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Binding Energy and Mass Defect Binding Energy and Mass defect Particle Relative Electric Charge Relative Mass Mass (kg) Charge (C) (u) -19 -4 -31 Electron -1 -1.60 x 10 5.485779 x 10 9.109390 x 10 Proton +1 +1.60 x 10-19 1.007276 1.672623 x 10-27 -27 Neutron 0 0 1.008665 1.674929 x 10 1u = 1.6605 x 10-27 kg -19 1eV = 1.60 x 10 Joules 1u is converted into 931.5 MeV Solutions. 1) 1. a) Mass of component parts m = 2p+2n = 2(1.672623 x 10-27) + 2(1.674929 x 10-27) m= 6.6950 x 10-27kg Mass defect = 6.6950 x 10-27kg - 6.6447x 10-27kg = 5.03 x 10-29kg b) Binding energy using E =mc2 = [5.03 x 10-29kg] x [3 x 108]2 E = 4.53 x 10-12 Joules c) Binding energy = 4.53 x 10-12 x 1.60 x 10-19 = 2.83 x 107 eV [= 28.3 MeV ] 2) 238 234 4 a) 92U 90Th + 2α b) First calculate mass change 238.0508u - (234.0426u + 4.0026u) mass change = 5.6 x 10-3u Convert to kg = 5.6 x 10-3u x 1.6605 x 10-27kg Mass defect = 9.2988 x 10-30 Energy released E=mc2 = 9.2988 x 10-30x (3x108)2 = 8.36892 x 10-13 J 10 10 3) Calculate the mass defect and binding energy the nuclide 5B where the mass of 5B atom = 10.0129 u 10 5B has 5 protons and 5 neutrons Total mass of nucleons = mass of protons + mass of neutrons = 5 [1.007276u] + 5 [1.008665u] = 5.03638u + 5.043325 = 10.079705u 10 Mass defect = Mass of nucleons - mass of 5B nucleus = 10.079705u - 10.0129 u = 0.066805 Mass defect in Kg = 1.1093 x 10 -28 Kg Binding Energy E = mc2 = 1.1093 x 10 -28 x (3 x 108)2 = 9.9836 x 10-12 J Binding Energy in eV = 9.9836 x 10-12 J / 1.6 x 10-19 7 = 6.2398 x 10 eV = 624 MeV 17 4) O-17 8 O has 8 protons in the nucleus and 9 neutrons Total mass of nucleons = mass of protons + mass of neutrons = 8 [1.007276u] + 9 [1.008665u] = 8.058208u + 9.077985u = 17.136193u Mass defect = Mass of nucleons - mass of O17 nucleus = 17.136193u - 17.00454u = 0.131653u Mass defect in Kg = 0. 131653 x 1.6605 x 10-27 = 2.186 x 10 -28 Kg Binding Energy E = mc2 = 2.186 x 10 -28 x (3 x 108)2 = 1.9675 x 10-11 J Binding Energy in eV = 1.9675 x 10-11 J / 1.6 x 10-19 = 1.2297 x 108 eV = 123 MeV 5) Write out the reaction first (words will do here) Thorium Radium + alpha particle Calculate mass of products and reactants in terms of u Reactants Products 232.038u 228.031 + 4.003 232.038u 232.034 Calculate the difference = 232.038 - 232.034 = 0.004u Energy released E = mc2 = 0.004 x 1.66 x 10-27 x (3 x 10 8) 2 = 5.976 x 10-13 J ) (a) 2 3 4 1 1퐻 + 1퐻 → 2퐻푒 + 0푛 (b) Calculate mass of products and reactants in Kg Reactants Products 3.345 x 10-27 + 5.008 x 10-27 Kg 6.647 x 10-27 Kg + mass of neutron 8.353 x 10-27 6.647 x 10 -27 + 1.67493 x 10 -27 Mass difference = 8.353 x 10-27 - 8.321888 x 10-27 = 3.1112 x 10-29 Energy released E = mc2 = 3.1112 x 10-29 x (3x108)2 = 2.80 x 10-12 J MARK SCHEME! 1. (a) Deduct [1] for each error or omission, stop at zero 2 max Property Effect on rate of decay increase decrease stays the same temperature of sample pressure on sample amount of sample 4 4 (b) (i) 2 He/ 2 α ; 222 86 Rn ; 2 -3 (ii) mass defect = 5.2 × 10 u; 2 energy = mc –3 –27 16 = 5.2 × 10 × 1.661 × 10 × 9.00 × 10 / 1 u = 930 MeV; –13 = 7.77 × 10 J / 4.86 MeV; 3 max (c) (i) (linear) momentum must be conserved; momentum before reaction is zero; so equal and opposite after (to maintain zero total); 3 (ii) 0 = mv + mRnvRn; v mRn vRn mα 222 = – = –55.5; 3 4 Ignore absence of minus sign. 2 (iii) kinetic energy of α-particle = ½mv ; V 2 1 222 kinetic energy of radon nucleus = 2 m α ; 4 55.5 this is 1 / 55.5 of kinetic energy of the α-particle; 3 max Accept alternative approaches up to [3 max]. (d) eg (-ray) photon energy or radiation; 1 (e) (i) two (light) nuclei; combine to form a more massive nucleus; with the release of energy / with greater total binding energy; 3 (ii) high temperature means high kinetic energy for nuclei; so can overcome (electrostatic) repulsion (between nuclei); to come close together / collide; high pressure so that there are many nuclei (per unit volume); so that chance of two nuclei coming close together is greater; 5 10 2. (a) (i) fission: nucleus splits; into two parts of similar mass; radioactive decay: nucleus emits; a particle of small mass and / or a photon; 4 235 1 (ii) 92 U 0 n ; 90 142 1 38 Sr 54 Xe 40 n ; 2 Allow ecf for RHS if LHS is incorrect. (iii) mass number unchanged; atomic number increases by +1; 2 (b) (i) kinetic energy of neutrons; and energy of gamma ray photons; 2 Accept other valid possibilities but do not accept “heat”. p 2 (ii) use of Ek = / equivalent; m –11 correct conversion of MeV to joule (1.63 × 10 J); –25 correct conversion of mass to kilogram (1.50 × 10 kg) –18 momentum = 2.2 × 10 N s; 4 (iii) total momentum after fission must be zero; must consider momentum of neutrons (and photons); 2 (iv) xenon not opposite to strontium but deviation < 30°); arrow shorter / longer; 2 dN (c) (i) probability of decay / constant in expression = –N; dt dN per unit time / and N explained; 2 dt ln 2 (ii) = (note: substitution is essential) (28 365 24 3600) –10 –1 = 7.85 × 10 s ; 1 10 N 0 exp(7.85 10 t) 6 (d) (i) = 1.2 × 10 ; N 0 exp(0.462t) 6 exp(0.462t) = 1.2 × 10 ; t = 30.3 s; 3 (ii) activity of the strontium will be much greater than that of the xenon; and extent of health hazard depends on activity; 2 [26] 11 3. (a) the nuclei of different isotopes of an element have the same number of protons; but different numbers of neutrons; 2 Look for a little more detail than say just “same atomic (proton) number, different mass (nucleon) number”. (b) Z for iodine = 53; + antineutrino; (accept symbol) 2 Do not accept neutrino or gamma or energy, etc. (c) 6.4×105 6.0 5.0 4.0 3.0 2.0 1.0 0 0 5.0 10 15 20 25 time / days shown on graph at least the 0, 8 and 16 day data points; exponential shape; scale on y-axis / goes through 24 day point; 3 0.69 (d) λ = ; (accept ln 2 for 0.69) T –1 –1 –6 –1 = 0.086 d / 0.87 d / 1.0 × 10 s ; 2 –0.086t (e) 0.5 = 6.4e ; 6 6 to give t = 30d / 2.6 × 10 s / 29d / 2.5 × 10 s; 2 [11] 4. (a) (i) an atom or nucleus that is characterized by the constituents of its nucleus / a particular type of atom or nucleus / OWTTE; (in particular) by its proton (atomic) number and its nucleon number / number of protons and number of neutrons; 2 (ii) nuclides that have the same proton number but different nucleon number / same number of protons different number of neutrons; 1 24 24 (b) (i) 11 Na 12 Mg β v – – 0 / e / 1 e ; v ; 2 (ii) 5.00216 MeV is equivalent to 0.00537 u; 23.99096 = 23.98504 + 0.00537 + rest mass of particle; rest mass= 0.00055 u; 3 No credit given for bald correct answer. (c) sodium-24 has more nucleons; and more nucleons (usually) means greater (magnitude of) binding energy; or sodium-23 has less nucleons; and less nucleons (usually) means less (magnitude of) binding energy; 2 12 5. (a) (i) there is more uranium-238 present than uranium-235; neutron capture is more likely in U-238 with high energy neutrons; if the neutrons are slowed they are more likely to produce fission in U-235 than neutron capture in U-238; 3 (ii) control rate at which reactions take place; by absorbing neutrons; 2 (b) (i) fuel enrichment means that the amount of uranium-235 present in the fuel is increased / OWTTE; this means that more U-235 available for fission; therefore the reaction can be sustained; 3 (ii) enriched fuel can be used in the manufacture of nuclear weapons; so possibly threatening World peace; 2 5 5 (c) (i) (energy released) = 2.1895 10 (1.3408 + 0.83749 + 0.0093956) 10 ; = 181.44 180MeV 1 (ii) kinetic; 1 121000 (d) (i) number of atoms in 1 kg of carbon = and number in N A 2351000 1 kg of U-235 = ; N A 124 2351.8108 energy per kg carbon = keV and per kg U-235 = N A N A keV; 8 therefore ratio = 8.8 10 ; 3 (ii) a much higher energy density implies that uranium will produce more energy per kg / smaller quantity of uranium needed to produce same amount of energy / OWTTE; 1 (e) (i) half-life: time for the activity to decrease by half / OWTTE; isotope: isotopes of elements are chemically identical but have different atomic masses / OWTTE / same number of protons in the nucleus but different number of neutrons / OWTTE; 2 239 239 (ii) 92 U 93Np v 239 93 Np ; ; v ; 3 (iii) advantage: plutonium is another fissionable element / may be used as nuclear fuel; and is readily produced in reactors that use uranium as a fuel; disadvantage: -particles are harmful to living organisms / OWTTE; and the plutonium lasts for a very long-time / OWTTE; 4 [25] 13 6.
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