Two identical hypothetical nuclei, each with mass ( 12 - .0064) amu, fuse to form a nucleus with mass ( 23 - .00629) amu and a neutron, whose mass is about 1.00867 amu. How much energy would be given off if 1.76 kg of these nuclei fused?

The total mass of the original particles is 2 ( 12 - .0064 ) amu, while the total mass of the resulting nucleus and neutron is ( 23 - .00629) amu + (1 + .00867 ) amu.

The total mass defect is therefore

• mass defect = 2 ( 12 - .0064 ) amu - [ ( 23 - .00629 ) amu - ( 1 + .00867 ) amu ] = .00856 amu = 13.69 * 10^-30 kg.

The corresponding energy is therefore

• energy = 13.69 * 10^-30 kg ( 3 * 10^8 m/s) ^ 2 = 123.2 * 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 1.76 kg there will be approximately

• number of nuclei = 1.76 kg / ( 39.84 * 10^-27 kg) = 4.417 * 10^25.

These nuclei would therefore give off

• energy released = 4.417 * 10^25 nuclei * ( 123.2 * 10^-14 J / nucleus ) = 544.1 * 10^11 Joules.