Binding Energy and Mass defect Particle Relative Electric Charge Relative Mass Mass (kg) Charge (C) (u) Electron -1-1.60 x 10-19 5.485779 x 10-4 9.109390 x 10-31 Proton +1 +1.60 x 10-19 1.007276 1.672623 x 10-27 Neutron 0 0 1.008665 1.674929 x 10-27 1u = 1.6605 x 10-27 kg 1eV = 1.60 x 10-19 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 10-27 kg Mass defect = 6.6950 x 10-27 kg - 6.6447x 10-27 kg = 5.03 x 10-29 kg b) Binding energy using E =mc 2 = [5.03 x 10-29 kg] x [3 x 10 8 ] 2 E = 4.53 x 10-12 Joules c) Binding energy = 4.53 x 10-12 x 1.60 x 10-19 2) a) 238 92U 234 90Th + 4 2α = 2.83 x 10 7 ev [= 28.3 MeV ] b) First calculate mass change 238.0508u - (234.0426u + 4.0026u) mass change = 5.6 x 10-3 u Convert to kg = 5.6 x 10-3 u x 1.6605 x 10-27 kg Mass defect = 9.2988 x 10-30 Energy released E=mc 2 = 9.2988 x 10-30 x (3x10 8 ) 2 = 8.36892 x 10-13 J
3) Calculate the mass defect and binding energy the nuclide 10 5B where the mass of 10 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 Mass defect = Mass of nucleons - mass of 10 5B nucleus = 10.079705u - 10.0129 u = 0.066805 Mass defect in Kg = 1.1093 x 10-28 Kg Binding Energy E = mc 2 = 1.1093 x 10-28 x (3 x 10 8 ) 2 = 9.9836 x 10-12 J Binding Energy in ev = 9.9836 x 10-12 J / 1.6 x 10-19 = 6.2398 x 10 7 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 = mc 2 = 2.186 x 10-28 x (3 x 10 8 ) 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 10 8 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 = mc 2 = 0.004 x 1.66 x 10-27 x (3 x 10 8 ) 2 = 5.976 x 10-13 J ) (a) (b) 2 3 4 1 1H + 1H 2He + 0 n 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 = mc 2 = 3.1112 x 10-29 x (3x10 8 ) 2 = 2.80 x 10-12 J
MARK SCHEME! 1. (a) Deduct for each error or omission, stop at zero 2 max Property temperature of sample pressure on sample amount of sample Effect on rate of decay increase decrease stays the same 4 (b) (i) He/ 4 2 2α ; 222 86 Rn ; 2 mass defect = 5.2 10-3 u; energy = mc 2 = 5.2 10 3 1.661 10 27 9.00 10 16 / 1 u = 930 MeV; = 7.77 10 13 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 0 = mv + mrnvrn; v v Rn m Rn m α 222 = = 55.5; 3 4 Ignore absence of minus sign. (iii) kinetic energy of α-particle = ½mv 2 ; 2 V α ; 222 kinetic energy of radon nucleus = 1 2 m 4 55.5 this is 1 / 55.5 of kinetic energy of the α-particle; Accept alternative approaches up to [3 max]. 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 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 (iii) 235 U 1 0 92 n ; 90 142 Sr Xe 1 4 n ; 2 38 54 0 Allow ecf for RHS if LHS is incorrect. 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 use of Ek = / equivalent; m correct conversion of MeV to joule (1.63 10 11 J); correct conversion of mass to kilogram (1.50 10 25 kg) momentum = 2.2 10 18 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 (c) (i) probability of decay / constant in expression dn dt = N; dn per unit time / and N explained; 2 dt = ln 2 (28 365 24 3600) (note: substitution is essential) = 7.85 10 10 s 1 ; 1 (d) (i) N 0 exp( 7.85 10 N 0 10 exp( 0.462t) t) = 1.2 10 6 ; exp(0.462t) = 1.2 10 6 ; t = 30.3 s; 3 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 10 5 6.0 (d) λ = 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 ; (accept ln 2 for 0.69) T = 0.086 d 1 / 0.87 d 1 / 1.0 10 6 s 1 ; 2 (e) 0.5 = 6.4e 0.086t ; to give t = 30d / 2.6 10 6 s / 29d / 2.5 10 6 s; 2 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 [11] nuclides that have the same proton number but different nucleon number / same number of protons different number of neutrons; 1 (b) (i) 24 24 11 Na 12 Mg β v / e 0 / 1 e ; v ; 2 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 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 enriched fuel can be used in the manufacture of nuclear weapons; so possibly threatening World peace; 2 (c) (i) (energy released) = 2.1895 10 5 (1.3408 + 0.83749 + 0.0093956) 10 5 ; = 181.44 180MeV 1 kinetic; 1 (d) (i) number of atoms in 1 kg of carbon = 2351000 1 kg of U-235 = ; N A energy per kg carbon = kev; N A 121000 N A 124 kev and per kg U-235 = and number in 2351.810 N A 8 therefore ratio = 8.8 10 8 ; 3 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 92 U 239 Np 93 v 239 93 Np ; ; (iii) v ; 3 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. (a) (i) fission 1 max kinetic energy 1 max (b) the two neutrons can cause fission in two more uranium nuclei producing four neutrons so producing eight etc; OWTTE; 1 max 7. (a) (i) activity = ()N; 2 4.2510 18 1 λ 4.7810 s ; 2 19 8.9010 Allow 1.51 10 10 yr 1. ln 2 17 T 1 1.4510 s; 18 2 4.7810 = 4.60 10 9 years; 2 (b) eg activity would change during analysis to find N / rate of change of activity is too great to allow N(t) to be determined / OWTTE; 1 [5] 14
MARK SCHEME! 1. D 2. D 3. B 4. B 5. C 6. B 7. A 8. A 9. B 10. B 11. A 12. C 13. C 14. B 15. D 16. D 17. C 18. B 10
19. B 20. A 21. D 22. B 23. C 24. C 25. B 26. C 27. C 28. D 29. A 30. C 31. A 11
MARK SCHEME! 2. D 6. A 11. D 14. B 15. D 19. B 24. A 25. A 38. C