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प्रश्न
How long can an electric lamp of 100W be kept glowing by fusion of 2.0 kg of deuterium? Take the fusion reaction as
\[\ce{^2_1H + ^2_1H -> ^3_1He + n + 3.27 MeV}\]
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उत्तर
The given fusion reaction is:
\[\ce{^2_1H + ^2_1H -> ^3_1He + n + 3.27 MeV}\]
Amount of deuterium, m = 2 kg
1 mole, i.e., 2 g of deuterium contains 6.023 × 1023 atoms.
∴2.0 kg of deuterium contains = `(6.023 xx 10^23)/2 xx 2000 = 6.023 xx 10^26` atoms
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:
`"E" = 3.27/2 xx 6.023 xx 10^26 "MeV"`
`= 3.27/2 xx 6.023 xx 10^26 xx 1.6 xx 10^(-19) xx 10^6`
`= 1.576 xx 10^14 " J"`
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:
`(1.576 xx 10^14)/100`s
`(1.576 xx 10^14)/(100 xx 60 xx 60 xx 24 xx 365) ~~ 4.9 xx 10^4` year
संबंधित प्रश्न
Distinguish between nuclear fission and fusion. Show how in both these processes energy is released. Calculate the energy release in MeV in the deuterium-tritium fusion reaction :
`""_1^2H+_1^3H->_2^4He+n`
Using the data :
m(`""_1^2H`) = 2.014102 u
m(`""_1^3H`) = 3.016049 u
m(`""_2^4He`) = 4.002603 u
mn = 1.008665 u
1u = 931.5 MeV/c2
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