E = mc2 NUCLEAR ENERGY FORMULA E = Energy release,J or eV m = mass defect,kg (*Always in positive number) c = speed of light = 3 x 108 ms-1
E = mc2 NUCLEAR ENERGY m = mass defect Mass defect = the differences of mass before and after reaction Atomic mass is too small,therefore it is measured in Atomic mass unit (a.m.u) 1 a.m.u = 1/12 of mass of carbon-12 atom. = 1/12 x 1.993 x 10-27 kg = 1.66 x 10-27 kg
NUCLEAR ENERGY 1 a.m.u = 1.66 x 10-27 kg EXERCISE Calculate the mass of atoms below in kg Radium –225 = 222.0254 a.m.u Radon – 222 = 222.0175 a.m.u Helium-4 = 4.0026 a.m.u 1 a.m.u = 1.66 x 10-27 kg 1. 222.0254 x 1.66 x 10-27 = 3.68562164 x 10-25 kg 2. 222.0175 x 1.66 x 10-27 = 3.6854905 x 10-25 kg 3. 4.0026 x 1.66 x 10-27 = 6.44316 x 10-27 kg
Ra Rn He NUCLEAR ENERGY energy + E = mc2 = 8.798 x 10-30 x (3 x 108)2 CALCULATING NUCLEAR ENERGY Nuclear equation of Radium-226 decay as follows; Ra 226 88 Rn 222 86 + He 4 2 energy Calculate the total nuclear energy released in Joule and eV. Given : Ra-226 = 226.0254 a.m.u Rn-222 = 222.0175 a.m.u He-4 = 4.0026 a.m.u Mass before = 226.0254 a.m.u Mass after = 222.0175 + 4.0026 = 226.0201 a.m.u E = mc2 Mass defect = 226.0254 - 226.0201 = 0.0053 a.m.u = 8.798 x 10-30 x (3 x 108)2 = 0.0053 x 1.66 x 10-27 = 7.92 x 10-13 J = 8.798 x 10-30 kg
Ra Rn He NUCLEAR ENERGY energy + E = E/e E = 7.92 x 10-13/1.60 x 10-19 CALCULATING NUCLEAR ENERGY Nuclear equation of Radium-226 decay as follows; Ra 226 88 Rn 222 86 + He 4 2 energy Calculate the total nuclear energy released in Joule and eV. Given : Ra-226 = 226.0254 a.m.u Rn-222 = 222.0175 a.m.u He-4 = 4.0026 a.m.u E = E/e E = 7.92 x 10-13/1.60 x 10-19 E = mc2 = 4.88 x 106 eV = 8.798 x 10-30 x (3 x 108)2 = 7.92 x 10-13 J
U Xe n Sr NUCLEAR ENERGY + energy X 1. Determine the value of X EXERCISE Below is a nuclear equation for nuclear reaction; U 235 92 Xe 140 54 n 1 Sr 94 38 + energy X Given : U-235 = 235.043924 a.m.u n = 1.008665 a.m.u Xe-140 = 139.921620 a.m.u Sr-94 = 93.915367 a.m.u 1. Determine the value of X 2. Calculate the total nuclear energy released in Joule and eV
U Xe n Sr NUCLEAR ENERGY energy + X 1. Determine the value of X EXERCISE Below is a nuclear equation for nuclear reaction; U 235 92 Xe 140 54 n 1 Sr 94 38 + energy X Given : U-235 = 235.043924 a.m.u n = 1.008665 a.m.u Xe-140 = 139.921620 a.m.u Sr-94 = 93.915367 a.m.u 1. Determine the value of X Nucleon number 235 + 1 = 140 + 94 + 1X 236 = 234 + 1X X = 2
U Xe n Sr NUCLEAR ENERGY energy + 2 EXERCISE Below is a nuclear equation for nuclear reaction; U 235 92 Xe 140 54 n 1 Sr 94 38 + energy 2 Given : U-235 = 235.043924 a.m.u n = 1.008665 a.m.u Xe-140 = 139.921620 a.m.u Sr-94 = 93.915367 a.m.u 2. Calculate the total nuclear energy released in Joule and eV. Mass before = 235.043924 + 1.008665 = 236.052589 a.m.u Mass after = 139.921620 + 93.915367 + 2(1.008665) = 235.854317 a.m.u
U Xe n Sr NUCLEAR ENERGY energy + 2 EXERCISE Below is a nuclear equation for nuclear reaction; U 235 92 Xe 140 54 n 1 Sr 94 38 + energy 2 Mass defect = 236.052589 – 235.854317 = 0.198272 a.m.u = 0.198272 x 1.66 x 10-27 = 0.32913152 x 10-27 kg E = mc2 = 0.32913152 x 10-27 x ( 3 x 108)2 = 2.962 x 10-11 J = 1.9 x 108 eV
PAST YEARS QUESTION 4 86 a. Complete the equation for the reaction above b. The nuclear reaction of on nucleus of Radium-226 experience a mass defect of 8.61818 x 10-30 kg.Calculate the energy released in the nuclear reaction. E = mc2 = 8.61818 x 10-30 x (3 x 108)2 = 7.76 x 10-15 J