Two identical hypothetical nuclei, each with mass ( 13 - .0072) amu, fuse to form a nucleus with mass ( 25 - .00707) amu and a neutron, whose mass is about 1.00867 amu. How much energy would be given off if .44 kg of these nuclei fused?

The total mass of the original particles is 2 ( 13 - .0072 ) amu, while the total mass of the resulting nucleus and neutron is ( 25 - .00707) amu + (1 + .00867 ) amu.

The total mass defect is therefore

• mass defect = 2 ( 13 - .0072 ) amu - [ ( 25 - .00707 ) amu - ( 1 + .00867 ) amu ] = .00854 amu = 13.66 * 10^-30 kg.

The corresponding energy is therefore

• energy = 13.66 * 10^-30 kg ( 3 * 10^8 m/s) ^ 2 = 122.9 * 10^-14 Joules.

The mass of each of the resulting particles is very close to 2 * 13 amu = 43.16 * 10^-27 kg, so in .44 kg there will be approximately

• number of nuclei = .44 kg / ( 43.16 * 10^-27 kg) = 1.019 * 10^25.

These nuclei would therefore give off

• energy released = 1.019 * 10^25 nuclei * ( 122.9 * 10^-14 J / nucleus ) = 125.2 * 10^11 Joules.