A helium-4 nucleus has a binding energy of 4.53 x 10^-12 J. Using c = 3.00 x 10^8 m/s, which weighs more: the assembled nucleus, or its two separate protons and two separate neutrons?

1. Releasing binding energy when the nucleus forms means the nucleus ends up with less mass than its separate parts, following E0 = m c^2. 2. Rearranging for mass gives m = E0 divided by c^2 = (4.53 x 10^-12 J) divided by (9.00 x 10^16). 3. Dividing gives a mass difference of about 5.03 x 10^-29 kg. 4. Because the nucleus released this energy to form, the separate protons and neutrons together weigh more than the assembled nucleus by this amount. 5. The two are not equal, and 4.53 x 10^-12 kg wrongly reuses the energy value as if it already were a mass. 6. So the separate nucleons weighing about 5.03 x 10^-29 kg more is correct. _Source: OpenStax Physics (CC BY 4.0), Ch 10 "Special Relativity", section 10.2 Consequences of Special Relativity_