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In the fusion reaction `._1^2H+_1^2Hrarr_2^3He+_0^1n`, the masses of deuteron, helium and neutron expressed in amu are `2.015, 3.017 and 1.009` respectively. If `1 kg` of deuterium undergoes complete fusion, find the amount of total energy released. 1 amu `=931.5 MeV//c^2`. |
Answer» `._(1)^(2)H+._(1)^(2)Herarr._(2)^(3)He+._(0)^(1)n` Mass defect: `Deltam = [(2.015xx2)-(3.017 +1.009)]` `=4 xx 10^(-3)"amu"` Equivalent enegry `= 4xx10^(-3) xx 931.5 = 3.726MeV` Number of atoms in `1kg` of `._(1)^(2)H= (1000)/(2) xx 6.023 xx 10^(23)` In one reaction, two atoms of `._(1)^(2)H` are used. Total energy released `=(1)/(2) ((100)/(2)xx6.023xx10^(23)) xx3.726xx1.6xx10^(-13)` `=9xx10^(13) J` |
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