 4.001 - 1.007 - 1.008) amu
= 0.026 amu
Energy released = 0.026 × 931 MeV
= 0.026 × 931 × 1.6 × 10-13 J
= 3.87 × 10-12 J
This is the energy produced by the consumption of three deutron atoms.
∴ Total energy released by 1040 deutrons
× 3.87 × 1012 J = 1.29 × 1028 J The average power radiated is P = 1016 W or 1016 J/s.
Therefore, total time to exhaust all deutrons of the star will be
1.29 × 1012s 1012s Create a free account to view solution
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