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Problem 5: Find the energy released in the following fission reaction; 0n1 +92U235 ------→ 36Kr92 + 56Ba141 + 3 0n1 + Q

Solution

This is an example of the fission reaction. U235 is bombarded with neutron resulting in the formation of daughter nuclei of Ba and Kr, 3 neutrons and release of energy. The energy released will be equal to the difference in the masses on both sides of the equation converted to energy.

Sum of the masses on LHS;

Mass of one 0n1 (neutron) = 1.0086654 u

Mass of one 92U235 = 235.0439299 u

Total mass on LHS = 236.0525953 u

Mass of 36Kr92 = 91.92615621 u

Mass of 56Ba141 = 140.9144110 u

Mass of 3 neutrons = 3 × 1.0086654 = 3.0259962 u

Total mass on RHS = 235.8665634 u

Difference in masses on both sides (mass defect) = Δm = 236.0525953 – 235.8665634

Δm = 0.18603189 u

Now we know that 1 u = 931.5 MeV

Therefore,   0.18603189 u = 0.18603189 × 931.5 = 173.2887 MeV

3 Comments

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