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Question
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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Solution
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
RELATED QUESTIONS
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
Calculate the height of the potential barrier for a head on collision of two deuterons.
(Hint: The height of the potential barrier is given by the Coulomb repulsion between the two deuterons when they just touch each other. Assume that they can be taken as hard spheres of radius 2.0 fm.)
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