in the fusion reaction `""_(1)H^(2)+""_(1)H^(2)to ""_(2)He^(3) +""_(0)N^(1),` are masses of deuteron and neutron expressed in amu are 2.015, 3.017 and 1.0009, respectively , if 1 kg of deuterium undergoes complete fusion , then find the amount of total energy released .
1 amu = 931.5 ME V/`c^(2)`.

Options

A. `~~6.02 xx10^(13)J`
B. `~~5.6 xx10^(13)J`
C. `~~9.0 xx10^(13)J`
D. `~~0.9 xx10^(13)J`

Answer

c

Explanation

Mass defect,
`Delta m =2(2.0015)-(3.017 +1.009)=0.004 a.m.u.`
As `1 a.m.u. =931. 5MeV//c^(2)`, energy released will be `0.004 xx 931.5 =3.726 MeV`
Energy released per deuterons is
`(3.726)/(2)=1.863 MeV`
Number of molecules in `1 kg` deuteron is
`(6.02 xx 10^(26))/(2)=3.01 xx 10^(26)`
Therfore, energy released per kg of deuterium fusion
`=(3.01 xx 10^(26) xx 1.863)`
`= 5.6 xx 10^(26) MeV ~~=9.0 xx10^(13) J`.