Chapter 44 Solutions. So protons and neutrons are nearly equally numerous in your body, each contributing mass (say) 35 kg:

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1 Chapter 44 Solutions *44. An iron nucleus (in hemoglobin) has a few more neutrons than protons, but in a typical water molecule there are eight neutrons and ten protons. So protons and neutrons are nearly equally numerous in your body, each contributing mass (say) 35 kg: 35 kg nucleon kg ~ 08 protons and ~ 0 8 neutrons The electron number is precisely equal to the proton number, ~ 0 8 electrons 44. mv q( V) and mv r qvb m( V) qr B r m ( V ) qb 000 V ( C) 0.00 T m r (a) For C: m u and m kg m m 7 r kg m 7.89 cm kg For 3 m 7 C: r kg m 8. cm kg With r m( V) qb and r m( V), qb the ratio gives r r m m. r 7.89 cm r 8. cm 0.96 and m m u 0.96 so they do agree. 3 u 000 by Harcourt, Inc. All rights reserved.

2 566 Chapter 44 Solutions *44.3 (a) F k e Q Q r ( N m /C ) ()(6)( C) ( m) 7.6 N a F m 7.6 N kg m/s away from the nucleus.. (c) U k e Q Q r ( N m /C ) ()(6)( C) ( m) J.73 MeV 44.4 E α 7.70 MeV (a) d min 9 9 4kZe e kze e ( )( 79)( ) 3 mv E 7. 70( ) The de Broglie wavelength of the α is α m 9.5 fm λ h m α v α h m α E α m 5.8 fm ( )7.70( ) (c) Since λ is much less than the distance of closest approach, the α may be considered a particle (a) The initial kinetic energy of the alpha particle must equal the electrostatic potential energy at the distance of closest approach. K i U f k e qq r min r min k e qq K i ( N m C ) ( 79) ( C) ( MeV) J MeV m Since K i m α v i k e qq r min, v i k e qq m α r min ( 79) ( C) ( 4.00 u) ( kg u) ( m) N m C ms

3 Chapter 44 Solutions 567 Goal Solution (a) Use energy methods to calculate the distance of closest approach for a head-on collision between an alpha particle having an initial energy of MeV and a gold nucleus ( 97 Au) at rest. (Assume the gold nucleus remains at rest during the collision.) What minimum initial speed must the alpha particle have in order to get as close as 300 fm? G : The positively charged alpha particle (q +e) will be repelled by the positive gold nucleus (Q +79e), so that the particles probably will not touch each other in this electrostatic collision. Therefore, the closest the alpha particle can get to the gold nucleus would be if the two nuclei did touch, in which case the distance between their centers would be about 6 fm (using r r 0 A /3 for the radius of each nucleus). To get this close, or even within 300 fm, the alpha particle must be traveling very fast, probably close to the speed of light (but of course v must be less than c). O : At the distance of closest approach, r min, the initial kinetic energy will equal the electrostatic potential energy between the alpha particle and gold nucleus. qq qq A : (a) K α U k e and r min k e ( N m /C )()(79)( C) r min K α (0.500 MeV)( fm J / MeV) Since K α mv k e qq r min v k eqq ( N m /C )()(79)( C) mr min 4( kg)( m) m/s L : The minimum distance in part (a) is about 00 times greater than the combined radii of the particles. For part, the alpha particle must have more than 0.5 MeV of energy since it gets closer to the nucleus than the 455 fm found in part (a). Even so, the speed of the alpha particle in part is only about % of the speed of light, so we are justified in not using a relativistic approach. In solving this problem, we ignored the effect of the electrons around the gold nucleus that tend to screen the nucleus so that the alpha particle sees a reduced positive charge. If this screening effect were considered, the potential energy would be slightly reduced and the alpha particle could get closer to the gold nucleus for the same initial energy. *44.6 It must start with kinetic energy equal to K i U f k e qq r f. Here r f stands for the sum of the radii of the 4 He and 79 Au nuclei, computed as 3 3 r f r 0 A + r 0 A ( m) ( ) m Thus, K U i f 9 9 ( N m C )( 79) C m J 5.6 MeV 000 by Harcourt, Inc. All rights reserved.

4 568 Chapter 44 Solutions 44.7 (a) r ra / 3 0 ( m)(4) / m r ra / 3 0 ( m)(38) / m *44.8 From r ra / 3 0, the radius of uranium is r U r 0 (38) /3. Thus, if r r U then r 0 A /3 r 0(38) /3 from which A The number of nucleons in a star of two solar masses is A.99 ( 0 30 kg) kg nucleon nucleons Therefore r r 0 A /3 ( m) ( ) km *44.0 V 4 3 πr m 3 m ρv ( kg / m 3 )( m 3 ) kg and F G m m r ( Nm /kg ) (9.6 0 kg) (.00 m) N toward the other ball. 44. The stable nuclei that correspond to magic numbers are: Z magic: He 8 O 0 Ca 8 Ni 50 Sn 8 Pb 6 N magic: T 3, He 4, N 5 7, O 6 8, Cl 37 7, K 39 9, Ca 40 0, V 5 3, Cr 5 4, Sr 88 38, Y 89 39, Zr 90 40, Xe 36 54, Ba 38 56, La 39 57, Ce 40 58, Pr 4 59, Nd 4 60, Pb 08 8, Bi , Po 84

5 Chapter 44 Solutions Of the 0 stable nuclei listed in Table A.3, (a) Even Z, Even N 48 Even Z, Odd N 6 (c) Odd Z, Even N 44 (d) Odd Z, Odd N (a) 44.4 (a) f n µb h (.935)( J/T)(.00 T) J s 9. MHz f p (.798)( J/T)(.00 T) J s 4.6 MHz (c) In the Earth's magnetic field, f p (.798)( )( ) khz 44.5 Using atomic masses as given in Table A.3, (a) For H, ( ) + ( ) 000 by Harcourt, Inc. All rights reserved.

6 570 Chapter 44 Solutions E b ( u) 93.5 MeV u. MeV/nucleon

7 Chapter 44 Solutions 57 For 4 He, ( ) + ( ) E b u 7.07 MeV/nucleon (c) For 56 Fe 6, 30( ) + 6( ) u E b u 8.79 MeV/nucleon (d) For 38 U 9, 46( ) + 9( ) u E b u 7.57 MeV/nucleon 44.6 M Zm H + Nm n M BE A M(93.5) A Nuclei Z N M in u M in u BE/A in MeV 55 M n Fe Co Fe has a greater BE/A than its neighbors. This tells us finer detail than is shown in Figure (a) The neutron-to-proton ratio, ( A Z)/ Z is greatest for 39 55Cs and is equal to.53. (c) 39 La has the largest binding energy per nucleon of MeV. 39 C s with a mass of u. We locate the nuclei carefully on Figure 44.3, the neutronproton plot of stable nuclei. Cesium appears to be farther from the center of the zone of stability. Its instability means extra energy and extra mass Use Equation The 3 Na, and for 3 Mg, E b A E b A 8. MeV/nucleon 7.90 MeV/nucleon The binding energy per nucleon is greater for 3 Na by 0.0 MeV. (There is less proton repulsion in Na 3.) 000 by Harcourt, Inc. All rights reserved.

8 57 Chapter 44 Solutions [ ] 44.9 The binding energy of a nucleus is E b ( MeV) ZM( H)+ Nm n M A Z X [ ] MeV u [ + 8( u) u] MeV u ( MeV u ) For 5 8O: E b 8( u)+ 7( u) u.96 MeV For 5 7 N: E b u Therefore, the binding energy of 5 7 N is larger by 3.54 MeV MeV 44.0 (a) The radius of the 40 Ca nucleus is: R r 0 A 3 ( m) ( 40) m The energy required to overcome electrostatic repulsion is U 3k e Q 5R 0( C) 54.0 ( 0 5 m) N m C The binding energy of 40 0 Ca is [ ] [ ] 93.5 MeV u J 84. MeV E b 0( u)+ 0( u) u 34 MeV (c) The nuclear force is so strong that the binding energy greatly exceeds the minimum energy needed to overcome electrostatic repulsion. 44. Removal of a neutron from 43 0 Ca would result in the residual nucleus, 4 0 Ca. If the required separation energy is S n, the overall process can be described by + + mass Ca S n mass 0Ca mass( n) S n ( ) u ( u)(93.5 MeV/u) 7.93 MeV 44. (a) The first term overstates the importance of volume and the second term subtracts this overstatement. For spherical volume 4 3 πr 3 4πR R 3 For cubical volume R 3 6R R 6 The maximum binding energy or lowest state of energy is achieved by building "nearly" spherical nuclei.

9 Chapter 44 Solutions E b E bf E bi For A 00, so E b 7.4 MeV A E bi 00( 7.4 MeV) 480 MeV For A 00, E b A 8.4 MeV so E bf ( 00) ( 8.4 MeV) 680 MeV E b E bf E bi E b 680 MeV 480 MeV 00 MeV 44.4 (a) "Volume" term: E C A (5.7 MeV)(56) 879 MeV "Surface" term: E C A /3 (7.8 MeV)(56) /3 60 MeV "Coulomb" term: E 3 C 3 Z(Z ) (6)(5) /3 (0.7 MeV) /3 MeV A (56) "Asymmetry" term: E 4 C 4 E b 49 MeV (A Z) A (3.6 MeV) (56 5) MeV E 79%; E b E 53. 0%, E b E3 4. 6%; E b E4 37. % E b 44.5 dn dt λ N so λ N dn dt ( )( ) s 5 4 ln T / λ s ( 9.3 min) ln t *44.6 R R e e ( 40. d) λ 804. d 640. mci 640. mci e ln 0 5 ( 6.40 mci) 0.00 mc i by Harcourt, Inc. All rights reserved.

10 574 Chapter 44 Solutions 44.7 (a) From R R e t 0 λ, R0 0.0 λ ln ln t R 4.00 h h s ln T / λ.4 h N 0 R Ci / s 5 λ / s Ci atoms λt 0 (c) R R e ( 0. 0 mci ) exp ( ).87 mc i Goal Solution A freshly prepared sample of a certain radioactive isotope has an activity of 0.0 mci. After 4.00 h, its activity is 8.00 mci. (a) Find the decay constant and half-life. How many atoms of the isotope were contained in the freshly prepared sample? (c) What is the sample's activity 30.0 h after it is prepared? G : Over the course of 4 hours, this isotope lost 0% of its activity, so its half-life appears to be around 0 hours, which means that its activity after 30 hours (~3 half-lives) will be about mci. The decay constant and number of atoms are not so easy to estimate. O : From the rate equation, R R 0 e λt, we can find the decay constant λ, which can then be used to find the half life, the original number of atoms, and the activity at any other time, t. A : (a) λ t ln R o R 4.00 h ( 60.0 s / h) 0.0 mci ln 8.00 mci s T / ln λ h h The number of original atoms can be found if we convert the initial activity from curies into becquerels (decays per second): Ci Bq R mci ( Ci) ( Bq / Ci) Bq Since R 0 λn 0, N 0 R 0 λ decays / s atoms s (c) R R o e λt (0.0 mci)e ( h )(30.0 h).87 mci L : Our estimate of the half life was about 0% short because we did not account for the non-linearity of the decay rate. Consequently, our estimate of the final activity also fell short, but both of these calculated results are close enough to be reasonable. The number of atoms is much less than one mole, so this appears to be a very small sample. To get a sense of how small, we can assume that the molar mass is about 00 g/mol, so the sample has a mass of only m ( atoms) ( 00 g / mol) ( atoms / mol) µg This sample is so small it cannot be measured by a commercial mass balance! The problem states that this sample was freshly prepared, from which we assumed that all the atoms within the sample are initially radioactive. Generally this is not true, so that N 0 only accounts for the formerly radioactive atoms, and does not include additional atoms in the sample that were not radioactive. Realistically then, the sample mass should be significantly greater than our above estimate.

11 Chapter 44 Solutions R R e λt 0 where λ ln 6.0 h 0.066/h R R e λt so ln (0.00) λ t h t t 86.4 h 44.9 The number of nuclei which decay during the interval will be N N N 0 e λ t e λ t First we find λ : λ ln T / h h s and N 0 R 0 λ (40.0 µci)( cps / µci) s nuclei Substituting these values, N N e (0.007 h )(0.0 h) e (0.007 h )(.0 h) Hence, the number of nuclei decaying during the interval is N N nuclei The number of nuclei which decay during the interval will be N N N 0 e λ t e λ t First we find λ : λ ln / T so e λ t e ln ( t/t / ) t/t / and N 0 R 0 λ R 0 T / ln Substituting in these values N N R 0T / ln e λ t e λ t R 0 T / ln t /T / t /T / ln.00 g 44.3 R λn 5.7 yr g / mol R. 3 0 decays yr yr s Bq 000 by Harcourt, Inc. All rights reserved.

12 576 Chapter 44 Solutions 44.3 (a) 65 8Ni * 8Pb (c) 7 55 Co (d) (e) 0 e H ( or p) Q M 38 U M 34 Th M4 He ( 93.5 MeV u ) Q ( )u ( 93.5 MeV u) 4.7 MeV 0.00 g N C.0 g mol molecules mol 9 N C ( N C.05 0 carbon atoms) of which in is a 4 C atom., λ 4 C ln 5730 yr. 0 4 yr s R λ N λ N 0 e λ t At t 0, R 0 λ N 0 ( s - )( ) s At time t, R decays week 0.88 week Taking logarithms, ln R λt so t R 0 λ ln R R 0 decays week t. 0 4 yr ln yr In the decay 3 H 3 He + 0 e+ ν, the energy released is E ( m) c M M c 3 3 since H He the antineutrino is massless and the mass of the electron is accounted for in the masses of 3 H and 3 He. Thus, E [ u u] ( 93.5 MeV u) MeV 8.6 kev

13 Chapter 44 Solutions (a) For e + decay, [ ]( 93.5 MeV u) Q ( M X M Y m e )c u u u Q.34 MeV Since Q < 0, the decay cannot occur spontaneously. For alpha decay, Q ( M X M α M Y )c [ u u u] ( 93.5 MeV u) Q.4 MeV Since Q < 0, the decay cannot occur spontaneously. (c) For alpha decay, Q ( M X M α M Y )c [ u u u] ( 93.5 MeV u) Q.9 MeV Since Q > 0, the decay can occur spontaneously (a) e + p n +ν For nuclei, 5 O + e 5 N + ν Add seven electrons to both sides to obtain 8O atom N atom +ν. (c) From Table A.3, m( 5 O) m( 5 N)+ Q c m u u u Q (93.5 MeV/u)( u).75 MeV 000 by Harcourt, Inc. All rights reserved.

14 578 Chapter 44 Solutions (a) Let N be the number of 38 U nuclei and N' be 06 Pb nuclei. Then N N 0 e λ t and N 0 N + N so N ( N + N )e λ t or e λ t + N N Taking logarithms, λ t ln + N N where λ ( ln )/T /. Thus, t T ln ln + N N If N N for the U Pb chain with T yr, the age is: t yr ln ln yr From above, e λ t + N N. Solving for N N gives N N e λ t e λ t With t yr and T yr for the λ t ln T t ln yr yr U Pb chain, and N N With t yr and T 4. 0 yr for the λ t ln yr yr and N N Th Pb chain, 44.39

15 Chapter 44 Solutions 579 *44.40 (a) Ci Bq pci L L Ci m 3 L 48 Bq m 3 N R λ R T Bq ln 48 m d ln s d atoms m 3 (c) mass atoms mol g m atoms mol g m Since air has a density of.0 kg m 3, the fraction consisting of radon is g m fraction 3.0 kg m *44.4 Number remaining: N N 0 e ( ln )t T Fraction remaining: (a) With T 3.8 d and t 7.00 d, When t.00 yr d, N e λ t e ( ln )t T N 0 N e ( ln ) ( 7.00) ( 3.8) 0.8 N 0 N e ( ln ) ( 365.5) ( 3.8) N 0 (c) Radon is continuously created as one daughter in the series of decays starting from the long-lived isotope 38 U Q [ M 7 Al + M α M 30 P m n] c Q [ ]u ( 93.5 MeV u) 64. MeV (a) 97 79Au + n Au * Hg + e + ν Consider adding 79 electrons: 97 79Au atom + 0 n Hg atom + ν + Q Q M 97 Au + m n M 98 Hg c Q [ ]u ( 93.5 MeV u) 7.89 MeV 000 by Harcourt, Inc. All rights reserved.

16 580 Chapter 44 Solutions *44.44 (a) For X, A and Z + 0 0, so X is 0 Ne A and Z , so X is Xe (c) A 0 and Z +, so X must be a positron. As it is ejected, so is a neutrino: X 0 e + and X 0 0 ν *44.45 Neglect recoil of product nucleus, (i.e., do not require momentum conservation). The energy balance gives K emerging K incident + Q. To find Q: Q [( M H + M Al ) ( M Si + m n )]c Q [( ) ( ) ]u 93.5 MeV u Thus, K emerging 66. MeV 56. MeV.00 MeV 56. MeV 0 *44.46 (a) 5B+ He 6C+ H 4 3 The product nucleus is 3 6C C+ H B+ He The product nucleus is 0 5B Be MeV 4 8 Be + 0 n, so M 4 8 Be M 4 9 Be Q c m n M 4 8 Be (.666 MeV ) u u u 93.5 MeV u 4 9 Be + 0 n 0 4Be MeV, so M 0 Be M 4 9 Be + m n Q c M 0 4Be u u 6.80 MeV 93.5 MeV u u 4

17 Chapter 44 Solutions 58 Goal Solution Using the Q values of appropriate reactions and from Table 44.5, calculate the masses of 8 Be and 0 Be i n atomic mass units to four decimal places. G : O : The mass of each isotope in atomic mass units will be approximately the number of nucleons (8 or 0), also called the mass number. The electrons are much less massive and contribute only about 0.03% to the total mass. In addition to summing the mass of the subatomic particles, the net mass of the isotopes must account for the binding energy that holds the atom together. Table 44.5 includes the energy released for each nuclear reaction. Precise atomic masses values are found in Table A.3. A : The notation 9 Be ( γ, n) 8 Be with Q.666 MeV means 9 Be + γ 8 Be + n.666 MeV Therefore m( 8 Be) m( 9 Be) m n MeV 93.5 MeV / u m( 8 Be) u The notation 9 Be ( n, γ ) 0 Be with Q 6.80 MeV means 9 Be + n 0 Be + γ MeV m( 0 Be) m( 9 Be)+ m n MeV 93.5 MeV / u m( 0 Be) u L : As expected, both isotopes have masses slightly greater than their mass numbers. We were asked to calculate the masses to four decimal places, but with the available data, the results could be reported accurately to as many as six decimal places U Rb + 55Cs + 30 n, so Q M 36 9U M Rb M43 55Cs 3m n c From Table A.3, Q [ ( ) ]u ( 93.5 MeV u) 65 MeV N N N 0 N 0 e λ T h / N 0 e λ T h / N 0 e λ T h / ln / e e ln / e ln / 000 by Harcourt, Inc. All rights reserved.

18 58 Chapter 44 Solutions *44.50 H + 7 3Li 7 4Be + 0 n Q [(M H + M Li ) (M Be + M n )](93.5 MeV / u) Q [( u u) ( u u) ](93.5 MeV / u) Q ( u)(93.5 MeV / u).644 MeV Thus, KE min + m incident projectile m target nucleus Q (.644 MeV).88 MeV 44.5 (a) N 0 mass mass per atom.00 kg ( u) kg u ln ln λ T yr s yr /.5 04 ( ) R 0 λ N 0 ( s )( ) 9. 0 Bq s (c) R R 0 e λ t, so t λ ln R R 0 λ ln R 0 R t s ln.9 0 Bq 0.00 Bq 3.38 yr 03 s s yr 44.5 (a) Co 6Fe + + e ν [ e] Co Fe The Q value for this positron emission is Q M M m c [ ] Q ( )... u 93.5 MeV u MeV Since Q < 0, this reaction cannot spontaneously occur C 7N + e ν The Q value for this e decay is Q [ M 4 C M4 N] c. Q [ ]u ( 93.5 MeV u) 0.56 MeV 56 kev Since Q > 0, the decay can spontaneously occur. (c) The energy released in the reaction of is shared by the electron and neutrino. Thus, K e can range from zero to 56 kev.

19 Chapter 44 Solutions (a) r r 0 A / A /3 m. When A, r m F k e (Z )e r ( N m /C )(Z )( C) r When Z 6 and r m, F 5 N (c) U k e q q r k e (Z )e ( )( ) (Z ) r r When Z 6 and r m, U J.6 MeV (d) A 38; Z 9, r m F 379 N and U.8 0 J 7.6 MeV (a) ln(counts/min) vs time 9.00 ln(cpm) t (h) A least-square fit to the graph yields: λ slope ( h ) h, and ln cpm t 0 intercept 830. λ 0.50 h h 60.0 min min T ln λ ln min.77 h min (c) From (a), intercept ln (cpm) Thus, (d) N 0 R 0 λ ( cpm) 0 λ Eff counts min min 0.00 ( cpm) 0 e 8.30 counts min counts min atoms 000 by Harcourt, Inc. All rights reserved.

20 584 Chapter 44 Solutions (a) Because the reaction p n + e + + ν would violate the law of conservation of energy, m p u m n u m e u Note that m n + m e + > m p The required energy can come from the electrostatic repulsion of protons in the nucleus. (c) Add seven electrons to both sides of the reaction for nuclei 3 7N 3 6 C + e + + ν to obtain the reaction for neutral atoms 3 7N atom 3 6 C atom + e + + e + ν [ ] [ 0] u Q c m( 3 N) m( 3 C) m e + m e m ν Q (93.5 MeV / u) Q (93.5 MeV/u)( u).0 MeV (a) If we assume all the 87 Sr came from 87 Rb, then N N 0 e λ t yields t λ ln N T N 0 ln ln N 0 N, where N N 87 Rr and N 0 N 87 Sr + N 87 Rb. ( t yr) ln ln yr It could be no longer. The rock could be younger if some 87 Sr were originally present (a) Let us assume that the parent nucleus (mass M p ) is initially at rest, and let us denote the masses of the daughter nucleus and alpha particle by M d and M α, respectively. Applying the equations of conservation of momentum and energy for the alpha decay process gives M d v d M α v α () M p c M d c + M α c + M αv α + M d v d () The disintegration energy Q is given by Q ( M M M ) c M v + α α α M v p d d d (3) Eliminating v d from Equations () and (3) gives Q M M α d v α M d Q K α + M α M d + M α v α v α + M M α v α M α v α + M α d K α + M α M α 4.87 MeV 4.78 MeV + ( 4 / ) M d M d

21 Chapter 44 Solutions (a) The reaction is 6 Pm 59Pr + α Q (M Pm M α M Pr )93.5 ( ) MeV (c) The alpha and daughter have equal and opposite momenta p α p d E α α p m α E d p d m d E α E α E tot E α + E d p α m α p α + p α m a m d m α m + d m d + m 4 97.% or.6 MeV α m α m d This is carried away by the alpha (a) If E is the energy difference between the excited and ground states of the nucleus of mass M, and h f is the energy of the emitted photon, conservation of energy gives E h f + E r () Where E r is the recoil energy of the nucleus, which can be expressed as E r Mv (Mv) M () Since momentum must also be conserved, we have Mv h f c (3) Hence, E r can be expressed as E r (hf) Mc. When h f << Mc, we can make the approximation that h f E, so E r ( E) Mc E r ( E) Mc where E MeV and Mc (57 u)(93.5 MeV/u) MeV Therefore, E r (.44 0 MeV) ()( MeV) ev 000 by Harcourt, Inc. All rights reserved.

22 586 Chapter 44 Solutions *44.60 (a) One liter of milk contains this many 40 K nuclei: N.00 g nuclei mol λ ln T 39. g mol nuclei ln yr yr s s R λ N ( s )( ) 6.8 Bq For the iodine, R R 0 e λ t with λ ln d t λ ln R 0 R 8.04 d 000 ln ln d *44.6 (a) ln ln d For cobalt-56, λ 3.8 yr. T 77. d yr The elapsed time from July 054 to July 000 is 946 yr. R R 0 e λ t implies R R 0 e λ t e ( 3.8 yr )( 946 yr) e 306 ln 0 e 349 ~ For carbon-4, λ R R 0 e λ t e ln 5730 yr. 0 4 yr (. 0 4 yr )( 946 yr) e *44.6 We have N 35 N 0, 35 e λ 35t Taking logarithms, and N 38 N 0, 38 e λ 38t N 35 N e ln ( t T h, 35 + ( ln )t T h, 38 ) ln yr + ln yr t or yr yr 493. t ( yr. yr ) ln ( ln ) t

23 Chapter 44 Solutions (a) Add two electrons to both sides of the reaction to have it in energy terms: 4 H atom 4 He atom + Q [ ] c Q mc 4 M H M 4 He Q [ 4( u) u] 93.5 MeV u J MeV J kg N kg atom atoms protons (c) The energy that could be created by this many protons in this reaction is: J ( 9. 0 protons) protons 45 J P E t so t E P J W s 07 billion years (a) Q [ M 9 Be + M 4 He M C m n] c Q [ u u u u] ( 93.5 MeV u) 5.69 MeV [ H He ] Q M M3 m n [ ] Q (. 04 0) u 93.5 MeV u 3.7 MeV (exothermic) E µ B so the energies are E +µb and E µb µ.798µ n and µ n J/T E µb (.798)( J/T)(.5 T) J ev 000 by Harcourt, Inc. All rights reserved.

24 588 Chapter 44 Solutions (a) λ ln ln yr T 5.7 yr s s t 30.0 months.50 yr R R 0 e λ t ( λ N 0 )e λ t so N 0 R λ eλ t N nuclei s yr ( 0.0 Ci ) Bq Ci s s ( e 9 s )( s) g mol Mass 3. 0 atoms atoms mol. 3 0 g.3 mg. We suppose that each decaying nucleus promptly puts out both a beta particle and two gamma rays, for Q MeV.8 Mev P QR (.8 Mev) ( J MeV) ( s ) 0.66 W For an electric charge density ρ Ze 4 3 πr 3 Using Gauss's Law inside the sphere, E 4πr 4 3 πr 3 e 0 Ze 4 3 πr 3 : E 4π e 0 Zer R 3 (r R) E 4π e 0 Ze r (r R) We now find the electrostatic energy: U e 0 E 4πr dr r 0 U e 0 0 R Z e r 4π e 0 R 6 4πr dr + e Z e 0 R 4π e 0 r 4 4πr dr Z e R 5 8π e 0 5R 6 + R 3 Z e 0 π e 0 R

25 Chapter 44 Solutions (a) For the electron capture, 93 43Tc + 0 e 93 4 Mo + γ The disintegration energy is Q [ M 93 Tc M 93 Mo] c. Q [ ] u ( 93.5 MeV u ) 3. 7 MeV >.44 MeV Electron capture is allowed to all specified excited states in 93 4 Mo. For positron emission, 93 43Tc 93 4 Mo e + γ [ Tc Mo e] The disintegration energy is Q M M m c. [ ] Q u 93.5 MeV u. 4 MeV Positron emission can reach the.35,.48, and.03 MeV states but there is insufficient energy to reach the.44 MeV state. The daughter nucleus in both forms of decay is 93 4Mo K mv, so v K m JeV ev kg ms The time for the trip is t x v m 36. s ms The number of neutrons finishing the trip is given by N N 0 e λ t. N ( ln ) tt 3 6 The fraction decaying is ( ln )(. s 64 s e e ) N % 000 by Harcourt, Inc. All rights reserved.

26 590 Chapter 44 Solutions (a) At threshold, the particles have no kinetic energy relative to each other. That is, they move like two particles which have suffered a perfectly inelastic collision. Therefore, in order to calculate the reaction threshold energy, we can use the results of a perfectly inelastic collision. Initially, the projectile M a moves with velocity v a while the target M X is at rest. We have from momentum conservation: M a v a ( M a + M X )v c The initial energy is: E i M a v a The final kinetic energy is: E f ( M a + M X )v c M a + M X M a v a M a + M X M a E ( M a + M X ) i From this, we see that E f is always less than E i and the loss in energy, E i E f, is given by M E i E f a M a + M E i M X X M a + M E i X In this problem, the energy loss is the disintegration energy Q and the initial energy is the threshold energy E th. Therefore, M Q X M a + M E th X or E th Q M X + M a M Q + M a X M X First, calculate the Q value for the reaction: Q [ M 4 N + M 4 He M7 O M H] c Q [ ]u ( 93.5 MeV u).9 MeV Then, E th Q M X + M a M.9 MeV X MeV R R 0 exp ( λ t) ln R ln R 0 λ t (the equation of a straight line) slope λ The logarithmic plot shown in Figure P44.7 is fitted by ln R t. If t is measured in minutes, then the decay constant λ is 0.6 per minute. The half-life is ln ln T λ 0. 6/min.64 min The reported half-life of 37 Ba is.55 min. The difference reflects experimental uncertainties.

Chapter 44. Nuclear Structure

Chapter 44. Nuclear Structure Chapter 44 Nuclear Structure Milestones in the Development of Nuclear Physics 1896: the birth of nuclear physics Becquerel discovered radioactivity in uranium compounds Rutherford showed the radiation