Question
In the fission reaction , the masses are: U-235 = 235.0439 u, Ba-141 = 140.9144 u, Kr-92 = 91.8973 u, neutron = 1.0087 u. Calculate the energy released per fission in MeV.
(JEE Main 2023, similar pattern)
Solution — Step by Step
Mass of reactants: u
Mass of products: u
Mass defect: u
Using the conversion: 1 u = 931.5 MeV/
This is approximately 200 MeV per fission — the standard value you should remember.
A single chemical reaction (like burning coal) releases about 4 eV of energy. A single nuclear fission releases about eV — that is 50 million times more energy per reaction.
This is why 1 kg of uranium can produce as much energy as about 3,000 tonnes of coal.
Why This Works
Einstein’s mass-energy equivalence tells us that mass and energy are interchangeable. In nuclear reactions, the total mass of products is slightly less than the total mass of reactants. This “missing” mass has been converted to energy.
The binding energy per nucleon curve explains why fission of heavy nuclei (like U-235) releases energy: the products (Ba, Kr) have higher binding energy per nucleon than uranium. The system moves to a more tightly bound state, releasing the excess as kinetic energy of fragments and neutrons.
The three neutrons released can trigger further fissions — this is the chain reaction that powers nuclear reactors and bombs.
Alternative Method
Instead of computing mass defect, you can use binding energies directly. Energy released = (total BE of products) - (total BE of reactants). But binding energy data is not always given in problems, while atomic masses usually are. The mass defect method is more universally applicable.
For JEE and CBSE: memorise that 1 u = 931.5 MeV. This conversion factor appears in nearly every nuclear physics calculation. Also remember: fission of U-235 releases about 200 MeV, and fusion of hydrogen to helium releases about 26 MeV (but per nucleon, fusion releases MORE energy).
Common Mistake
Students sometimes forget to account for the neutrons in the product side. The reaction produces 3 neutrons, each with mass 1.0087 u. Missing even one neutron changes significantly. Always balance the nuclear equation first and count all particles on both sides before computing masses.