Two identical hypothetical nuclei, each with mass ( 12 - .0034) amu, fuse to form a nucleus with mass ( 23 - .00327) amu and a neutron, whose mass is about 1.00867 amu. How much energy would be given off if .73 kg of these nuclei fused?

The total mass of the original particles is 2 ( 12 - .0034 ) amu, while the total mass of the resulting nucleus and neutron is ( 23 - .00327) amu + (1 + .00867 ) amu.

The total mass defect is therefore

• mass defect = 2 ( 12 - .0034 ) amu - [ ( 23 - .00327 ) 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 * 12 amu = 39.84 * 10^-27 kg, so in .73 kg there will be approximately

• number of nuclei = .73 kg / ( 39.84 * 10^-27 kg) = 1.832 * 10^25.

These nuclei would therefore give off

• energy released = 1.832 * 10^25 nuclei * ( 122.9 * 10^-14 J / nucleus ) = 225.1 * 10^11 Joules.