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

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.

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

To calculate the energy released during a fission reaction, we need to use Einstein's mass-energy equivalence equation: E = mc².

Step 1: Convert the mass of uranium-235 to kilograms.
We know that 1 gram is equal to 0.001 kilograms. Therefore, 0.5 grams is equal to 0.5 × 0.001 = 0.0005 kilograms.

Step 2: Calculate the energy released using the mass-energy equivalence equation.
E = (0.0005 kg) × (speed of light)²
The speed of light, c, is approximately 3 × 10^8 meters per second.
E = 0.0005 × (3 × 10^8)²

Step 3: Calculate the energy.
E = 0.0005 × (9 × 10^16)
E = 4.5 × 10^13 Joules

Therefore, when 0.5 grams of uranium-235 undergoes a fission reaction, approximately 4.5 × 10^13 Joules of energy are released.