Physics exercises_solution: Chapter 44

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Physics exercises_solution: Chapter 44

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Bộ tài liệu bài tập tham khảo môn vật lý bậc đại học bằng tiếng anh Chapter 44

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Nội dung Text: Physics exercises_solution: Chapter 44

  1.  1  44.1: a) K  mc 2   1  0.1547mc 2  1  v2 c2    m  9.109  10 kg, so K  1.27  10 14 J 31 b) The total energy of each electron or positron is E  K  mc 2  1.1547mc 2  9.46  1014 J. The total energy of the electron and positron is converted into the total energy of the two photons. The initial momentum of the system in the lab frame is zero (since the equal-mass particles have equal speeds in opposite directions), so the final momentum must also be zero. The photons must have equal wavelengths and must be traveling in opposite directions. Equal λ means equal energy, so each photon has energy 9.46  10 14 J. c) E  hc λ so λ  hc E  hc (9.46  1014 J )  2.10 pm The wavelength calculated in Example 44.1 is 2.43 pm. When the particles also have kinetic energy, the energy of each photon is greater, so its wavelength is less. 44.2: The total energy of the positron is E  K  mc 2  5.00 MeV  0.511 MeV  5.51 MeV. We can calculate the speed of the positron from Eq. 37.38 2 2 mc 2 v  mc 2   0.511 MeV  E   1   E   1   5.51 MeV   0.996.    1  v2 2 c     c 44.3: Each photon gets half of the energy of the pion 1 1 1 Eγ  m c 2  (270 me )c 2  (270)(0.511 MeV)  69 MeV 2 2 2 E (6.9  10 eV)(1.6  10 19 J eV) 7  f   34  1.7  10 22 Hz h (6.63  10 J  s) c 3.00  108 m s λ   1.8  10 14 m gamma ray. f 1.7  10 Hz 22 hc hc h (6.626  10 34 J  s) 44.4: a) λ     E mμ c 2 mμ c (207)(9.11  10  31 kg)(3.00  108 m s)  1.17  10 14 m  0.0117 pm. In this case, the muons are created at rest (no kinetic energy). b) Shorter wavelengths would mean higher photon energy, and the muons would be created with non-zero kinetic energy. 44.5: a) m  m   m   270 me  207 me  63 me  E  63(0.511 MeV)  32 MeV. b) A positive muon has less mass than a positive pion, so if the decay from muon to pion was to happen, you could always find a frame where energy was not conserved. This cannot occur.
  2. 44.6: a) The energy will be the proton rest energy, 938.3 MeV, corresponding to a frequency of 2.27  10 23 Hz and a wavelength of 1.32  10 15 m. b) The energy of each photon will be 938.3 MeV  830 MeV  1768 MeV, with frequency 42.8  10 22 Hz and wavelength 7.02  10 16 m. 44.7: E  (m)c 2  (400 kg  400 kg )(3.00  10 8 m s) 2  7.20  1019 J. 44.8: 4 He 9 Be  12 C 0 n 2 4 6 1 We take the masses for these reactants from Table 43.2, and use Eq. 43.23 Q  (4.002603 u  9.012182 u  12.000000 u  1.008665 u)(931.5 MeV u )  5.701 MeV. This is an exoergic reaction. 44.9: 1 0n 10 B  7 Li 4 He 5 3 2 m(0 n 10 B)  1.008665 u  10.012937 u  11.021602 u 1 5 m(7 Li  4 He)  7.016004 u  4.002603 u  11.018607 u 3 2 m  0.002995 u; (0.002995 u)(931.5 MeV u )  2.79 MeV The mass decreases so energy is released and the reaction is exoergic. 44.10: a) The energy is so high that the total energy of each particle is half of the available energy, 50 GeV. b) Equation (44.11) is applicable, and Ea  226 MeV. qB m 2mf 44.11: a)   B  m q q 2 (2.01 u)(1.66  10 27 kg u )(9.00  10 6 Hz) B 1.60  10 19 C  B  1.18 T q 2 B 2 R 2 (1.60  10 19 C) 2 (1.18 T) 2 (0.32 m) 2 b) K    27  5.47  10 13 J 2m 2(2.01 u )(1.66  10 kg u )  3.42  106 eV  3.42 MeV 2K 2(5.47  1013 J) and v    1.81  107 m s . m (2.01 u)(1.66  10 27 kg u )  eB eBR 44.12: a) 2 f    3.97  10 7 s. b) R   3.12  10 7 m s. c) For three-  m m figure precision, the relativistic form of the kinetic energy must be used, (γ  1 )mc 2 eV  (γ  1 )mc 2 , so eV  (γ  1 )mc 2 , so V   5.11  106 V. e
  3. 44.13: a) E a2  2mc 2 ( E m  mc 2 ) Ea2  Em  2  mc 2 2mc The mass of the alpha particle is that of a 4 He atomic mass, minus two electron masses. 2 But to 3 significant figures this is just M ( 4 He)  4.00 u  (4.00 u)(0.9315 GeV u )  3.73 GeV. 2 (16.0 GeV) 2 So E m   3.73 GeV  30.6 GeV. 2(3.73 GeV) b) For colliding beams of equal mass, each has half the available energy, so each has 8.0 GeV. 1000  103 MeV 44.14: a) γ   1065.8, so v  0.999999559c. 938.3 MeV b) Nonrelativistic: eB   3.83  10 8 rad s . m Relativistic: eB 1   3.59  105 rad s . m γ Ea2 44.15: a) With E m  mc 2 , E m  Eq. (44.11). 2mc 2 [2(38.7 GeV)]2 So E m   3190 GeV. 2(0.938 GeV) b) For colliding beams the available energy E a is that of both beams. So two proton beams colliding would each need energy of 38.7 GeV to give a total of 77.4 GeV. 44.16: The available energy E a must be (mη0  2m ρ )c 2 , so Eq. (44.10) becomes (mη0  2m p ) 2 c 4  2m p c 2 ( Et  2m p c 2 ), or (mη0  2m p ) 2 c 2 Et   2m p c 2 2m p (547.3 MeV  2(938.3 MeV)) 2   2(938.3 MeV)  1254 MeV. 2(938.3 MeV) 44.17: Section 44.3 says m( Z0 )  91.2 GeV c 2 . E  91.2  109 eV  1.461  108 J; m  E c 2  1.63  10 25 kg m(Z0 ) m(p)  97.2
  4. 44.18: a) We shall assume that the kinetic energy of the 0 is negligible. In that case we can set the value of the photon’s energy equal to Q. Q  (1193  1116) MeV  77 MeV  E photon . b) The momentum of this photon is E photon (77  10 6 eV)(1.60  10 18 J eV) p   4.1  10  20 kg  m s c (3.00  10 m s) 8 To justify our original assumption, we can calculate the kinetic energy of a 0 that has this value of momentum p2 E2 (77 MeV) 2 K 0     2.7 MeV  Q  77 MeV. 2m 2mc 2 2(1116 MeV) Thus, we can ignore the momentum of the 0 without introducing a large error. 44.19: m  M (  )  m p  m 0 . Using Table (44.3):  E  (m)c 2  1189 MeV  938.3 MeV  135.0 MeV  116 MeV. 44.20: From Table (44.2), (m  me  2mv )c 2  105.2 MeV. 44.21: Conservation of lepton number. a)    e   ve  v   Lu : 1  1, Le : 0  1  1 so lepton numbers are not conserved. b) τ   e   ve  vτ  Le : 0  1  1 Lτ : 1  1 so lepton numbers are conserved. c)    e   γ. Lepton numbers are not conserved since just one lepton is produced from zero original leptons. d) n  p  e   γ e  Le : 0  1  1, so the lepton numbers are conserved. 44.22: a) Conserved: Both the neutron and proton have baryon number 1, and the electron and neutrino have baryon number 0. b) Not conserved: The initial baryon number is 1 +1 = 2 and the final baryon number is 1. c) Not conserved: The proton has baryon number 1, and the pions have baryon number 0. d) Conserved: The initial and final baryon numbers are 1+1 = 1+1+0.
  5. 44.23: Conservation of strangeness: a) K      v  . Strangeness is not conserved since there is just one strange particle, in the initial states. b) n  K   p   0 . Again there is just one strange particle so strangeness cannot be conserved. c) K   K    0   0  S : 1  1  0, so strangeness is conserved. d) p  K   0   0  S : 0  1  1  0, so strangeness is conserved. 44.24: a) Using the values of the constants from Appendix F, e2 1  7.29660475  10 3  , 4 0 c 137.050044 or 1 137 to three figures. b) From Section 38.5 e2 v1  2 ε0 h  e2   h  but notice this is just   4ε c  c as claimed  rewriting  as 2π .   0     f 2 (J  m) 44.25:    1 1  c  (J  s)(m  s ) f2 and thus is dimensionless. (Recall f 2 has units of energy times distance.) c
  6. 44.26: a) The   particle has Q  1 (as its label suggests) and S  3. Its appears as a “hole”in an otherwise regular lattice in the S  Q plane. The mass difference between each S row is around 145 MeV (or so). This puts the   mass at about the right spot. As it turns out, all the other particles on this lattice had been discovered already and it was this “hole” and mass regularity that led to an accurate prediction of the properties of the ! 2 1 1 b) See diagram. Use quark charges u   , d  , and s  as a guide. 3 3 3 2  1  1 44.27: a) uds : Q e          0; 3  3  3 1 1 1 B     1; 3 3 3 S  0  0  (1)  1 C  0  0  0  0. Q 2 2 b) cu :    0; e 3 3 1  1 B      0; 3  3  S  0  0  0; C  1  0  1. Q   1 1  3   1; B  3   1; c) ddd: e  3   3 S  3(0)  0; C  3(0)  0. Q 1   2  1  1     1; B      0; d) dc : e 3  3  3  3  S  0  0  0; C  0  (1)  1.
  7. 44.28: a) S  1 indicates the presence of one s antiquark and no s quark. To have baryon number 0 there can be only one other quark, and to have net charge + e that quark must be a u, and the quark content is us . b) The particle has an s antiquark, and for a baryon number of –1 the particle must consist of three antiquarks. For a net charge of –e, the quark content must be d d s . c) S  2 means that there are two s quarks, and for baryon number 1 there must be one more quark. For a charge of 0 the third quark must be a u quark and the quark content is uss. 44.29: a) The antiparticle must consist of the antiquarks so: n  u dd . b) So n  udd is not its own antiparticle c) ψ  cc so ψ  c c  ψ so the ψ is its own antiparticle. 44.30: (mγ  2mτ )c 2  (9460 MeV  2(1777 MeV))  5906 MeV (see Sections 44.3 and 44.4 for masses). 44.31: In   decay, 1 p   1 e  1 n  ve 0 1 0 1 1p  uud , 0 n  udd , so in   decay a u quark changes to a d quark. 1 44.32: a) Using the definition of z from Example 44.9 we have that (λ  λ s ) λ 0 1 z 1 0  . λ0 λs Now we use Eq. 44.13 to obtain cv 1 v 1  1 z   c  . cv 1 cv 1  b) Solving the above equation for  we obtain (1  z ) 2  1 1.5 2  1    0.3846. (1  z ) 2  1 1.5 2  1 Thus, v  0.3846 c  1.15  108 m s . c) We can use Eq. 44.15 to find the distance to the given galaxy, v (1.15  108 m s) r   1.6  103 Mpc H 0 (7.1  10 4 (m s) Mpc) 44.33: a) v  H 0 r  (20 (km s) Mly)(5210 Mly)  1.04  10 5 km s . λ0 cv 3.0  105 km s  1.04  105 km s b)    1.44. λs cv 3.0  105 km s  1.04  105 km s c 3.00  10 8 m s 44.34: From Eq. (44.15), r    1.5  10 4 Mly. b) This distance H 0 20(km s) Mly represents looking back in time so far that the light has not been able to reach us.
  8.  (λ λ ) 2  1  658.5 590.02  1 44.35: a) v   0 s 2  c    (2.998  10 m s) 8  ( λ 0 λ s )  1  658.5 590.0  1 2  3.280  10 m s. 7 v 3.280  10 7 m s b) r    1640 Mly. H 0 2.0  10 4 m Mly 44.36: Squaring both sides of Eq. (44.13) and multiplying by c  v gives λ 0 (c  v)  λ 2 (c  v), and solving this for v gives Eq. (44.14). 2 s 44.37: a) m  M (1 H)  M (1 H)  M ( 3 He) where atomic masses are used to balance 1 2 2 electron masses.  m  1.007825 u  2.014102 u  3.16029 u  5.898  10 3 u  E  (m)c 2  (5.898  10 3 u)(931.5 Me V u )  5.494 MeV. b) m  mn  M ( 3 He)  M ( 4 He) 2 2  1.0086649 u  3.016029 u  4.002603 u  0.022091 u  E  (m)c 2  (0.022091 u)(931.5 Me V u)  20.58 MeV. 44.38: 3m( 4 He)  m(12 C)  7.80  10 3 u, or 7.27 MeV. 44.39: m  me  mp  mn  mve so assuming mve  0, m  0.0005486 u  1.007276 u  1.008665 u  8.40  10 4 u  E  (m)c 2  (8.40  10  4 u)(931.5 Me V u)  0.783 MeV and is endoergic. 44.40: m12 C  m 4 He  m16 O  7.69  10 3 u, or 7.16 MeV, an exoergic reaction. 6 2 8 44.41: For blackbody radiation λ mT  2.90  103 m  K , so λ m1 T1  λ m 2 T2  λ m1  2.728 K (1.062  10 3 m)  9.66  10 7 m. 3000 K 44.42: a) The dimensions of  are energy times time, the dimensions of G are energy times time per mass squared, and so the dimensions of G / c 3 are 1/2 2  (E  T)(E  L M 2 )   E  T  L T  2 2              L.  (L T) 3  M   L  T   L  1 12  G  2  (6.626  10 J  s)(6.673  10 N  m kg )   34 11 2 2 b)  3        1.616  10 35 m. c   2π (3.00  108 m s)3  44.43: a) E a  2(7 TeV)  14 TeV b) Fixed target; equal mass particles, 2 Ea (1.4  10 7 MeV) 2 Em   mc  2  938.3MeV 2mc 2 2(938.3 MeV)  1.04  1011 MeV  1.04  10 5 TeV.
  9. hc hc 44.44: K  mp c 2  ,K   mp c 2  652 MeV. λ λ 44.45: The available energy must be the sum of the final rest masses: (at least) Ea  2me c 2  mπ 0 c 2  2(0.511 MeV)  135.0 MeV  136.0 MeV. Ea2 (136.0 MeV) 2 For alike target and beam particles: Eme   me c 2   0.511 2me c 2 2(0.511 MeV) MeV  1.81  10 4 MeV. So K  (1.81  10 4 MeV)  me c 2  1.81  10 4 MeV. 44.46: In Eq.(44.9), Ea  ( m0  mK 0 )c 2 , and with M  mp , m  mπ  and Em  ( mπ  )c 2  K , Ea2  (m  c 2 ) 2  (mpc 2 ) 2 K 2  (m  )c 2 2mp c (1193 MeV  497.7 MeV)2  (139.6 MeV)2  (938.3 MeV)2   139.6MeV 2(938.3 MeV)  904 MeV. 44.47: The available energy must be at least the sum of the final rest masses. Ea  (m0 )c 2  (mK  )c 2  (mK  )c 2  1116 MeV  2(493.7 MeV)  2103 MeV.Ea2  2(mp )c ((mp )c 2 ) 2  ((mK  )c 2 ) 2 . Ea2  ((mp )c 2 ) 2  ((mK  )c 2 ) 2 (2103) 2  (938.3) 2  (493.7) 2 So EK    MeV 2(mp )c 2 2(938.3)  EK   1759 MeV  (mK  )c 2  K . So the threshold energy K = 1759 MeV– 493.7 MeV=1265 MeV. 44.48: a) The decay products must be neutral, so the only possible combinations are  0 0 0 or  0   b) mη0  3m 0  142.3Me V c 2 , so the kinetic energy of the  0 mesons is 142.3 MeV. For the other reaction, K  (m 0  m 0  m   m  )c 2  133.1 MeV. 44.49: a) If the   decays, it must end in an electron and neutrinos. The rest energy of  (139.6 MeV) is shared between the electron rest energy (0.511 MeV) and kinetic energy (assuming the neutrino masses are negligible). So the energy released is 139.6 MeV – 0.511 MeV = 139.1 MeV. b) Conservation of momentum leads to the neutrinos carrying away most of the energy.
  10.  (1.054  10 34 J  s) 44.50:  19  1.5  10 22 s. E (4.4  10 eV)(1.6  10 J/eV) 6 44.51: a) E  (m)c 2  (mp )c 2  (mK  )c 2  (mK  )c 2  1019.4 MeV  2(493.7 MeV)  32.0 MeV. Each kaon gets half the energy so the kinetic energy of the K  is 16.0 MeV. b) Since the  0 mass is greater than the energy left over in part (a), it could not have been produced in addition to the kaons. c) Conservation of strangeness will not allow   K     or   K     . 44.52: a) The baryon number is 0, the charge is  e , the strangeness is 1, all lepton numbers are zero, and the particle is K  . b) The baryon number is 0, the charge is  e , the strangeness is 0, all lepton numbers are zero, and the particle is   . c) The baryon numbers is –1, the charge is 0, the strangeness is zero, all lepton numbers are 0, and the particle is an antineutron. d) The baryon number is 0, the charge is  e , the strangeness is 0, the muonic lepton number is –1, all other lepton numbers are 0, and the particle is  .  1.054  1034 J  s 44.53: t  7.6  10 21 s  E    1.39  1014 J  87 keV t 7.6  10 21 s E 0.087 MeV 2   2.8  10 5. mψ c 3097 MeV 44.54: a) The number of protons in a kilogram is  6.023  10 23 molecules mol  (1.00 kg )   3  (2 protons molecule)  6.7  10 25.   18.0  10 kg mol  Note that only the protons in the hydrogen atoms are considered as possible sources of proton decay. The energy per decay is mpc 2  938.3 MeV  1.503  10 10 J, and so the  ln (2)  10 energy deposited in a year, per kilogram, is (6.7  1025 )  1.0  1018 y (1 y) (1.50  10 J)     3 7.0  10 Gy  0.70 rad. b) For an RBE of unity, the equivalent dose is (1) (0.70 rad) = 0.70 rem.
  11. 44.55: a) E  (m)c 2  (mΞ  )c 2  (m0 )c 2  (mπ  )c 2  1321 MeV  1116 MeV  139.6 MeV  E  65 MeV. b) Using (nonrelativistic) conservation of momentum and energy: P0  0  Pf  m 0  m0 v0  mπ   vπ   vπ      v0 . m    π  Also K 0  K π   E from part (a). 1 1m 0   m0 So K 0  mπ  vπ   K 0     m0 v0  K 0 1    2 2  m  2 2  mπ      π  E 65 MeV  K 0    7.2 MeV m0 1116 Me V 1 1 m 139.6 MeV π  K π   65  7.2 MeV  57.8 MeV. 7.2 So the fractions of energy carried off by the particles are  0.11 for the 0 and 65 0.89 for the   . dR dR dt HR 44.56: a) For this model,  HR, so   H , presumed to be the same for dt R R dr dR all points on the surface. b) For constant  ,    HR  Hr. c) See part (a), dt dt dR dt dR H0  . d) The equation  H 0 R is a differential equation, the solution to R dt which, for constant H 0 is R(t )  R0e H 0 t , where R0 is the value of R at t  0 . This equation dR dt d may be solved by separation of variables, as  ln ( R)  H 0 and integrating both R dt sides with respect to time. e) A constant H 0 would mean a constant critical density, which is inconsistent with uniform expansion.
  12. r 44.57: From Pr.(44.56): r  R  R  .  dR 1 dr r dθ 1 dr dθ So   2  since  0. dt θ dt θ dt θ dt dt 1 dR 1 dr 1 dr dr  1 dR  So   v   r  H 0 r. R dt Rθ dt r dt dt  R dt  dv d  r dR  d  dR  Now 0   θ  dθ dθ  R dt  dθ  dt  dR   K where K is a constant. dt dR K K d    R   t since  0. dt    dt 1 dR  K 1  H0    . R dt Kt  t 1 So the current value of the Hubble constant is where T is the present age of the T universe.
  13. v0  vcm 44.58: a) For mass m, in Eq. (37.23) u  vcm , v  v0 , and so vm  . 1  v0 vcm c 2 For mass M , u  vcm , v   0, so v M  vcm . b) The condition for no net momentum in the center of mass frame is mγm vm  MγM vM  0, where γm and γM correspond to the velocities found in part (a). The algebra reduces to  m γm  (  0   )γ0 γM , where v v  0  0 ,     cm , and the condition for no net momentum becomes c c m(  0   ) γ 0 γ M  0 m mv0 M γM , or     0 , and vcm  . M m  M 1  0 2 m  M 1  (v0 / c) 2 1 mγ0 c) Substitution of the above expression into the expressions for the velocities found in part (a) gives the relatively simple forms M m vm  v0 γ0 ,vM  v0 γ0 . m  M γ0 mγ0  M After some more algebra, m  Mγ M  mγ 0 γm  0 , γM  , from which m  M  2mM γ 0 2 2 m  M 2  2mM γ 0 2 m m  Mγ M  m 2  M 2  2mMγ 0 . This last expression, multiplied by c 2 , is the available energy E a in the center of mass frame, so that Ea2  (m 2  M 2  2mMγ0 )c 4  (mc 2 )2  (Mc 2 )2  ( 2Mc 2 )(mγ0c 2 )  (mc 2 ) 2  ( Mc 2 ) 2  2Mc 2 Em , which is Eq. (44.9).
  14. 44.59: 0  n   0 a) E  (m)c 2  (m0 )c 2  (mn )c 2 (m 0 )c 2  1116 MeV  939.6 MeV  135.0 MeV  41.4 MeV. b) Using conservation of momentum and kinetic energy; we know that the momentum of the neutron and pion must have the same magnitude, p n  p K n  En  mn c 2  (mn c 2 ) 2  ( pn c) 2  mn c 2  (mn c 2 ) 2  ( pn c) 2  mn c 2  (mn c 2 ) 2  K2  2m c 2 K   mn c 2  K   K n  K   (mn c 2 ) 2  K2  2m c 2 K   mn c 2  E. ( mn c 2 ) 2  K 2  2m c 2 K   E 2  (mn c 2 ) 2  K 2  2 Em c 2  2 EK   2mn c 2 K  . Collecting terms we find : K  (2m c 2  2 E  2mn c 2 )  E 2  2 Emn c 2 (41.4 MeV) 2  2(41.4 MeV)(939.6 MeV)  K  . 2(135.0 MeV)  2(41.4 MeV)  2(939.6 MeV)  K   35.62 MeV. 35.62 So the fractional energy carried by the pion is  0.86, and that of the neutron is 0.14. 41.4
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