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

A, C

Explanation

`Deltam=2(2.015)-(3.017+1.009)=0.004`amu
`:. Energy released `=(0.004xx931.5)MeV=3.726MeV`
Energy released per deuteron `=(3.726)/(2)=1.863MeV`
Number of deuterons in `1kg=(6.02xx10^26)/(2)=3.01xx10^26`
:. Energy released per kg of deuterium fusion `=(3.01xx10^26xx1.863)=5.6xx10^26MeV`
`=9.0xx10^13J`