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प्रश्न
Calculate the Q-values of the following fusion reactions :-
(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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उत्तर
(a) `Q = [2 xx m""_1^2"H" - (m(""_3^3"H") + m(""_1^3"H"))]c^2`
`= (4.028204 - 4.023874) xx 931` MeV
= 4.05 MeV
(b) `Q = [2 xx m ""_1^2"H" - (m_2^3"H" + m_n)]c^2`
`= [4.028204 - 4.024694] xx 931`
`= 0.00351 xx 931`
= 3.25 MeV
(c) `Q = [m ""_1^2"H" + m _1^3"H" - m _1^4"He"- m_n)]c^2`
`= (2.014102 + 3.016049 - 4.002603 - 1.008665) xx 931`
= 17.57 MeV
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संबंधित प्रश्न
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`
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 :
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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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