CHAPTER 31 RADIOACTIVITY AND NUCLEAR PHYSICS

OpenStax College Physics Instructor Solutions Manual Chapter 31 CHAPTER 31: RADIOACTIVITY AND NUCLEAR PHYSICS 31.2 RADIATION DETECTION AND DETECTORS Solution The energy of 30.0 eV is required to ionize a molecule of the gas inside a Geiger tube, thereby producing an ion pair Suppose a particle of ionizing radiation deposits 0.500 MeV of energy in this Geiger tube What maximum number of ion pairs can it create? ( 0.500 MeV) (1.00 × 10 eV/MeV) # pairs = = 1.67 × 10 pairs 30.0 eV/pair A particle of ionizing radiation creates 4000 ion pairs in the gas inside a Geiger tube as it passes through What minimum energy was deposited, if 30.0 eV is required to create each ion pair? Solution ( 4000 pairs)( 30.0 eV/pair) = 120 keV (a) Repeat Exercise 31.2, and convert the energy to joules or calories (b) If all of this energy is converted to thermal energy in the gas, what is its temperature increase, assuming 50.0 cm of ideal gas at 0.250-atm pressure? (The small answer is consistent with the fact that the energy is large on a quantum mechanical scale but small on a macroscopic scale.) Solution  1.60 × 10 −19 J   = 1.92 × 10 −15 J (1.20 × 10 eV) eV   (a) 3 2∆E nRT ⇒ ∆E = nR∆T , so ∆T = Also PV = nRT 2 3nR T 2T∆E ⇒ = ⇒ ∆T = nR PV 3PV E= (b) Assume T = 20° C = 293 K OpenStax College Physics Instructor Solutions Manual Chapter 31 (2 3)( 293 K )(1.92 × 10 −15 J) ∆T = (0.250 atm )(1.01 × 10 Pa atm )(50 cm )(1 m × 10 cm ) = 2.97 × 10 −13 K Suppose a particle of ionizing radiation deposits 1.0 MeV in the gas of a Geiger tube, all of which goes to creating ion pairs Each ion pair requires 30.0 eV of energy (a) The applied voltage sweeps the ions out of the gas in 1.00 µs What is the current? (b) This current is smaller than the actual current since the applied voltage in the Geiger tube accelerates the separated ions, which then create other ion pairs in subsequent collisions What is the current if this last effect multiplies the number of ion pairs by 900? Solution I= (a) (b)  1.00 × 10 eV  1.60 × 10 −19 C  Q Ne   = 1.07 × 10 −8 A = = 2 −6 t t  30.0 eV  1.00 × 10 s  I = (900)(1.067 × 10 −8 ) = 9.601.00 × 10 −6 A = 9.60 μA 31.3 SUBSTRUCTURE OF THE NUCLEUS Solution Solution 17 Verify that a 2.3 × 10 kg mass of water at normal density would make a cube 60 km on a side, as claimed in Example 31.1 (This mass at nuclear density would make a cube 1.0 m on a side.) M M = ρv = ρa ⇒ a =   ρ     2.3 × 1017   =    1000 kg/m  = 6.1× 10 m = 61 km Find the length of a side of a cube having a mass of 1.0 kg and the density of nuclear 17 matter, taking this to be 2.3 × 10 kg/m M M = ρv = ρa ⇒ a =   ρ      1.0 kg  =  17  × 10 kg/m   What is the radius of an α particle? = 1.6 ì10 m = 1.6 àm OpenStax College Physics Solution Solution r = r0 A Instructor Solutions Manual ( ) = 1.2 ×10 −15 m ( ) 3 Chapter 31 = 1.9 ×10 −15 m = 1.9 fm 238 238 Find the radius of a Pu nucleus Pu is a manufactured nuclide that is used as a power source on some space probes r = r0 A = (1.2 × 10 −15 m )( 238) (a) Calculate the radius of 58 = 7.4 × 10 −15 m = 7.4 fm N i , one of the most tightly bound stable nuclei (b) What 58 258 is the ratio of the radius of N i to that of Ha , one of the largest nuclei ever made? Note that the radius of the largest nucleus is still much smaller than the size of an atom Solution (a) rNi = r0 ANi ( = (1.2 × 10 −15 m )( 58) ) rHa = 1.2 × 10 −15 m ( 258) 3 = 4.6 × 10 −15 m = 4.6 fm = 7.6 × 10 −15 m = 7.6 fm, so that rNi 4.645 × 10 −15 m = = 0.61 to 7.639 × 10 −15 m (b) rHa 10 −27 The unified atomic mass unit is defined to be u = 1.6605 ×10 kg Verify that this amount of mass converted to energy yields 931.5 MeV Note that you must use fourq digit or better values for c and e Solution E = mc = 1.6605 × 10 −27 kg 2.998 × 10 m/s = 1.49246× 10 −10 J  (1.49246×10 −10 J ) 1.60221 eV  = 9.315×10 eV = 931.5 MeV −19 ×10 J   11 What is the ratio of the velocity of a β particle to that of an α particle, if they have the same nonrelativistic kinetic energy? ( )( ) OpenStax College Physics Solution Chapter 31 v  mβ 1 2 KE = mv ⇒ m β v β = mα vα ⇒  α  = ⇒ v  2 mα  β vα  m β   = v β  mα  vβ vα 12 Instructor Solutions Manual =   9.11 × 10 −31 kg =  − 27  ( 4.002 ) 1.6605 × 10 kg  ( ) = 1.17 × 10 − = 85.4 to 1.17 × 10 − If a 1.50-cm-thick piece of lead can absorb 90.0% of the γ rays from a radioactive source, how many centimeters of lead are needed to absorb all but 0.100% of the γ rays? Solution If there is a 90.0% reduction for each 1.50 cm, then (0.100) x will remain, where x = number of distances of 1.50 cm ( 0.00100 ) = ( 0.100) x ⇒ ln( 0.00100 ) = x ln( 0.100) ln( 0.00100 ) x= = 3.00 ⇒ d = x(1.50 cm ) = 4.50 cm ln( 0.100 ) 13 The detail observable using a probe is limited by its wavelength Calculate the energy −16 of a γ -ray photon that has a wavelength of ×10 m , small enough to detect details about one-tenth the size of a nucleon Note that a photon having this energy is difficult to produce and interacts poorly with the nucleus, limiting the practicability of this probe Solution 14 hc ( 6.63 × 10 −34 J ⋅ s )( 3.00 × 108 m/s ) = = 1.989 × 10 −9 J λ 1.00 × 10 −16 m eV   = (1.989 × 10 −9 m/s )  = 1.24 × 10 eV = 12.4 GeV −16  1.60 × 10 J  E = hf = 1/ (a) Show that if you assume the average nucleus is spherical with a radius r = r0 A , and with a mass of A u, then its density is independent of A (b) Calculate that density in 56 u/fm3 and kg/m , and compare your results with those found in Example 31.1 for F e OpenStax College Physics Solution Instructor Solutions Manual r = r0 A , ρ = (a) ρ= (b) M M Au 3u = = = πr πr A 4πr0 V 3 ( ) 4π (1.2 fm ) ( ) ( )( −1 −27  u  1.381× 10 1.6605 × 10 kg = 1.381× 10 −1   =  fm  1.00 × 10 −15 m 3u ρ = 2.3 × 1017 kg/m3 15 Chapter 31 ( ) ) The answer is the same as Example 31.1 What is the ratio of the velocity of a 5.00-MeV β ray to that of an α particle with the same kinetic energy? This should confirm that β s travel much faster than α s even when relativity is taken into consideration (See also Exercise 31.11.) Solution KE rel  v2  1 −  γ = = (γ −1)mc , since  c  −1 v2   = 1 −  v = c 1− 2 γ  γ  , or ,c For the β particle: KE = 5.00 MeV = (γ − 1)(0.511 MeV) , so that (γ − 1) = 9.785 or γ = 10.785 Thus, vβ = c − 1 = ( 2.998 × 10 m s) − = 2.985 × 10 m s 2 γ (10.785)  931.5 MeV u  KE = 5.00 MeV = (γ − 1)( 4.0026 u ) c c2   so that For the α particle: γ = 1.00134 Thus, vα = (2.998 × 10 m s) − vβ the ratio of the velocities is given by vα 16 = = 1.551× 10 m s (1.00134) Finally, 2.985 × 10 m s = 19.3 to 1.55 × 10 m s (a) What is the kinetic energy in MeV of a β ray that is traveling at 0.998c ? This gives some idea of how energetic a β ray must be to travel at nearly the same speed as a γ ray (b) What is the velocity of the γ ray relative to the β ray? Solution (a) [ γ = − (v c ) ] −1 [ = − (0.998) ] −1 = 15.819, KE = (γ − 1)mc = (14.819)( 0.511 MeV) = 7.57 MeV OpenStax College Physics (b) Instructor Solutions Manual c c = = 1.002 vβ 0.998c Chapter 31 (4 significant figures used to show difference) 31.4 NUCLEAR DECAY AND CONSERVATION LAWS 17 β − decay of H (tritium), a manufactured isotope of hydrogen used in some digital watch displays, and manufactured primarily for use in hydrogen bombs Solution 18 β − decay of 40 K , a naturally occurring rare isotope of potassium responsible for some of our exposure to background radiation Solution 40 19 19 β + decay of Solution 50 25 20 β + decay of Solution 52 26 21 Electron capture by Be Solution 22 Solution H → 23 He1 + β − + ve − K 21 →40 20 Ca 20 + β + ve 50 Mn + Mn 25 →50 24 Cr26 + β + ve 52 Fe + Fe 26 →52 25 Mn 27 + β + ve Be3 + e − →73 Li + ve Electron capture by 106 49 106 In In 57 + e − →106 48 Cd 58 + ve OpenStax College Physics Instructor Solutions Manual Chapter 31 23 α decay of 210 Po , the isotope of polonium in the decay series of 238 U that was discovered by the Curies A favorite isotope in physics labs, since it has a short half-life and decays to a stable nuclide Solution 210 84 24 α decay of 226 Ra , another isotope in the decay series of 238 U , first recognized as a new element by the Curies Poses special problems because its daughter is a radioactive noble gas Solution 226 88 25 β − decay producing 137 Ba The parent nuclide is a major waste product of reactors and has chemistry similar to potassium and sodium, resulting in its concentration in your cells if ingested Solution 137 55 26 β − decay producing 90 Y The parent nuclide is a major waste product of reactors and has chemistry similar to calcium, so that it is concentrated in bones if ingested 90 ( Y is also radioactive.) Solution 90 38 27 α decay producing Po126 →206 82 Pb124 + He Ra 138 →222 86 Rn136 + He − Cs82 →137 56 Ba 81 + β + ve Sr52 →9039 Y51 + β − + ve 228 Ra The parent nuclide is nearly 100% of the natural element and is found in gas lantern mantles and in metal alloys used in jets ( radioactive) 228 Ra is also Solution 232 90 28 α decay producing 208 Pb The parent nuclide is in the decay series produced by the only naturally occurring isotope of thorium Th142 →228 88 Ra 140 + He 232 Th , OpenStax College Physics Instructor Solutions Manual Chapter 31 Solution 212 84 29 When an electron and positron annihilate, both their masses are destroyed, creating two equal-energy photons to preserve momentum (a) Confirm that the annihilation + − equation e + e → γ + γ conserves charge, electron family number, and total number Po128 →208 82 Pb126 + He of nucleons To this, identify the values of each before and after the annihilation (b) Find the energy of each γ ray, assuming the electron and positron are initially nearly at rest (c) Explain why the two γ rays travel in exactly opposite directions if the center of mass of the electron-positron system is initially at rest Solution (a) charge : ( + 1) + ( − 1) = 0; electron family number : ( + 1) + ( − 1) = 0; A : + = (b) P1 = − P2 since Ptot = 0, E1 = E2 and E1 + E2 = 2me c ( )( ) E1 = E2 = me c = 0.511 MeV/c c = 0.511 MeV (c) The two γ rays must travel in exactly opposite directions to conserve momentum, since initially there is zero momentum if the center of mass is initially at rest 30 Confirm that charge, electron family number, and the total number of nucleons are all A A− 4 conserved by the rule for α decay given in the equation Z X N → Z −2YN −2 + He To this, identify the values of each before and after the decay Solution Z = ( Z − 2) + before/after; efn : + = 0; A = ( A − ) + ; 31 Confirm that charge, electron family number, and the total number of nucleons are all A A − − conserved by the rule for β decay given in the equation Z X N →Z +1YN −1 + β +ν e To this, identify the values of each before and after the decay Solution Z = ( Z + 1) − before/after; efn : = ( + 1) + ( − 1); A = A 32 Confirm that charge, electron family number, and the total number of nucleons are all A A + − conserved by the rule for β decay given in the equation Z X N →Z −1YN +1 + β +ν e To this, identify the values of each before and after the decay Solution Z = ( Z − 1) + before/after; efn : = (1) + ( − 1) ; A = A OpenStax College Physics 33 Instructor Solutions Manual Chapter 31 Confirm that charge, electron family number, and the total number of nucleons are all conserved by the rule for electron capture given in the equation A − A Z X N + e →Z −1YN +1 + ν e To this, identify the values of each before and after the capture Solution 34 Z − = ( Z − 1) before/after;efn : ( + 1) = ( + 1) ; A = A 222 14 A rare decay mode has been observed in which Ra emits a C nucleus (a) The A decay equation is 222Ra →A X +14 C Identify the nuclide X (b) Find the energy 222 emitted in the decay The mass of Ra is 222.015353 u Solution 222 88 Ra 134 → ZA X N + 146 C8 , so we know that: (a) The decay is A = 222 − 14 = 208; Z = 88 − = 82 and N = A − Z = 208 − 82 = 126, so from the X= 208 82 Pb126 periodic table the element is lead and ∆m = m ( 222 88 ) Ra 134 − m ( 208 82 ) Pb126 − m ( 14 C8 ) = 222.015353u − 207.976627 u − 14.003241u = 3.5485× 10 −2 u (b) 35 Solution  931.5 MeV/c  c = 33.05 MeV E = ∆mc = 3.5485× 10 −2 u  u   ( ) (a) Write the complete α decay equation for decay (a) 226 88 226 Ra (b) Find the energy released in the Ra138 →222 86 Rn 136 + He ∆m = m ( 226 ) Ra − m ( 222 ) Rn − mα = ( 226.025402 − 222.017570 − 4.002603) u = 5.229 × 10 −3 u (b) 36  931.5 MeV/c  c = 4.87 MeV E = Δmc = 5.229 × 10 −3 u  u   ( ) (a) Write the complete α decay equation for decay 249 Cf (b) Find the energy released in the OpenStax College Physics Solution (a) 249 98 Instructor Solutions Manual Chapter 31 Cf151 →245 96 Cm149 + He ∆m = ( 249.074844 − 245.065483 − 4.002603) u = 6.758 × 10 −3 u  931.5 MeV/c  c = 6.295 MeV E = ∆mc = 6.758 × 10 −3 u  u   ( (b) ) 37 − (a) Write the complete β decay equation for the neutron (b) Find the energy released in the decay Solution −1 (a) n → p + β + ve ∆m = mn − m p − me = (1.008665 − 1.007276 − 0.00054858) u = 8.4042× 10 −4 u  931.5 MeV/c  c = 0.783 MeV E = ∆mc = 8.4042× 10 −4 u  u   ( (b) 38 Solution − 90 (a) Write the complete β decay equation for Sr , a major waste product of nuclear reactors (b) Find the energy released in the decay (a) 90 38 Sr 52 →9039 Y51 + β −1 + v e ∆m = m (b) 39 Solution ) ( 90 ) Sr − m ( 90 ) Y = ( 89.907738 − 89.907152) u = 5.86 × 10 −4 u  931.5 MeV/c  c = 0.546 MeV E = ∆mc = 5.86 × 10 − u  u   ( ) + 22 Calculate the energy released in the β decay of Na , the equation for which is given 22 22 in the text The masses of Na and Ne are 21.994434 and 21.991383 u, respectively ∆m = m ( 22 ) Na − m ( 22 ) Ne − 2me = ( 21.994434− 21.991383 − 2(0.00054858) u = 1.954 × 10 −3 u  931.5 MeV/c  c = 1.820 MeV E = ∆mc = 1.954 × 10 −3 u  u   ( ) OpenStax College Physics Solution Instructor Solutions Manual Chapter 31 0.693 N ( 0.693 )( 6.02 × 10 23 ) m R= = t1 t1 t1 = 2 (0.693)(6.0 × 10 23 ) m (0.693)(6 02 ×10 23 )(1.00 ×10 −3 g) = MR (235 g )(80.0 Bq) 1y   = 2.219 ×1016 s = 7.03 × 10 y   3.156 ×10 s  52 Solution 50 V has one of the longest known radioactive half-lives In a difficult experiment, a 50 researcher found that the activity of 1.00 kg of V is 1.75 Bq What is the half-life in years? R= [ ] 0.693 N (0.693) (6.02 × 10 23 atoms mol) / M m = t1 t1 2 From the periodic table, M = 50.94 g/mol , so t1 (0.693)(6.02 ×10 23 atoms mol)(1000 g ) = (50.94 g mol)(1.75 Bq) 1y   = 4.681× 10 24 s = 1.48× 1017 y   3.156 ×10 s  53 You can sometimes find deep red crystal vases in antique stores, called uranium glass because their color was produced by doping the glass with uranium Look up the natural isotopes of uranium and their half-lives, and calculate the activity of such a vase assuming it has 2.00 g of uranium in it Neglect the activity of any daughter nuclides OpenStax College Physics Solution Instructor Solutions Manual Chapter 31  6.02 × 10 23  (2.00 g )(0.00720) = 3.689 × 1019 ; N 235 =   235 g   6.02 × 10 23  (2.00 g )(0.99274) = 5.022 × 10 21 N 238 =   238 g  R= 0.693N (0.693)(3.689 × 10 Bq) ; R235 = = 1.151× 10 Bq t1 (7.04 × 10 y)(3.156 × 10 s y) R238 = (0.693)(3.689 × 10 Bq) = 2.468 × 10 Bq (4.468 × 10 y)(3.156 × 10 s y) R = R235 + R238 = 2.58×10 Bq 54 A tree falls in a forest How many years must pass before the the tree’s carbon drops to 1.00 decay per hour? Solution  6.02 × 10 23   6.02 × 10 23   m = (1.3 × 10 −12 ) (1.00 g) = 6.522× 1010 N = (1.3 × 10 −12 ) M 12 g     14 C activity in 1.00 g of ln( R0 R ) ln( R0 R ) t 12 R0 = e ⇒ t = = λ 0.693 −1 ln (2.499 × 10 d s) (1 d 3600 s) (5730 y) = = 5.6 × 10 y 0.693 − λt [ 55 Solution 56 What fraction of the today? ] 40 K that was on Earth when it formed 4.5 × 10 years ago is left 9 N −0.693 t t1 = e −λt = e = e −[ ( 0.693 )( 4.5×10 y ) ] (1.28×10 y ) = 0.087 N0 60 A 5000-Ci Co source used for cancer therapy is considered too weak to be useful when its activity falls to 3500 Ci How long after its manufacture does this happen? OpenStax College Physics Solution Instructor Solutions Manual Chapter 31 ln( R0 R ) t1 R 0.693t R  = e −λt ⇒ ln  = λt = ⇒t = R0 t1 0.693  R = [ ln(5000 Ci 3500 Ci)](5.2714 y) = 2.71 y 0.693 57 235 238 Natural uranium is 0.7200% U and 99.27% U What were the percentages of 235 U and 238 U in natural uranium when Earth formed 4.5 × 10 years ago? Solution Assume we have a 10,000-particle sample and take it backwards in time N 235 ≈ 72 and N 238 ≈ 9927 N 235 = ( N 235 ) e = 72e % N 235 =     - 693 / t1 / 58 -0.693/ t1/  − 0.693× 4.5×10 y /( 7.038×10 y)  = 6049 ( N 235 ) ×100% = 23.27% = 23%; ( N 235 ) + ( N 238 ) ( N 238 ) = 9927 e % N 238 = ⇒ ( N 235 ) = N 235 e    0.693(4.5×109 y)  (4.468×108 y) = 19,950; ( N 238 ) ×100% = 76.73% = 77% ( N 235 ) + ( N 238 ) The β particles emitted in the decay of H (tritium) interact with matter to create light in a glow-in-the-dark exit sign At the time of manufacture, such a sign contains 15.0 Ci of H (a) What is the mass of the tritium? (b) What is its activity 5.00 y after manufacture? − Solution R= (a) 0.693 N ( 0.693)( m M ) = t1 t1 The half-life of tritium is 12.33 y, and the atomic  1.6605 × 10 −27 kg   = 5.0082 × 10 − 27 kg atom M = 3.016050 u  1u   mass is , so Rt1 M m= 0.693 or 2 OpenStax College Physics m= Instructor Solutions Manual (15.0 Ci)(12.33 y)(5.0082 × 10 −27 kg ) M 0.693 Chapter 31  3.70 × 1010 Bq  3.156 × 10 s     Ci y    = 1.56 × 10 −6 kg = 1.56 mg    0.693(5.00 y)  0.693   = 11 Ci R = R0 e − λt = R0 exp − t  = (15.0 Ci) exp − 12.33 y   t1     (b) 59 World War II aircraft had instruments with glowing radium-painted dials (see Figure 31.2) The activity of one such instrument was 1.0 × 10 Bq when new (a) What mass 226 of Ra was present? (b) After some years, the phosphors on the dials deteriorated chemically, but the radium did not escape What is the activity of this instrument 57.0 years after it was made? Solution R= (a) 0.693 N ( 0.693)( m M ) = t1 t1 The half-life of radium is 1.60 × 10 y and the atomic  1.6605 × 10 −27 kg   = 3.7532× 10 −25 kg atom M = 226.025402 u  1u   mass is , so Rt1 M m= 0.693 or 2 (1.00 × 10 Bq)(1.60 × 10 y)(3.7532× 10 −25 kg )  3.156× 10 s    m= 0.693 y   = 2.73 × 10 −6 kg = 2.73 μg  0.693   0.693(57.0 y)   R = R0 e −λt = R0 exp − t  = (1.00 ×105 ) exp −  t1  60 × 10 y     (b) = 9.76 ×10 Bq 60 210 (a) The Po source used in a physics laboratory is labeled as having an activity of 1.0 µCi on the date it was prepared A student measures the radioactivity of this source with a Geiger counter and observes 1500 counts per minute She notices that the source was prepared 120 days before her lab What fraction of the decays is she observing with her apparatus? (b) Identify some of the reasons that only a fraction of the α s emitted are observed by the detector OpenStax College Physics Solution Instructor Solutions Manual Chapter 31 R0 = (1.00 ×10 −6 Ci)(3.70 ×1010 Bq Ci) = 3.70 ×10 Bq; 1500 = 25.0 Bq = observed 60.0 s R = R0 e −λt = (3.70 ×10 Bq)e −[ ( 0.693 )(120 d ) ] (138.4 d ) = 2.03 ×10 Bq R' = R' 25.0 Bq = = 1.23 ×10 −3 R 03 × 10 Bq (a) (b) Only part of the emitted radiation goes in the direction of the detector Only a fraction of that causes a response in the detector Some of the emitted radiation (mostly α particles) is observed within the source Some is absorbed within the source, some is absorbed by the detector, and some does not penetrate the detector 61 Armor-piercing shells with depleted uranium cores are fired by aircraft at tanks (The high density of the uranium makes them effective.) The uranium is called depleted 235 238 because it has had its U removed for reactor use and is nearly pure U Depleted uranium has been erroneously called non-radioactive To demonstrate that this is 238 wrong: (a) Calculate the activity of 60.0 g of pure U (b) Calculate the activity of 60.0 g of natural uranium, neglecting the Solution 234 U and all daughter nuclides  6.02 ×10 23  (60.0 g ) = 1.517 × ×10 23 ; N =   238.05 g  0.693 N (0.693)(1.517 ×10 23 Bq) R= ; R235 = = 7.457×105 Bq t1 (4.468×10 y)(3.156 ×10 s y) (a) = 7.46 ×105 Bq N 235 R238  6.02 × 10 23  (60.0 g ) = 1.106 × ×10 21; = (0.0072)  235.05 g  = (0.992745)( 7.457× 105 Bq) = 7.403× 10 Bq R235 = (b) 62 (0.693)(1.517 × 10 23 Bq) = 3.451× 10 Bq (7.038× 10 y)(3.156 × 10 s y) Rtot = R235 + R238 = 7.75 × 105 Bq The ceramic glaze on a red-orange Fiestaware plate is U O and contains 50.0 grams OpenStax College Physics of 238 Instructor Solutions Manual U , but very little 235 Chapter 31 U (a) What is the activity of the plate? (b) Calculate the total 238 energy that will be released by the U decay (c) If energy is worth 12.0 cents per kW ⋅h , what is the monetary value of the energy emitted? (These plates went out of production some 30 years ago, but are still available as collectibles.) Solution R0 = (a) 0.693(m M ) 0.693(0.0500 kg )(1 u 1.6605 × 10 −27 kg) = t1 (238.050784 u )( 4.47 × 109 y)(3.156 × 107 s 1y) = 6.21× 10 Bq = 1.68 ×10 −5 Ci (b) From Appendix B, the energy released per decay is 4.27 MeV, so  4.27 × 10 eV  1.602 × 10 −19 J  1u  decay     E = 0.0500 kg  − 27 decay eV  238.050784 u  1.6605 × 10 kg    = 8.65 × 1010 J (c) The monetary value of the energy is $0.12  h  (8.75 ×1010 J ) 1×10   = $2.9 ×10 W ⋅ h  3600 s   63 238 Large amounts of depleted uranium ( U ) are available as a by-product of uranium processing for reactor fuel and weapons Uranium is very dense and makes good 238 counter weights for aircraft Suppose you have a 4000-kg block of U (a) Find its activity (b) How many calories per day are generated by thermalization of the decay energy? (c) Do you think you could detect this as heat? Explain Solution R0 = (a) 0.693(m M ) 0.693(4000 kg )(1 u 1.6605 × 10 −27 kg ) = t1 (238.050784 u )( 4.47 × 109 y)(3.156 × 107 s 1y) = 4.97 ×1010 Bq = 1.34 Ci E 5.00 ×1010 decay  4.27 MeV  8.64 ×10 s  1.602 ×10    = day s day eV  decay   = 8.95 ×10 −3 (b) J  cal  − cal   = 7.1×10 day  4.186 J  day (c) This will not be noticeable J   OpenStax College Physics 64 Instructor Solutions Manual Chapter 31 The Galileo space probe was launched on its long journey past several planets in 1989, 238 with an ultimate goal of Jupiter Its power source is 11.0 kg of Pu , a by-product of nuclear weapons plutonium production Electrical energy is generated thermoelectrically from the heat produced when the 5.59-MeV α particles emitted in 238 each decay crash to a halt inside the plutonium and its shielding The half-life of Pu 238 is 87.7 years (a) What was the original activity of the Pu in becquerel? (b) What power was emitted in kilowatts? (c) What power was emitted 12.0 y after launch? You may neglect any extra energy from daughter nuclides and any losses from escaping γ rays Solution R0 = (a) 0.693 N 0.693(6.02 × 10 23 238.05 g)(11 × 103 g ) = = 6.97 × 1015 Bq t1 (87.7 y)(3.156 × 10 s y) ∆E 5.59 × 10 eV = 6.97 × 1015 counts s 1.60 × 10 −19 J eV ∆t count = 6.24 × 10 W = 6.24 kW ( P0 = (b) (c) 66 )( ) P = P0 e −λt = (6.22 kW )e −[ ( 0.693 )(12 y ) ] (87 y ) = 5.67 kW 236 Unreasonable Results A nuclear physicist finds 1.0  µg of U in a piece of uranium ore and assumes it is primordial since its half-life is 2.3 × 10 y (a) Calculate the 236 amount of U that would had to have been on Earth when it formed 4.5 × 10 y ago for 1.0  µg to be left today (b) What is unreasonable about this result? (c) What assumption is responsible? Solution M = M e − λt = M e M = Me (a) ( 0.693 ) t t1 − ( 0.693 ) t t1 = (1.0 × 10 −6 g )e (0.693 )( 4.5×10 y ) ( 2.3×107 y ) = 7.67 × 10 52 g = 7.7 × 10 49 kg (b) This mass is impossibly large; it is greater than the mass of the entire Milky Way galaxy 236 (c) The assumption that the U was primordial is unreasonable because its half-life is too short compared to the age of the earth 67 Unreasonable Results (a) Repeat Exercise 31.57 but include the 0.0055% natural OpenStax College Physics Instructor Solutions Manual Chapter 31 U with its 2.45 × 10 y half-life (b) What is unreasonable about this 234 result? (c) What assumption is responsible? (d) Where does the U come from if it is not primordial? abundance of Solution 234 (a) Calculate the ratio of the original ( N 234 ) = e ( 0.693 )( 4.5×10 y ) N 234 ( 2.34 ×105 y ) 234 U abundance to its present value = e127 28 ≅ 105500 (b) This is an absurdly large ratio, since it implies more the universe 234 U than the known mass of 234 (c) The unreasonable assumption is that U existed primordially and was not created as a daughter nucleus from an alpha process (d) See Figure 31.27 68 234 U is in the decay chain of 238 U Unreasonable Results The manufacturer of a smoke alarm decides that the smallest current of α radiation he can detect is 1.00 µA (a) Find the activity in curies of an α emitter that produces a 1.00 µA current of α particles (b) What is unreasonable about this result? (c) What assumption is responsible? Solution I = 1.00 μA = 1.00 × 10 −6 C s ; qα = 2e = 2(1.60 × 10 −19 C) ∴R = (a) 1.00 × 10 −6 C s = 3.125 × 1012 s −1 −19 2(1.60 × 10 C) Ci = 3.70 × 1010 s −1 ⇒ R = 3.125 × 1012 s −1 = 84.5 Ci 3.70 × 1010 s −1Ci −1 (b) This is much too hot a source (c) The 1.00 μA current is unreasonably large The α –radiation is detected by another, more sensitive, method 31.6 BINDING ENERGY 69 H is a loosely bound isotope of hydrogen Called deuterium or heavy hydrogen, it is stable but relatively rare—it is 0.015% of natural hydrogen Note that deuterium has OpenStax College Physics Instructor Solutions Manual Chapter 31 Z = N , which should tend to make it more tightly bound, but both are odd numbers Calculate BE / A , the binding energy per nucleon, for H and compare it with the approximate value obtained from the graph in Figure 31.27 Solution [ ( ) ( )] BE Zm H + Nmn − m H c = A A  931.5 MeV c  c = (1.007825 u + 1.008665 u − 2.014102 u ) u   BE = 1.112 MeV ≈ graph's value A 70 56 Fe is among the most tightly bound of all nuclides It is more than 90% of natural 56 iron Note that Fe has even numbers of both protons and neutrons Calculate 56 Fe and compare it with the approximate value obtained from the graph in Figure 31.27 BE / A , the binding energy per nucleon, for Solution [ ( ) BE Zm H + Nmn − m = A A ( 56 26 )] Fe30 c  931.5 MeV c  c    = [ 26(1.007825 u ) + 30(1.008665 u ) − 55.93439u ] u   56  BE = 8.790 MeV ≈ graph's value A 71 Solution 209 Bi is the heaviest stable nuclide, and its BE / A is low compared with medium209 mass nuclides Calculate BE / A , the binding energy per nucleon, for Bi and compare it with the approximate value obtained from the graph in Figure 31.27 [ ( ) ( )] BE Zm H + Nmn − m 209 83 Bi126 c = , so A A BE [ 83(1.007825 u ) + 126(1.008665 u ) − 208.980374 u ]  931.5 MeV c   c = A 209 u   = 7.848 MeV nucleon This binding energy per nucleon is approximately the value given in the graph OpenStax College Physics 72 (a) Calculate BE / A for (b) Calculate BE / A for Instructor Solutions Manual Chapter 31 235 U , the rarer of the two most common uranium isotopes 238 U (Most of uranium is 238 U ) Note that numbers of both protons and neutrons Is the BE / A of 235 from that of U ? [ ( ) Solution 238 238 U has even U significantly different )] ( BE Zm H + Nmn − m 235 92 U143 c = A A [ 92(1.007825 u ) + 143(1.008665 u ) − 235.043924 u ]  931.5 MeV/c c =   235 u   BE = 7.591 MeV ≈ graph's value (a) A BE [ 92(1.007825 u ) + 143(1.008665 u ) − 235.043924 u ]  931.5 MeV/c   c = A 238 u   (b) = 7.570 MeV This is a significant difference given that these are isotopes 73 (a) Calculate BE / A for 12 C Stable and relatively tightly bound, this nuclide is most of 14 12 natural carbon (b) Calculate BE / A for C Is the difference in BE / A between C 14 and C significant? One is stable and common, and the other is unstable and rare Solution [ ] BE Zm( H ) + Nmn − m( 126 C ) c = A A  931.5 MeV  = [ 6(1.007825 u ) + 6(1.008665 u ) − 12.000000 u ]  12   BE = 7.680 MeV ≈ graph's value (a) A [ ] BE Zm( H ) + Nmn − m( 126 C ) c = A A  931.5 MeV  = [ 6(1.007825 u ) + 8(1.008665 u ) − 14.003241u ]  12   BE = 7.520 MeV (b) A OpenStax College Physics Instructor Solutions Manual Chapter 31 12 This is not significantly different from the value for C , but it is sufficiently lower to allow decay into another nuclide that is more tightly bound 74 Solution The fact that BE / A is greatest for A near 60 implies that the range of the nuclear force is about the diameter of such nuclides (a) Calculate the diameter of an A = 60 58 90 nucleus (b) Compare BE / A for Ni and Sr The first is one of the most tightly bound nuclides, while the second is larger and less tightly bound (a) d = 2r = 2r0 A1 = 2(1.2 ×10 −15 m)( 60)1 = 9.40 ×10 −15 m [ ( ) Zm H + Nmn − m  BE    = A  A  Ni ( 58 28 )] Ni 30 c  931.5 MeV  = [ 28(1.007825 u ) + 30(1.008665 u ) − 57.935346u ]  58   (b) BE = 8.732 MeV A [ ( ) Zm H + Nmn − m  BE    = A  A Sr ( 58 28 )] Ni30 c  931.5 MeV  = [ 38(1.007825 u ) + 52(1.008665 u ) − 89.907738 u ]  90   BE = 8.696 MeV A 75 The purpose of this problem is to show in three ways that the binding energy of the electron in a hydrogen atom is negligible compared with the masses of the proton and electron (a) Calculate the mass equivalent in u of the 13.6-eV binding energy of an electron in a hydrogen atom, and compare this with the mass of the hydrogen atom obtained from Appendix A (b) Subtract the mass of the proton given in Table 31.2 from the mass of the hydrogen atom given in Appendix A You will find the difference is equal to the electron’s mass to three digits, implying the binding energy is small in comparison (c) Take the ratio of the binding energy of the electron (13.6 eV) to the energy equivalent of the electron’s mass (0.511 MeV) (d) Discuss how your answers confirm the stated purpose of this problem OpenStax College Physics Solution Instructor Solutions Manual Chapter 31  E  13.6 eV   u = = 1.46 × 10 −8 u  2 2 c  c   931.5 MeV c  (a) , compared to 1.00782 u for H E = mc ⇒ m = (b) m(1 H) − m p = 1.007825 u − 1.007276 u = 0.000549 u 13.6 eV = 2.66 × 10 −5 (c) 5.11 × 10 eV (d) In part (a), since the mass equivalent of the BE of the electron is orders of magnitude smaller than the mass of the hydrogen atom, the BE of the electron is negligible compared to the masses of the proton and electron In part (b), since the mass difference between the proton and the hydrogen atom equals the mass of the electron, there is negligible BE of the electron And finally, in part (c), since the BE is orders of magnitude smaller than the energy equivalent of the electron’s mass, it is clearly negligible compared to the mass of the electron 76 Unreasonable Results A particle physicist discovers a neutral particle with a mass of 2.02733 u that he assumes is two neutrons bound together (a) Find the binding energy (b) What is unreasonable about this result? (c) What assumptions are unreasonable or inconsistent? Solution  931.5 MeV c  c BE = [ 2mn − m(particle)] c = [ 2(1.008665 u ) − 2.02733 u ] u   (a) = − 9.315 MeV (b) The binding energy cannot be negative; the nucleons would not stay together (c) The particle cannot be made from two neutrons 31.7 TUNNELING 77 Derive an approximate relationship between the energy of α decay and half-life using t the data in Table 31.3 It may be useful to graph the log of against E α to find some straight-line relationship OpenStax College Physics Solution Nuclide 216 Instructor Solutions Manual Eα (MeV) Ra 9.5 t1 0.18 μs ln(t1 in s) − 15.53 194 Po 240 Cm 6.4 226 Ra 4.9 27 d 1600 y 232 Th 4.1 1.4 × 1010 y 40.63 7.0 s Chapter 31 − 0.357 14.66 24.64 ln t1 Eα (MeV ) ln t1 = mEα + b −1 Regression analysis ⇒ m = slope = −10.14(MeV) , b = intercept = 7.75 ×101 = 77.5 ⇒ ln t1 ≅ 78 − 10.14 Eα ( MeV) + 77.5 MeV Integrated Concepts A 2.00-T magnetic field is applied perpendicular to the path of charged particles in a bubble chamber What is the radius of curvature of the path of a 10 MeV proton in this field? Neglect any slowing along its path Solution Since the energy of the proton (10.0MeV) is substantially less than the rest mass energy of the proton (938MeV), we know the velocity is non-relativistic and that E = mv2 Therefore, 12 12  ⋅ 10.0 MeV   = (0.1460)( 2.998 × 10 m s) So , =    938.27 MeV c  mv (1.6726 × 10 −27 kg )( 4.377 × 10 m s) r= = = 0.228 m = 22.8 cm qB (1.602 × 10 −19 C)( 2.00 T)  2E  v=   m  OpenStax College Physics 79 Solution Instructor Solutions Manual Chapter 31 235 (a) Write the decay equation for the α decay of U (b) What energy is released in this decay? The mass of the daughter nuclide is 231.036298 u (c) Assuming the residual nucleus is formed in its ground state, how much energy goes to the α particle? (a) 235 92 U143 →23190 Th141 + 42 He [ ] E = ∆mc = m( 231 U) − m( 231 Th) − mα c (b)  931.5 MeV/c  c = 4.679 MeV = (235.043924 u − 231.036298 u − 4.002603 u ) u   (c) The energy released is much smaller than the rest energy of the decay products, so it is safe to use non-relativistic formulas for conservation of energy and momentum The available energy goes into kinetic energy of the two particles and, of course, momentum is conserved The original momentum is zero (decaying particle is at rest), so the magnitudes of the momenta for each decay particles will be equal m 2 1 2 m1v1 = m2 v ⇒ v1 = 2 v E = m1v1 + m2 v m1 2 Thus: and  m2 1 m E + m2 v = v m2  + 1 ⇒ v m2 = 2 m1v 2 2 + (m2 m1 )  m1  4.6789 MeV Eα = = 4.599 MeV + (4.0026 u 231.04 u ) E= 80 48 Unreasonable Results The relatively scarce naturally occurring calcium isotope Ca 16 has a half-life of about × 10 y (a) A small sample of this isotope is labeled as 48 having an activity of 1.0 Ci What is the mass of the Ca in the sample? (b) What is unreasonable about this result? (c) What assumption is responsible? Solution R= m= (a) 0.693(6.02 ×10 23 M ) m t1 RMt1 0.693(6.02 ×10 23 ) = = ×1012 g = ×10 kg (3.7 × 10 10 Bq)( 48.0 g )( × 1016 y)(3.156×10 s y) 0.693(6.02 × 10 23 ) OpenStax College Physics Instructor Solutions Manual Chapter 31 (b) The mass is much too large for such a small sample (c) The activity is much too high to be reasonable 81 Unreasonable Results A physicist scatters γ rays from a substance and sees evidence –13 of a nucleus 7.5 × 10  m in radius (a) Find the atomic mass of such a nucleus (b) What is unreasonable about this result? (c) What is unreasonable about the assumption? Solution  R   7.5 × 10 −13 m   ⇒ A = (625) = 2.4 × 108 R = R0 A ⇒ A =   =  −15 R × 10 m   0  (a) 3 13 (b) The greatest known values of A are around 260 This is unreasonably large (c) The assumed radius is much too large to be reasonable 82 Unreasonable Results A frazzled theoretical physicist reckons that all conservation laws are obeyed in the decay of a proton into a neutron, positron, and neutrino (as in β + decay of a nucleus) and sends a paper to a journal to announce the reaction as a possible end of the universe due to the spontaneous decay of protons (a) What energy is released in this decay? (b) What is unreasonable about this result? (c) What assumption is responsible? Solution p → n + β + + ve E = ( m p − m n − me ) c (a)  931.5 MeV/c  c = − 1.805 MeV = [1.007276 u − 1.008665 u − 0.0005485 u ] u   (b) Negative energy implies energy input is necessary and the reaction cannot be spontaneous (c) Although all conversation laws are obeyed, energy must be supplied, so the assumption of spontaneous decay is incorrect This file is copyright 2016, Rice University All Rights Reserved ... 7.57 MeV OpenStax College Physics (b) Instructor Solutions Manual c c = = 1.002 vβ 0.998c Chapter 31 (4 significant figures used to show difference) 31. 4 NUCLEAR DECAY AND CONSERVATION LAWS 17... radius r = r0 A , and with a mass of A u, then its density is independent of A (b) Calculate that density in 56 u/fm3 and kg/m , and compare your results with those found in Example 31. 1 for F e ... College Physics Instructor Solutions Manual Chapter 31 23 α decay of 210 Po , the isotope of polonium in the decay series of 238 U that was discovered by the Curies A favorite isotope in physics