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Question
Calculate the energy in fusion reaction:
`""_1^2H+_1^2H->_2^3He+n`, where BE of `""_1^2H`23He=7.73MeV" data-mce-style="position: relative;">=2.2323He=7.73MeV MeV and of `""_2^3He=7.73 MeV`
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Solution
Given:
Binding energy of `""_1^2H`" data-mce-style="position: relative;" data-mce-tabindex="0"> E1 = 2.23 MeV
Binding energy of `""_2^3H`E2= 7.73 MeV
Energy in the fusion reaction is given by
∆E=E2−2E1=7.73 − (2×2.23)=3.27 MeV
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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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(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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