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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
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\[\ce{^2_1H + ^2_1H -> ^3_1He + n + 3.27 MeV}\]
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(a) `""_1^2H + ""_1^2H → ""_1^3H + ""_1^1H`
(b) `""_1^2H + ""_1^2H → ""_2^3H + n`
(c) `""_1^2H + ""_1^3H → _2^4H + n`.
Atomic masses are `m(""_1^2H) = 2.014102 "u", m(""_1^3H) = 3.016049 "u", m(""_2^3He) = 3.016029 "u", m(""_2^4He) = 4.002603 "u".`
(Use Mass of proton mp = 1.007276 u, Mass of `""_1^1"H"` atom = 1.007825 u, Mass of neutron mn = 1.008665 u, Mass of electron = 0.0005486 u ≈ 511 keV/c2,1 u = 931 MeV/c2.)
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