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. `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