13. How long can an electric lamp of 100W be kept glowing by fusion of 2Kg of deuterium? Take fusion reaction is 1H2 + 1H2 = 2He4 + n+ 3.27MeV
Asked by Sudheer Nair
| 28th Feb, 2012,
02:19: PM
Expert Answer:
The given fusion reaction is:
Amount of deuterium, m = 2 kg
1 mole, i.e., 2 g of deuterium contains 6.023 x 1023 atoms.
And 2.0 kg of deuterium contains 
It can be inferred from the given reaction that when two atoms of deuterium fuse, 3.27 MeV energy is released.
Total energy per nucleus released in the fusion reaction:
Power of the electric lamp, P = 100 W = 100 J/s
Hence, the energy consumed by the lamp per second = 100 J
The total time for which the electric lamp will glow is calculated as:
The given fusion reaction is:
Amount of deuterium, m = 2 kg
1 mole, i.e., 2 g of deuterium contains 6.023 x 1023 atoms.
And 2.0 kg of deuterium contains
It can be inferred from the given reaction that when two atoms of deuterium fuse, 3.27 MeV energy is released.
Total energy per nucleus released in the fusion reaction:
Power of the electric lamp, P = 100 W = 100 J/s
Hence, the energy consumed by the lamp per second = 100 J
The total time for which the electric lamp will glow is calculated as:
Answered by
| 28th Feb, 2012,
09:08: PM
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