Question

Calculate the energy released when 1.00 kg of uranium-235 undergoe the following fission process. 592U136 539 9639Y4n Mass (amu) Particle lodine-136 135.8401 95.8629 Yttrium-96 Uranium-235 234.9935 Neutron 1.00867

Answer 1

In nuclear reaction energy is released due to mass defect. The mass of the left side does not become equal to that of right side. From the mass defect we can calculate the energy released by the equation, E = mc².

Mass of left side = (mass on 'n' + mass of U) = (1.00867 + 234.9935) amu = 236.00217 amu

Mass of right side = (mass of I + mass of Y + 4 x mass of 'n') = (135.8401 + 95.8629 + 4x1.00867) amu = 235.73768 amu

So, mass defect = (236.00217 - 235.73768) amu = 0.26449 amu = [ 0.26449 / (6.023 x 10²³) ] g = 4.3913 x 10^-25 g (per mole of U²³⁵ nuclei)

Now, this much mass defect is observed for one mole U²³⁵ nuclei.

Now, 1 Kg U²³⁵ = 1000 g U²³⁵ = (1000 / 234.9935) mol U²³⁵ = 4.2554 mol U²³⁵

So, for 1 Kg U²³⁵ mass defect will be (4.2554 x 4.3913 x 10^-25) g = 1.8687 x 10^-24 g = 1.8687 x 10^-27 Kg

Thus, released energy = [ 1.8687 x 10^-27 x (3 x 10⁸)² ] J = 1.682 x 10^-10 J

In MeV unit (divide energy in J by 1.602 x 10^-19) the released energy is 1050 MeV.