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

`._(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`