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`.

Options

A. `9.0 xx 10^(13)`J
B. `20 xx 10^(5)` J
C. `5 xx 10^(16)`J
D. `8 xx 10^(5)`J

Answer

A

Explanation

`Deltam = 2(2.0915) - (3.017 + 1.009) = 0.004 amu`
`therefore` Energy released = `(0.004 xx 931.5)`MeV = 3.726 MeV
Number of deuterons in 1 Kg = (6.02 xx 10^(26))/2 = 3.01 xx 10^(26)`
`therefore Energy released per kg of deuterium fusion
`(3.01 xx 10^(26) xx 1.863)`
`=5.6 xx 10^(26)` MeV `=9.0 xx 10^(13)`J