In the nuclear fusion reaction deuterium plus deuterium producing tritium, hydrogen-1, and E, the rest masses are 2.014 u for each deuterium, 3.016 u for tritium, and 1.008 u for hydrogen-1. Using 1 u = 1.66 x 10^-27 kg and c = 3.00 x 10^8 m/s, what is the energy E released?

1. The mass deficit is twice the deuterium mass minus the combined product masses: 2(2.014) minus (3.016 + 1.008) u. 2. Twice 2.014 is 4.028, and the products add to 4.024, so the deficit is 4.028 minus 4.024, which is 0.004 u. 3. Convert to kilograms: (0.004 u)(1.66 x 10^-27 kg/u) = 6.64 x 10^-30 kg. 4. Apply E = m c^2: (6.64 x 10^-30 kg)(3.00 x 10^8 m/s)^2. 5. Multiplying gives E of about 5.98 x 10^-13 J, which rounds to 6 x 10^-13 J. 6. 6 x 10^-14 J is ten times too small, and 6 x 10^-12 J and 6 x 10^-11 J are ten and a hundred times too large. 7. So 6 x 10^-13 J is correct. _Source: OpenStax Physics (CC BY 4.0), Ch 10 "Special Relativity", section 10.2 Consequences of Special Relativity_