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`.

Explanation

given , fusion reaction , `""_(1)H^(2)+""_(1)H^(2) to ""_(2)He^(3) +""_(0)n^(1) `
Mass defect , `Delta m=2xxm_(1)(""_(1)H^(2))-[m(""_(2)He^(4))+M_(n)]`
`=[2.015xx2)-(3.017+1.009)]`
`=4xx10^(-3) am u`
Equivalent energy,`E=Delta Mc^(2)=Delta mxx931.5 MeV //c^(2)`
`=4xx10^(-3)xx931.5=3.726 Me V `
number of atoms in 1 kg of `""_(1)H^(2)`
`=("given mass")/("Atomic mass")xxAvogadro number `
`(1000)/(2)xx6.023xx10^(23)`
in one reaction , two atoms of `""_(1)^(2)H` are used .
So total energy released
`=(1)/(2)((1000)/(2)xx6.023xx10^(23))xx3.726xx1.6xx10^(-13)J`







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