Calculate the energy released in joules in the following nuclear reaction : `""_(1)^(2)H + ""_(1)^(2)H to ""_(2)^(3)He + ""_(0)^(1)N` Assume that the masses of `""_(1)^(2)H, ""_(2)^(3)He` and neutron are 2.0141, 3.0160 and 1.0087 amu respectively.

Options

A. `5.22 xx 10^(-13) ` J
B. `5.22 xx 10^(-12)` J
C. `5.22 xx 10^(-14)` J
D. `5.22 xx 10^(-11)` J

Answer

A

Explanation

`Deltam = 2 xx m ""_(1)^(2)H - m""_(2)^(3)He + m_(n) = (2 xx 2.0141) - (3.016 + 1.0087) = 0.0035` amu
`DeltaE = Deltam xx 931.48 "Mev" = 0.0035 xx 931.48 = 3.26 "Mev" = 5.223 xx 10^(-13)J (because 1"Mev" = 1.60 xx 10^(-13))`