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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
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
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