What is the nuclear binding energy in joules for an atom of helium–4? assume the following: mass defect = 5.0531 × 10-29 kilograms. use e = mc2, with c = 3 × 108 m/s.

The equation given, E = mc², can be directly used to determine the unknown amount of nuclear binding energy. Substitute the values given to the equation,
               E = (5.0531 x 10^-29 kg) x (3 x 10^8 m/s)² = 4.548 x 10^-12 J
Thus, the nuclear binding energy is equal to 4.548 x 10^-12 J. 

The correct answer is 4.53 * 10-12 joules