PhysicsAndMathsTutor.com 1 1. (a) (i) proton number 82 and nucleon number 214 (ii) Pb 2 (b) (i) kinetic energy [or electrostatic potential energy] (ii) m = 8.6 E 2 c 1 10 = 8 2 (10 ) = 9.6 10 0 kg [5] 2. 100y = 100 65 24 600 (=.15 10 9 s) energy needed =.15 10 9 00 10 (= 9.46 10 12 J) 12 number of disintegrations = 9.46 10 (= 2.96 10 2 ) 11.2 10 2 number of moles needed = 2.96 10 (= 0.49) 2 6.0210 molar mass = 0.29kg mass needed = 0.49 0.29 = 0.117 kg [7]. (a) N 120 100 80 60 40 20 0 0 20 40 60 80 Z straight line between (Z = 0, N = 0) to (Z = 20, N = 20) curving upwards to Z = 80; N = 110 10 2
PhysicsAndMathsTutor.com 2 (b) (i) A = any region below the line of stability but N > 80and Z > 60 (ii) B = any region above and close to the line of stability (iii) C = any region below and close to the line of stability (c) mode of decay change in proton number, Z change in neutron number, N emission 2 2 emission +1 1 + emission 1 +1 e capture 1 +1 p emission 1 0 n emission 0 1 lose one mark for each row in error [8] 4. (a) energy needed to separate nucleus into constituent nucleons 2 (b) (i) mass defect = 26 1.00728 + 0 1.00867 + 26 0.00055 55.949 = 0.529(u) binding energy = 0.529 91 = 492 (MeV) 492 binding energy per nucleon = 8.8 (MeV) 56 (ii) mass defect = 0.529 1.66 10 27 = 8.8 10 28 (kg) 6 [8] 5. (a) (i) = 0.69 = t 1 2 = 6.4 10 12 (s 1 ) 0.69 420 65 24 600 1 dn (ii) N = = 45010 dt 12 6.410 = 7.0 10 16 4
PhysicsAndMathsTutor.com (b) (i) pass through with no [or very small] deflection (ii) volume of nucleus << volume of atom (*) [or nucleus small and atom mostly empty space] (*) most of mass in nucleus (*) nucleus has positive charge (*) size of nucleus << separation (*) (*) any two [7] 6. (a) (i) which atom decays at what time is chance (ii) isotopes are (different forms) of same element same proton number, Z, different nucleon number, A [or with same number of protons but different number of neutrons] half-life is time for number of nuclei to halve [or to halve activity] for a particular isotope dn = N dt is constant of proportionality [or probability of decay] [or is probability of decay in unit time ] max 6 ln 2 (b) (i) = 8.04 24 600 = 1.0 10 6 s 1 (ii) N = 5.0 10 1.0 10 4 6 = 5.0 10 10 N (iii) ln 0 = t N 10 5.4 10 ln 10 5.0 10 t = = 7.7 10 4 s 6 1.0 10 4 7.7 10 = = 21(.4) (hour) max 6 600 [12]
PhysicsAndMathsTutor.com 4 7. (a) (i) (inner) orbiting electron [or electron surrounding the nucleus] captured by a proton (in the nucleus) converted into a neutron QWC (ii) daughter nuclide/nucleus/atom might be excited and energy given up as electromagnetic radiation [or orbiting electrons drop down to fill space (left by captured electron)] QWC (iii) 20Bi 20 8 82 Pb + 1 + + (e) (+Q) [allow 0 e 1 for 1 + ] max 5 (b) (i) (use of N = N o e t and N activity gives) 290 = 1200 exp( 24 60 60) ln(1200/ 290) = 24 60 60 (= 1.64 10 5 s 1 ) ln 2 (ii) (use of T ½ = ln 2/ gives) T ½ = 1.6410 5 = 4.2() 10 4 s (= 11.(7) hr) (use of = 1.6 10 5 s 1 gives T ½ = 4. 10 4 s or 12 hr) (iii) N (use of = N gives) ( )1200 = ( )1.64 10 5 N t N = 7.(2) 10 7 (nuclei) (use of = 1.6 10 5 s 1 gives N = 7.5 10 7 (nuclei)) max 5 [10] 8. (a) (i) alpha (ii) two different track lengths short range particles have lower energy than long range particles particles in each range have same energy 4
PhysicsAndMathsTutor.com 5 (b) (i) 29 25 92 94 Pu U + (+Q) (ii) 25 92 U 21 90Th + (+Q) (iii) U 25 because of the inverse relationship between half life and alpha particle energy (iv) because the Th 90 nucleus is neutron rich compared with U 25 [or Pu 29] 5 [9] 9. (a) (R = R0 A) plot R against A with axes labelled units on axes scales chosen to use more than 50% of page element R/10 15 m A R /10 45 m carbon 2.66 12 18.8 silicon.4 28 40.4 iron 4.5 56 82. tin 5.49 120 165.5 lead 6.66 208 295 calculate data for table plot data lose one mark for each error calculation of gradient 0010 e.g. gradient = 21 1 r 0 (= gradient) = (1.41 10 45 1 ) 45 (= 1.41 10 45 m ) = 1.1(2) 10 15 m
PhysicsAndMathsTutor.com 6 alternative: plot R against A 1/ with axes labelled units on axes scales chosen to use more than 50% of page element R/10 15 m A A 1/ carbon 2.66 12 2.29 silicon.4 28.04 iron 4.5 56.8 tin 5.49 120 4.9 lead 6.66 208 5.9 calculate data for table plot data lose one mark for each error calculation of gradient 15 e.g. gradient = 6.7210 = (1.1(2) 10 45 m ) 6.0 r 0 = gradient = 1.1(2) 10 15 m [or plot lnr against lna...] max 8 (b) assuming the nucleus is spherical ignoring the gaps between nucleons assuming all nuclei have same density assuming total mass is equal to mass of constituent nucleus any one assumption M = 4 R 4 M R 0 a 27 m 4R = 1.6710 15 0 4 (1.1210 ) = 2.8 10 17 kgm 4 [12]
PhysicsAndMathsTutor.com 7 10. (a) intensity of scattered electrons diffraction angle graph shows a minimum which does not touch the axis 2 (b) the (de Broglie) wavelength of high energy electrons is comparable to nuclear radii [or not subject to the strong nuclear force] 1 (c) nuclear density is constant separation of neighbouring nucleons is constant [or nucleons are close-packed] 2 R (d) A correct curve 1 (e) R = r 0 A 1 R 0 1 A0 R c =.04 10 15 Ac 16 12 1 R 0 =.5 10 15 m [9] 25 26 11. (a) (i) 1 0 n + 92 U > 92 U (ii) 26 U 145 87 92 56 Ba + 6 Kr + 4 1 0 n 2
PhysicsAndMathsTutor.com 8 (b) (m = m u mba mkr 4m n, electron masses balance) m = 26.0457 144.92694 86.9140 4 1.00867 = 0.17071u Q(= 0.17071 91.MeV)= 159(MeV) [5] 12. (a) (i) T 1/2 = 50 s (from graph) (ii) = ln 2 50 = 0.014s 1 (iii) N = A 0 = 2.4 10 0. 014 5 = 1.7 10 7 5 (b) (i) elapsed time = 50s = 1 half-life N 0 = N 0 e 0 = 1.71 10 7 e 0 0.014 = 1.12 10 7 no. decayed from t = 0s to t = 80s is 1.1210 2 7 = 056 10 7 [alternative (b)(i) N 0 = N 0 e 0 and N 80 = N 0 e 80 give 1.12 10 7 and 0.56 10 7 number decayed (=1.12 10 7 0.56 10 7 ) = 0.56 10 7 ] (ii) energy released = 0.56 10 7 1.0 10 12 = 5.6 10 6 J max 4 [9] 1. (a) (i) a neutron strikes the nucleus nucleus splits into two fragments (ii) some electrostatic Ep converted to E k of fragments some electrostatic Ep used to overcome strong interaction some electrostatic Ep used to increase surface energy (iii) fission fragments repel and collide with other atoms in fuel rod high energy fission neutrons enter moderator [or collide with moderator atoms] atoms gain Ek due to collisions (and vibrate more) temperature depends on the average E k of (vibrating) atoms a chain reaction occurs max 8 (b) energy from fuel per year at 100% efficiency
PhysicsAndMathsTutor.com 9 = 1600(MW).2 10 7 s 5.0 10 16 (J) energy supplied from fuel per year at 25% efficiency 4 5.0 10 16 2.0 10 17 (J) energy released per kilogram of fuel = 200 1.6 10 1 6.0 10 2 1 0. 28 0.0 2.4 10 12 (J) mass of fuel needed per year = 2.0 10 2.4 10 17 12 8 10 4 kg 5 [1] 14. (a) (i) 28 92 U 4 2 + 24 90 Th (ii) m = 28.05076 4.00260 24.0457 = 0.00459(u) Q = 91 0.00459 (MeV) = 4.MeV 5 (b) (i) overall change in proton number (= 92 82) = 10 change in proton number due to particles (= 8 2) = 16 therefore Z = 6 for the particles corresponding to the six particles (ii) proton changes to a neutron plus a positron [or p n + + (+ v e + Q)] Pb-206 has a lower neutron to proton ratio than U-28 alpha emission raises the neutron to proton ratio slightly emission lowers the ratio (more) + emission increases neutron to proton ratio positron emission competes with emission but is energetically less favourable max 6 [11] 15. (a) time for half of (active) nuclei (of radioactive substance) to decay 1 (b) t/minute 0 10 20 0 40 50 60 number of counts in 0s, C 60 42 5 2 18 14 10 ln C 4.094.78.555.15 2.890 2.69 2.0 Correct values of ln C above 1 (c) (i) seven points correctly plotted [six points correct ]
PhysicsAndMathsTutor.com 10 4.5 graph of InC against time 4.5 In C In C Linear (In C) Linear (In C) 2.5 y = 0.2941x + 4.7 2 0 1 2 t/min 5 6 7 8 (ii) best straight line through points sensible scale (iii) from sensible triangle on graph gradient = 0.00 [0.294] (min 1 ) (min 1 ) max 5 (d) (i) C = C 0 e t, ln C ln C 0 = t hence using y = mx + c, = ( )gradient (ii) half life = ln 2 0.69 = 2 min 0.0 (e) count over longer period than half minute [or repeat experiment] use stronger source use background count correctly max 2
PhysicsAndMathsTutor.com 11 (f) for 14 C, = 0.69 T 1 2 =1.21 10 4 (year) R 5.2 t ln ln = 0.22 R 0 6. 5 0.22 t = = 1840(year) 4 4 1.2110 [16] 16. (a) (i) 25 92 U + 1 0 n 98 15 8Sr + 54 Xe + 1 0 n(+q) (ii) three correct positions to within 2 on x-axis (one mark if two correct) (iii) estimate of energy released: binding energy of U-25 nucleus = (25 7.5) = 176(15)(MeV) binding energy of Sr-98 = (98 8.6) = 84( 15)(MeV) binding energy of Xe-15 = (15 8.4) = 114( 15)(MeV) binding energy released = 114 + 84 176 = 214MeV (±40MeV) max 6 (b) (i) 25g of U-25 releases 6 10 2 214 1.6 10 1 J = 2.1 10 1 (J) 1.0 kg of uranium containing % U-25 contains 0g of U-25 energy from 1.0kg of uranium = 2.1 10 0 25 = 2.6 10 12 J [[1.6 10 25 MeV]]
PhysicsAndMathsTutor.com 12 (ii) advantage: less mass of fuel used because more energy per kilogram [alternative: less harm to environment because does not generate greenhouse gases or any statement argued ] disadvantage: hazardous waste because fission products are radioactive [alternative: long term responsibility because waste needs to be stored for many years or any statement argued ] max 6 [12] 17. (a) m = 4.0026 1.66 10 27 (kg) (= 6.6 10 27 kg electron masses are not significant) kinetic energy (= 2 1 m 2 ) = 0.5 6.65 10 27 (2.00 10 7 ) 2 (= 1. 10 12 J) 2 (b) loss in k.e. = gain in p.e. loss of ke. [or 1. 10 12 ] = Qq 4 0 R 2 2Ze 4 0R 19 2 R = 2 79 (1.6 10 ) 12 12 4 8.8510 1.10 =2.7 10 14 m 4 (c) any valid point including: strong force complicates the process (*) scattering caused by distribution of protons not whole nucleon distribution (*) particles are massive causing recoil of nucleus which complicates results (*) (*) any one 1 [7] 18. (a) (i) proportion of U-25 is greater than in natural uranium (ii) induced fission more probable with U-25 than with U-28 2
PhysicsAndMathsTutor.com 1 (b) (i) for steady rate of fission, one neutron per fission required to go on to produce further fission each fission produces two or three neutrons on average some neutrons escape [or some absorbed by U-28 without fission] control rods absorb sufficient neutrons (to maintain steady rate of fission) (ii) neutrons need to pass through a moderator to slow them (in order to cause further fissions or prevent U-28 absorbing them) neutrons that leave the fuel rod (and pass through the moderator) are unlikely to re-enter the same fuel rod makes it easier to replace the fuel in stages max 5 [7] 19. (a) (i) most alpha particles undeflected some through small angles (very) small (but significant) number deflected through > 90 max 2 (b) atom mostly empty space positive charge concentrated in a volume much less than total volume [or radius] max 2 [4] 20. (a) (i) proton number = 6 neutron number = 56 (ii) krypton 100 (b) one-fifth efficiency so total output (= 10 = 50(MW) 20 energy in one day = 50 10 6 24 600(J) (4.2 10 12 J) 12 fission atoms per day = 4.2 10 =1.5 10 2 11.2 10 [6]