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Homework 16 Ch30: P 1, 9, 29, 39, 45; Ch31: P 21, 31; Ch32: P 1, 9, 11 Problems (Ch30): 1. (I) A pi meson has a mass of 139 Me V c 2 . What is this in atomic mass units? Solution 139 MeV c 2 m We convert the units: 9. (I) Estimate the total binding energy for 931.5MeV uc 40 20 Ca, 2 0.149 u. using Fig. 30–1. Solution From Figure 30–1, we see that the average binding energy per nucleon at A 40 is 8.6MeV. Thus the total binding energy for 40 408.6MeV 340MeV. Ca is 29. (II) How much energy is released in electron capture by beryllium: 7 4 Be 0 7 1 e 3 Li v? Solution For the electron capture 74 Be 01 e 73 Li v, we see that if we add three electron masses to both sides to use the atomic mass for Li, we use the captured electron for the atomic mass of Be. We find the Q value from Q m 7 Be m 7 Li c2 7.016928u 7.016003u c 2 931.5MeV uc 2 0.861MeV. 39. (I) What fraction of a sample of after 3.0 yr? 68 32 Ge, whose half-life is about 9 months, will remain Solution We find the fraction remaining from N N0et ; 0.693 3.0 yr 12mo yr 9mo N e t e N0 0.0625. 45. (II) Calculate the mass of a sample of pure 40 2.0 105 s 1 . The half-life of 19 K is 1.28109 yr. Solution N 2.0 10 5 s 1 t T 1 1.28 10 9 yr 40 19 K N N t ln 2 T1 2 40 g / mol with an initial decay rate of N T12 N t ln 2 2 m? We find the number of nuclei 9 7 5 1 1.28 10 yr 3.16 10 s / yr N 2.0 10 s 1.7 10 22 nuclei ln 2 The mass is m 1.17 10 22 nuclei 40.0g / mol 6.02 1023 nuclei / mol 0.77 g. Problems (Ch31): 21. (II) How much mass of 238 92 U is required to produce the same amount of energy as burning 1.0 kg of coal (about 3 10 7 J )? Solution We assume that an average of 200 MeV is released per fission of a uranium nucleus (see Eq31-5). 0.238 kg 1MeV 1nucleus 7 238 3 107 J 1.601eV 3.7 10 kg 92 U 19 6 23 10 J 110 eV 200 MeV 6.02 10 nuclei 31. (II) If a typical house requires 950 W of electric power on average, what minimum amount of deuterium fuel would have to be used in a year to supply these electrical needs? Assume the reaction of Eq. 31–8b. Solution The reaction of Eq. 31-8b consumes 2 deuterons and releases 3.27 MeV of energy. The atomic mass of Deuterium is 2.01g. The amount of energy needed is the power times the elapsed time, and the energy can be related to the mass of deuterium by the reaction. 3 J 1eV 1MeV 2 d 2.01 10 kg 7 s 950 1 yr 3.156 10 s yr 1.60 1019 J 106 eV 3.27 MeV 6.02 1023 d 3.8 104 kg 0.38g Problems (Ch32): 1. (I) What is the total energy of a proton whose kinetic energy is 6.35 GeV? Solution The total energy is given by Eq. 26-7a. E m0 c KE 0.94 GeV 6.35GeV 7.29 GeV 2 9. (II) What is the wavelength ( maximum resolvable distance) of 7.0-TeV protons? Solution Because the energy of the protons is much greater than their rest mass, we have KE E pc . Combine that with Eq. 27-8 for the de Broglie wavelength. E pc ; p h E hc hc E 6.63 10 J s 3.0 10 m s 1.8 10 7.0 10 eV 1.60 10 J eV 34 8 19 12 19 m 11. (II) The 4.25-km-radius tunnel that will be used to house the magnets for the Large Hadron Collider (LHC) calls for proton beams of energy 7.0 TeV. What magnetic field will be required? Solution Because the energy of the protons is much greater than their rest mass, we have KE E pc . A relationship for the magnetic field is given in section 32-1. qBr E v mv qBr p qBr qBr m c 7.0 10 eV 1.60 10 J eV B 5.5 T qrc 1.60 10 C 4.25 10 m 3.00 10 m s 19 15 E 19 3 8