CALCULATE THE ENERGY RELEASED WHEN O.5 GRAMS OF URANIUM 235 UNDERGOES FISSION REACTION.

Answer 1

0.50g U-235 => (2.13E-3 mole)(6.02E23molecules/mole)(235nuclei/molecule)(7.59MeV/nucleon)* = 2.287E24Mev
= (2.287E24MeV)(1.60218E-16Kj/Mev) = 3.66E8Kj per 0.50g U-235.

*7.59MeV/nucleon = mass defect for U-235 mass to energy conversion equivalence. It is calculated as follows...
=> The sum of the masses of the 92 protons and 143 neutrons in U-235 is (92protons) (1.007825 a.m.u./proton) + (143neutrons) (1.008665 a.m.u./neutron) = 236.958995 a.m.u. (Theoretical Atomic Mass).
This is in excess of the 'actual' atomic mass of U-235 molecule => 235.043943 a.m.u. Therefore, the total nuclear binding energy is thus
(236.958995 – 235.043943)a.m.u. = 1.915052 a.m.u. (mass defect)
= (1.915052 a.m.u.) (931 MeV/a.m.u.) = 1782.9 MeV/molecule
The binding energy per nucleon is then ... (1782.9MeV/molecule)/(235nucleons/molecule) = 7.59 MeV/nucleon.