1 Our calculation is similar to that shown in Sample Problem 42-1 We set K = 5.30 MeV=U = (1/ 4πε )( qα qCu / rmin ) and solve for the closest separation, rmin: rmin −19 qα qCu kqα qCu ( 2e )( 29 ) (1.60 × 10 C )( 8.99 × 10 V ⋅ m/C ) = = = 4πε K 4πε K 5.30 ×106 eV = 1.58 ×10−14 m = 15.8 fm We note that the factor of e in qα = 2e was not set equal to 1.60 × 10– 19 C, but was instead allowed to cancel the “e” in the non-SI energy unit, electronvolt Kinetic energy (we use the classical formula since v is much less than c) is converted into potential energy (see Eq 24-43) From Appendix F or G, we find Z = for Lithium and Z = 90 for Thorium; the charges on those nuclei are therefore 3e and 90e, respectively We manipulate the terms so that one of the factors of e cancels the “e” in the kinetic energy unit MeV, and the other factor of e is set equal to its SI value 1.6 × 10–19 C We note that k = πε can be written as 8.99 × 109 V·m/C Thus, from energy conservation, we have K =U r = k q1q K c8.99 × 10 hc3 × 16 × 10 Chb90eg = V⋅ m C 3.00 × 106 eV which yields r = 1.3 × 10– 13 m (or about 130 fm) −19 The conservation laws of (classical kinetic) energy and (linear) momentum determine the outcome of the collision (see Chapter 9) The final speed of the α particle is vαf = mα − mAu vαi , mα + mAu and that of the recoiling gold nucleus is vAu, f = mα vαi mα + mAu (a) Therefore, the kinetic energy of the recoiling nucleus is K Au,f § 2mα · 4mAu m 1 mAu ă = mAu vAu, ¸ vα i = Kα i f = 2 ( mα + mAu ) © mα + mAu ¹ = ( 5.00MeV ) (197u )( 4.00u ) ( 4.00u+197u ) = 0.390MeV (b) The final kinetic energy of the alpha particle is Kα f 1 = mα vα2 f = mα 2 § mα − mAu · § mα − mAu · ă v i = K i ă â m + mAu â m + mAu Đ 4.00u 197u ã = ( 5.00MeV ) ă © 4.00u + 197u ¹ = 4.61MeV We note that K af + K Au, f = Kαi is indeed satisfied 2 (a) protons, since Z = for carbon (see Appendix F) (b) neutrons, since A – Z = 14 – = (see Eq 42-1) (a) Table 42-1 gives the atomic mass of 1H as m = 1.007825 u Therefore, the mass excess for 1H is ∆ = (1.007825 u – 1.000000 u)= 0.007825 u (b) In the unit MeV/c2, ∆ = (1.007825 u – 1.000000 u)(931.5 MeV/c2·u) = +7.290 MeV/c2 (c) The mass of the neutron is given in Sample Problem 42-3 Thus, for the neutron, ∆ = (1.008665 u – 1.000000 u) = 0.008665 u (d) In the unit MeV/c2, ∆ = (1.008665 u – 1.000000 u)(931.5 MeV/ c2·u) = +8.071 MeV/c2 (e) Appealing again to Table 42-1, we obtain, for 120Sn, ∆ = (119.902199 u – 120.000000 u) = – 0.09780 u (f) In the unit MeV/c2, ∆ = (119.902199 u – 120.000000 u) (931.5 MeV/ c2·u) = – 91.10 MeV/c2 (a) The atomic number Z = 39 corresponds to the element Yttrium (see Appendix F and/or Appendix G) (b) The atomic number Z = 53 corresponds to Iodine (c) A detailed listing of stable nuclides (such as the website http://nucleardata nuclear.lu.se/nucleardata) shows that the stable isotope of Yttrium has 50 neutrons (this can also be inferred from the Molar Mass values listed in Appendix F) (d) Similarly, the stable isotope of Iodine has 74 neutrons (e) The number of neutrons left over is 235 – 127 – 89 = 19 We note that the mean density and mean radius for the Sun are given in Appendix C Since ρ = M/V where V ∝ r , we get r ∝ ρ −1/ Thus, the new radius would be Fρ I F 1410 kg / m IJ r = R G J = c6.96 × 10 m h G H × 10 kg / m K H ρK 1/ s s 17 1/ = 1.3 × 10 m (a) Since U > , the energy represents a tendency for the sphere to blow apart (b) For obtain Pu, Q = 94e and R = 6.64 fm Including a conversion factor for J → eV we 239 c hc 94 1.60 × 10 −19 C 8.99 × 10 N ⋅ m / C 3Q U= = 20 πε r 6.64 × 10 −15 m c h h FG 1eV IJ H 1.60 × 10 J K −19 × 10 eV = 1.15GeV = 115 (c) Since Z = 94, the electrostatic potential per proton is 1.15 GeV/94 = 12.2 MeV/proton (d) Since A = 239, the electrostatic potential per nucleon is 1.15 GeV/239 = 4.81 MeV/nucleon (e) The strong force that binds the nucleus is very strong (a) For 55Mn the mass density is ρm = M 0.055kg/mol = = 2.3 ì1017 kg/m3 V ( / 3) ê 1.2 ×10−15 m ( 55 )1/ º 6.02 ì1023 / mol ) ) ơ( ẳ ( (b) For 209Bi, ρm = M 0.209 kg / mol = = 2.3 × 1017 kg / m 1/ 3 −15 23 V π / 1.2 × 10 m 209 6.02 × 10 / mol f c fc c (c) Since V ∝ r = r0 A1/ h h ∝ A, we expect ρ m ∝ A / V ∝ A / A ≈ const for all nuclides (d) For 55Mn the charge density is ( 25) (1.6 ×10−19 C ) Ze ρq = = = 1.0 × 1025 C/m3 1/ V ( / 3) ơê(1.2 ì1015 m ) ( 55) ¼º (e) For 209Bi a fc fc h 83 1.6 × 10 −19 C Ze ρq = = V π / 1.2 × 10 −15 m 209 a 1/ 3 f = 8.8 × 10 24 C / m Note that ρ q ∝ Z / V ∝ Z / A should gradually decrease since A > 2Z for large nuclides 10 (a) The mass number A is the number of nucleons in an atomic nucleus Since m p ≈ mn the mass of the nucleus is approximately Amp Also, the mass of the electrons is negligible since it is much less than that of the nucleus So M ≈ Am p (b) For 1H, the approximate formula gives M ≈ Amp = (1)(1.007276 u) = 1.007276 u The actual mass is (see Table 42-1) 1.007825 u The percentage deviation committed is then δ = (1.007825 u – 1.007276 u)/1.007825 u = 0.054% ≈ 0.05% (c) Similarly, for 31P, δ = 0.81% (d) For 120Sn, δ = 0.81% (e) For 197Au, δ = 0.74% (f) For 239Pu, δ = 0.71% (g) No In a typical nucleus the binding energy per nucleon is several MeV, which is a bit less than 1% of the nucleon mass times c2 This is comparable with the percent error calculated in parts (b) – (f) , so we need to use a more accurate method to calculate the nuclear mass 73 We note that hc = 1240 MeV·fm (see problem 83 of Chapter 38), and that the classical kinetic energy 12 mv can be written directly in terms of the classical momentum p = mv (see below) Letting p ∆p ∆h / ∆x h / r , we get ( hc ) = (1240 MeV ⋅ fm ) p2 E= 30 MeV 2m ( mc ) r 2 ( 938 MeV ) ª(1.2 fm )(100 )1/ ẳ 2 74 Adapting Eq 42-21, we have N Kr = § −9 ã M sam N A = ăă 20 ì 10 g áá ( 6.02 ì 1023 atoms mol ) = 1.3 ì 1014 atoms ă 92 g mol M Kr â Consequently, Eq 42-20 leads to R= c h × 1014 ln N ln 13 = = 4.9 × 1013 Bq 184 T1 s 75 Since R is proportional to N (see Eq 42-17) then N/N0 = R/R0 Combining Eq 42-14 and Eq 42-18 leads to t=− T1 ln ln FG R IJ = − 5730 y lnb0.020g = 3.2 × 10 y H R K ln 76 (a) The mass number A of a radionuclide changes by in an α decay and is unchanged in a β decay If the mass numbers of two radionuclides are given by 4n + k and 4n' + k (where k = 0, 1, 2, 3), then the heavier one can decay into the lighter one by a series of α (and β) decays, as their mass numbers differ by only an integer times If A = 4n + k, then after α-decaying for m times, its mass number becomes A = 4n + k – 4m = 4(n – m) + k, still in the same chain (b) For 235U, 235 = 58 × + = 4n + (c) For 236U, 236 = 59 × = 4n (d) For 238U, 238 = 59 × + = 4n + (e) For 239Pu, 239 = 59 × + = 4n + (f) For 240Pu, 240 = 60 × = 4n (g) For 245Cm, 245 = 61 × + = 4n + (h) For 246Cm, 246 = 61 × + = 4n + (i) For 249Cf, 249 = 62 × + = 4n + (j) For 253Fm, 253 = 63 × + = 4n + 77 Let A Z X represent the unknown nuclide The reaction equation is A Z X + 01 n→ −01 e +2 42 He Conservation of charge yields Z + = – + or Z = Conservation of mass number yields A + = + or A = According to the periodic table in Appendix G (also see Appendix F), lithium has atomic number 3, so the nuclide must be 73 Li 78 We note that 2.42 = 145.2 s We are asked to plot (with SI units understood) c ln R = ln R0 e − λt + R0′e − λ ′t h where R0 = 3.1 × 105, R0' = 4.1 × 106, λ = ln 2/145.2 and λ' = ln 2/24.6 Our plot is shown below We note that the magnitude of the slope for small t is λ' (the disintegration constant for Ag), and for large t is λ (the disintegration constant for 108Ag) 110 79 The lines that lead toward the lower left are alpha decays, involving an atomic number change of ∆Zα = – and a mass number change of ∆Aα = – The short horizontal lines toward the right are beta decays (involving electrons, not positrons) in which case A stays the same but the change in atomic number is ∆Zβ = +1 Fig 42-16 shows three alpha decays and two beta decays; thus, Z f = Zi + 3∆Zα + ∆Z β and A f = Ai + 3∆Aα Referring to Appendix F or G, we find Zi = 93 for Neptunium, so Zf = 93 + 3(– 2) + 2(1) = 89, which indicates the element Actinium We are given Ai = 237, so Af = 237 + 3(– 4) = 225 Therefore, the final isotope is 225Ac 80 (a) Replacing differentials with deltas in Eq 42-12, we use the fact that ∆N = – 12 during ∆t = 1.0 s to obtain ∆N = − λ∆t λ = 4.8 × 10 −18 / s N where N = 2.5 × 1018, mentioned at the second paragraph of §42-3, is used (b) Eq 42-18 yields T1/2 = ln 2/λ = 1.4 × 1017 s, or about 4.6 billion years 81 Eq 24-43 gives the electrostatic potential energy between two uniformly charged spherical charges (in this case q1 = 2e and q2 = 90e) with r being the distance between their centers Assuming the “uniformly charged spheres” condition is met in this instance, we write the equation in such a way that we can make use of k = 1/4 πε0 and the electronvolt unit: a2efa90ef = F 8.99 × 10 U=k H r V⋅m C I c3.2 × 10 Ch a90ef = 2.59 × 10 K r r −19 −7 eV with r understood to be in meters It is convenient to write this for r in femtometers, in which case U = 259/r MeV This is shown plotted below 82 We locate a nuclide from Table 42-1 by finding the coordinate (N, Z) of the corresponding point in Fig 42-4 It is clear that all the nuclides listed in Table 42-1 are stable except the last two, 227Ac and 239Pu 83 Although we haven’t drawn the requested lines in the following table, we can indicate their slopes: lines of constant A would have – 45° slopes, and those of constant N – Z would have 45° As an example of the latter, the N – Z = 20 line (which is one of “eighteen-neutron excess”) would pass through Cd-114 at the lower left corner up through Te-122 at the upper right corner The first column corresponds to N = 66, and the bottom row to Z = 48 The last column corresponds to N = 70, and the top row to Z = 52 Much of the information below (regarding values of T1/2 particularly) was obtained from the websites http://nucleardata.nuclear.lu.se/nucleardata and http://www.nndc.bnl.gov/ nndc/ensdf (we refer the reader to the remarks we made in the solution to problem 8) 118 Te 6.0 days 117 119 Te 16.0 h Sb 118 2.8 h 3.6 116 Sn 14.5% 115 In 95.7% 114 Cd 28.7% 117 Sb Sn 7.7% 116 In 14.1 s 115 Cd 53.5 h 120 Te 0.1% 119 Sb 38.2 s 118 Sn 24.2% 117 In 121 19.4 days 120 Cd 7.5% Sb 15.9 119 Sn 8.6% 118 43.2 116 Te In 5.0 s 117 122 Te 2.6% 121 Sb 57.2% 120 Sn 32.6% 119 In 2.4 Cd 118 2.5 h 50.3 Cd 84 The problem with Web-based services is that there are no guarantees of accuracy or that the webpage addresses will not change from the time this solution is written to the time someone reads this Still, it is worth mentioning that a very accessible website for a wide variety of periodic table and isotope-related information is http://www.webelements.com Two websites aimed more towards the nuclear professional are http://nucleardata.nuclear.lu.se/nucleardata and http://www.nndc.bnl.gov/nndc/ensdf, which are where some of the information mentioned below was obtained (a) According to Appendix F, the atomic number 60 corresponds to the element Neodymium (Nd) The first website mentioned above gives 142Nd, 143Nd, 144Nd, 145Nd, 146 Nd, 148Nd, and 150Nd in its list of naturally occurring isotopes Two of these, 144Nd and 150 Nd, are not perfectly stable, but their half-lives are much longer than the age of the universe (detailed information on their half-lives, modes of decay, etc are available at the last two websites referred to, above) (b) In this list, we are asked to put the nuclides which contain 60 neutrons and which are recognized to exist but not stable nuclei (this is why, for example, 108Cd is not included here) Although the problem does not ask for it, we include the half-lives of the nuclides in our list, though it must be admitted that not all reference sources agree on those values (we picked ones we regarded as “most reliable”) Thus, we have 97Rb (0.2 s), 98Sr (0.7 s), 99 Y (2 s), 100Zr (7 s), 101Nb (7 s), 102Mo (11 minutes), 103Tc (54 s), 105Rh (35 hours), 109In (4 hours), 110Sn (4 hours), 111Sb (75 s), 112Te (2 minutes), 113I (7 s), 114Xe (10 s), 115Cs (1.4 s), and 116Ba (1.4 s) (c) We would include in this list: 60Zn, 60Cu, 60Ni, 60Co, 60Fe, 60Mn, 60Cr, and 60V 85 (a) In terms of the original value of u, the newly defined u is greater by a factor of 1.007825 So the mass of 1H would be 1.000000 u, the mass of 12C would be (12.000000/1.007825) u = 11.90683 u (b) The mass of 238U would be (238.050785/ 1.007825) u = 236.2025 u 86 We take the speed to be constant, and apply the classical kinetic energy formula: t= mn r 2mc d d = = 2r = 2K c v K 2K / m (1.2 ×10 ≈ −15 m ) (100 ) 1/ 3.0 × 10 m/s ≈ ×10 −22 s ( 938MeV ) 5MeV 87 We solve for A from Eq 42-3: F r I F 3.6 fm IJ A=G J =G H r K H 1.2 fm K 3 = 27