what is the energy released when 0.5 grams of uranium 235 under goes a fission reaction?

Answer 1

ΔE = mc²
… m = 0.50g = 5 x 10ˉ⁴ kg
… c = 3 x 10⁸ m/s (speed of light in vacuum)
ΔE = (5 x 10ˉ⁴ Kg)(3 x 10⁸ m/s)² = 4.5 x 10¹³ kg∙m²/s² = 4.5 x 10¹³ joules
= 2.8 x 10²⁶ MeV

Answer 2

Published is ~200 MeV per nucleon fission energy of U-235 => 200 MeV/nucleon x (0.50/235)mole x 6.023E+23 nucleons/mole = 2.6E+23 MeV for fission of 0.50g U-235. Not sure why ΔE = mc² does not match this quantity. Both calculations are mass defect fission factors.