http://www.elsolucionario.blogspot.com LIBROS UNIVERISTARIOS Y SOLUCIONARIOS DE MUCHOS DE ESTOS LIBROS LOS SOLUCIONARIOS CONTIENEN TODOS LOS EJERCICIOS DEL LIBRO RESUELTOS Y EXPLICADOS DE FORMA CLARA VISITANOS PARA DESARGALOS GRATIS Chapter Solutions *1.1 With V = (base area) · (height) V = π r2 · h and ρ = m , we have V ρ= m kg 10 mm = π r2 h π (19.5 mm)2 39.0 mm m3 ρ = 2.15 × 104 kg/m3 1.2 ρ= ρ= 1.3 M M = V πR3 3(5.64 × 1026 kg) = 623 kg/m3 π (6.00 × 107 m) VCu = V0 − Vi = VCu = 3 π (ro – ri ) π [(5.75 cm)3 – (5.70 cm)3] = 20.6 cm3 5.7cm 0.05 cm ρ=m V m = ρV = (8.92 g/cm3)(20.6 cm3) = 184 g 1.4 V = Vo – Vi = 3 π (r2 – r1 ) 3 ρ= *1.5 (a) πρ (r2 – r1 ) m 3 , so m = ρV = ρ π (r2 – r1) = V 3 The number of moles is n = m/M, and the density is ρ = m/V Noting that we have mole, V1 mol = mFe nFe MFe (1 mol)(55.8 g/mol) = = = 7.10 cm3 ρFe ρFe 7.86 g/cm3 © 2000 by Harcourt College Publishers All rights reserved Chapter Solutions (b) In mole of iron are NA atoms: V1 atom = V1 mol 7.10 cm3 = = 1.18 × 10–23 cm3 NA 6.02 × 1023 atoms/mol = 1.18 × 10-29 m3 1.18 × 10–29 m3 = 2.28 × 10–10 m = 0.228 nm (c) datom = (d) V1 mol U = (1 mol)(238 g/mol) = 12.7 cm3 18.7 g/cm3 V1 atom U = V1 mol U 12.7 cm3 = = 2.11 × 10–23 cm3 NA 6.02 × 1023 atoms/mol = 2.11 × 10-29 m3 datom U = *1.6 1.7 r2 = r1 3 V1 atom U = 2.11 × 10–29 m3 = 2.77 × 10–10 m = 0.277 nm = (4.50 cm)(1.71) = 7.69 cm Use m = molar mass/NA and u = 1.66 × 10-24 g 4.00 g/mol = 6.64 × 10-24 g = 4.00 u 6.02 × 1023 mol-1 (a) For He, m = (b) For Fe, m = 55.9 g/mol = 9.29 × 10-23 g = 55.9 u 6.02 × 1023 mol-1 (c) For Pb, m = 207 g/mol -22 g = 207 u 23 -1 = 3.44 × 10 6.02 × 10 mol © 2000 by Harcourt College Publishers All rights reserved Chapter Solutions Goal Solution Calculate the mass of an atom of (a) helium, (b) iron, and (c) lead Give your answers in atomic mass units and in grams The molar masses are 4.00, 55.9, and 207 g/mol, respectively, for the atoms given Gather information: The mass of an atom of any element is essentially the mass of the protons and neutrons that make up its nucleus since the mass of the electrons is negligible (less than a 0.05% contribution) Since most atoms have about the same number of neutrons as protons, the atomic mass is approximately double the atomic number (the number of protons) We should also expect that the mass of a single atom is a very small fraction of a gram (~10–23 g) since one mole (6.02 × 1023) of atoms has a mass on the order of several grams Organize: An atomic mass unit is defined as 1/12 of the mass of a carbon-12 atom (which has a molar mass of 12.0 g/mol), so the mass of any atom in atomic mass units is simply the numerical value of the molar mass The mass in grams can be found by multiplying the molar mass by the mass of one atomic mass unit (u): u = 1.66 × 10–24 g Analyze: For He, m = 4.00 u = (4.00 u)(1.66 × 10–24 g/u) = 6.64 × 10–24 g For Fe, m = 55.9 u = (55.9 u)(1.66 × 10–24g/u) = 9.28 × 10–23 g For Pb, m = 207 u = (207 u)(1.66 × 10–24 g/u) = 3.44 × 10–22 g Learn: As expected, the mass of the atoms is larger for bigger atomic numbers If we did not know the conversion factor for atomic mass units, we could use the mass of a proton as a close approximation: 1u ≈ mp = 1.67 × 10–24 g *1.8 ∆n = ∆m 3.80 g – 3.35 g = = 0.00228 mol M 197 g/mol ∆N = (∆n)NA = (0.00228 mol)(6.02 × 1023 atoms/mol) = 1.38 × 1021 atoms ∆t = (50.0 yr)(365 d/yr)(24.0 hr/d)(3600 s/hr) = 1.58 × 109 s ∆N 1.38 × 1021 atoms = = 8.72 × 1011 atoms/s ∆t 1.58 × 109 s 1.9 (a) m = ρL3 = (7.86 g/cm3)(5.00 × 10-6 cm)3 = 9.83 × 10-16 g (b) N=m NA = (9.83 × 10-16 g)(6.02 × 1023 atoms/mol) Molar mass 55.9 g/mol = 1.06 × 107 atoms © 2000 by Harcourt College Publishers All rights reserved 1.10 Chapter Solutions (a) The cross-sectional area is 15.0 cm A = 2(0.150 m)(0.010 m) + (0.340 m)(0.010 m) 1.00 cm = 6.40 × 10-3 m2 36.0 cm The volume of the beam is V = AL = (6.40 × 10-3 m2)(1.50 m) = 9.60 × 10-3 m3 1.00 cm Thus, its mass is m = ρV = (7.56 × 103 kg/m3)(9.60 × 10-3 m3) = 72.6 kg (b) Presuming that most of the atoms are of iron, we estimate the molar mass as M = 55.9 g/mol = 55.9 × 10-3 kg/mol The number of moles is then n= m 72.6 kg = = 1.30 × 103 mol M 55.9 × 10-3 kg/mol The number of atoms is N = nNA = (1.30 × 103 mol)(6.02 × 1023 atoms/mol) = 7.82 × 1026 atoms *1.11 (a) n= m 1.20 × 103 g = = 66.7 mol, and M 18.0 g/mol 23 Npail = nNA = (66.7 mol)(6.02 × 10 molecules/mol) = 4.01 × 1025 molecules (b) Suppose that enough time has elapsed for thorough mixing of the hydrosphere Nboth = Npail mpail = (4.01 × 1025 molecules) 1.20 kg , or Mtotal 1.32 × 1021 kg Nboth = 3.65 × 104 molecules 1.12 r, a, b, c and s all have units of L (s – a)(s – b)(s – c) s = L×L×L = L L2 = L Thus, the equation is dimensionally consistent © 2000 by Harcourt College Publishers All rights reserved Chapter Solutions 1.13 The term s has dimensions of L, a has dimensions of LT -2, and t has dimensions of T Therefore, the equation, s = kamtn has dimensions of L = (LT- ) (T) m n or m n-2m L T =L T The powers of L and T must be the same on each side of the equation Therefore, L1 = Lm and m = Likewise, equating terms in T, we see that n – 2m must equal Thus, n = 2m = The value of k, a dimensionless constant, cannot be obtained by dimensional analysis 1.14 2π g 1.15 (a) This is incorrect since the units of [ax] are m2/s2, while the units of [v] are m/s (b) This is correct since the units of [y] are m, and cos(kx) is dimensionless if [k] is in m-1 1.16 l = L L/T = T2 = T Inserting the proper units for everything except G, kg m = G[kg]2 s2 [m]2 Multiply both sides by [m]2 and divide by [kg]2; the units of G are m3 kg · s2 1.17 One month is mo = (30 day)(24 hr/day)(3600 s/hr) = 2.592 × 106 s Applying units to the equation, V = (1.50 Mft3/mo)t + (0.00800 Mft3/mo2)t2 Since Mft3 = 106 ft3, V = (1.50 × 106 ft3/mo)t + (0.00800 × 106 ft3/mo2)t2 © 2000 by Harcourt College Publishers All rights reserved Chapter Solutions Converting months to seconds, V= 1.50 × 106 ft3/mo 0.00800 × 106 ft3/mo2 t + t 2.592 × 106 s/mo (2.592 × 106 s/mo)2 Thus, V[ft3] = (0.579 ft3/s)t + (1.19 × 10 -9 ft3/s 2)t2 *1.18 Apply the following conversion factors: in = 2.54 cm, d = 86400 s, 100 cm = m, and 109 nm = m -2 in/day (2.54 cm/in)(10 m/cm)(10 nm/m) = 9.19 nm/s 86400 s/day 32 This means the proteins are assembled at a rate of many layers of atoms each second! 1.19 Area A = (100 ft)(150 ft) = 1.50 × 104 ft2, so A = (1.50 × 104 ft2)(9.29 × 10-2 m2/ft2) = 1.39 × 103 m2 Goal Solution A rectangular building lot is 100 ft by 150 ft Determine the area of this lot in m2 G: We must calculate the area and convert units Since a meter is about feet, we should expect the area to be about A ≈ (30 m)(50 m) = 500 m2 O: Area = Length × Width Use the conversion: m = 3.281 ft A: A = L × W = (100 ft) 1m 1m (150 ft ) = 390 m2 3.281 ft 3.281 ft L: Our calculated result agrees reasonably well with our initial estimate and has the proper units of m2 Unit conversion is a common technique that is applied to many problems 1.20 (a) V = (40.0 m)(20.0 m)(12.0 m) = 9.60 × 103 m3 V = 9.60 × 103 m3 (3.28 ft/1 m)3 = 3.39 × 10 ft3 © 2000 by Harcourt College Publishers All rights reserved Chapter Solutions (b) The mass of the air is m = ρairV = (1.20 kg/m3)(9.60 × 103 m3) = 1.15 × 104 kg The student must look up weight in the index to find Fg = mg = (1.15 × 104 kg)(9.80 m/s2) = 1.13 × 105 N Converting to pounds, Fg = (1.13 × 105 N)(1 lb/4.45 N) = 2.54 × 104 lb *1.21 (a) Seven minutes is 420 seconds, so the rate is r= 30.0 gal = 7.14 × 10-2 gal/s 420 s Converting gallons first to liters, then to m3, (b) r = 7.14 × 10 -2 gal 3.786 L 10-3 m3 s gal L r = 2.70 × 10-4 m3/s At that rate, to fill a 1-m3 tank would take (c) m3 hr = 1.03 hr 2.70 × 10 m /s 3600 s t= 1.22 v = 5.00 furlongs 220 yd 0.9144 m fortnight day hr fortnight 1 furlong yd 14 days 24 hrs 3600 s = 8.32 × 10-4 m/s This speed is almost mm/s; so we might guess the creature was a snail, or perhaps a sloth 1.23 It is often useful to remember that the 1600-m race at track and field events is approximately mile in length To be precise, there are 1609 meters in a mile Thus, acre is equal in area to mi2 1609 m = 4.05 × 103 m2 640 acres mi (1 acre) © 2000 by Harcourt College Publishers All rights reserved Chapter Solutions 1.24 Volume of cube = L3 = quart (Where L = length of one side of the cube.) Thus, gallon 3.786 liters 1000 cm3 L3 = (1 quart) = 946 cm3, and quarts gallon liter L = 9.82 cm 1.25 The mass and volume, in SI units, are m = (23.94 g) kg = 0.02394 kg 1000 g V = (2.10 cm3)(10-2 m/cm)3 = 2.10 × 10-6 m3 Thus, the density is ρ= m 0.02394 kg = = 1.14 × 104 kg/m3 V 2.10 × 10-6 m Goal Solution A solid piece of lead has a mass of 23.94 g and a volume of 2.10 cm3 From these data, calculate the density of lead in SI units (kg/m3) G: From Table 1.5, the density of lead is 1.13 × 104 kg/m3, so we should expect our calculated value to be close to this number This density value tells us that lead is about 11 times denser than water, which agrees with our experience that lead sinks m O: Density is defined as mass per volume, in ρ = We must convert to SI units in the calculation V A: ρ = 23.94 g kg 100 cm = 1.14 × 104 kg/m3 2.10 cm3 1000 g m L: At one step in the calculation, we note that one million cubic centimeters make one cubic meter Our result is indeed close to the expected value Since the last reported significant digit is not certain, the difference in the two values is probably due to measurement uncertainty and should not be a concern One important common-sense check on density values is that objects which sink in water must have a density greater than g/cm3, and objects that float must be less dense than water © 2000 by Harcourt College Publishers All rights reserved Chapter Solutions 1.26 (a) We take information from Table 1.1: LY = (9.46 × 1015 m) AU = 6.31 × 104 AU 1.50 × 1011 m (b) The distance to the Andromeda galaxy is × 1022 m = (2 × 1022 m) AU = 1.33 × 1011 AU 11 1.50 × 10 m mSun 1.99 × 1030 kg = = 1.19 × 1057 atoms matom 1.67 × 10-27 kg 1.27 Natoms = 1.28 mi = 1609 m = 1.609 km; thus, to go from mph to km/h, multiply by 1.609 1.29 (a) mi/h = 1.609 km/h (b) 55 mi/h = 88.5 km/h (c) 65 mi/h = 104.6 km/h Thus, ∆v = 16.1 km/h (a) × 10 12 $ hr day yr = 190 years 1000 $/s 3600 s 24 hr 365 days (b) The circumference of the Earth at the equator is 2π (6378 × 103 m) = 4.01 × 107 m The length of one dollar bill is 0.155 m so that the length of trillion bills is 9.30 × 1011 m Thus, the trillion dollars would encircle the Earth 9.30 × 1011 m = 2.32 × 104 times 4.01 × 107 m Goal Solution At the time of this book’s printing, the U.S national debt is about $6 trillion (a) If payments were made at the rate of $1 000 per second, how many years would it take to pay off a $6-trillion debt, assuming no interest were charged? (b) A dollar bill is about 15.5 cm long If six trillion dollar bills were laid end to end around the Earth’s equator, how many times would they encircle the Earth? Take the radius of the Earth at the equator to be 378 km (Note: Before doing any of these calculations, try to guess at the answers You may be very surprised.) (a) G: $6 trillion is certainly a large amount of money, so even at a rate of $1000/second, we might guess that it will take a lifetime (~ 100 years) to pay off the debt O: Time to repay the debt will be calculated by dividing the total debt by the rate at which it is repaid © 2000 by Harcourt College Publishers All rights reserved Chapter 46 Solutions 46.21 (a) Λ0 → p + π – Strangeness: –1 → + (b) π – + p → Λ0 + K Strangeness: + → –1 + (0 = and strangeness is conserved ) (c) – – p + p → Λ + Λ0 Strangeness: + → +1 – (0 = and strangeness is conserved ) (strangeness is not conserved ) (d) π – + p → π – + Σ + Strangeness: + → – (0 ≠ –1: strangeness is not conserved ) Ξ – → Λ0 + π – Strangeness: –2 → –1 + (–2 ≠ –1 so strangeness is not conserved Strangeness: –2 → + (–2 ≠ so strangeness is not conserved ) (e) ) 46.22 (f) Ξ0 → p + π – (a) µ – → e– + γ Le : → + 0, (b) n → p + e– + ν e Le : → + + (c) Λ0 → p + π Strangeness: –1 → + 0, (d) p → e+ + π (e) *46.23 (a) Ξ0 → n + π0 and Lµ : → and charge: → +1 + Baryon number: +1 → + Strangeness: –2 → + π − + p → η violates conservation of baryon number as + → not allowed (b) K − + n → Λ0 + π − Baryon number = + → + Charge = –1 + → – Strangeness, – + → –1 + Lepton number, → The interaction may occur via the strong interaction since all are conserved (c) K− → π − + π Strangeness, –1 → + Baryon number, → Lepton number, → Charge, –1 → –1 + Strangeness is violated by one unit, but everything else is conserved Thus, the reaction can occur via the weak interaction , but not the strong or electromagnetic interaction (d) Ω − → Ξ− + π Baryon number, → + Lepton number, → Charge, –1 → –1 + Strangeness, –3 → –2 + May occur by weak interaction , but not by strong or electromagnetic (e) η → 2γ Baryon number, → Lepton number, → Charge, → Strangeness, → © 2000 by Harcourt, Inc All rights reserved 10 Chapter 46 Solutions No conservation laws are violated, but photons are the mediators of the electromagnetic interaction Also, the lifetime of the η is consistent with the electromagnetic interaction *46.24 (a) Ξ− → Λ0 + µ − + ν µ Baryon number: + → + + + Le : → + + Charge: –1 → – + Lµ : → + + Lτ : → + + Strangeness: –2 → –1 + + Conserved quantities are: (b) K 0S → 2π Baryon number: → Le : → Charge: → Lµ : → Lτ : → Strangeness: +1 → Conserved quantities are: (c) Charge: –1 + → + Lµ : + → + Lτ : + → + Strangeness: –1 + → –1 + Charge: → Lµ : → + Lτ : → + Strangeness: –1 → –1 + B, S, charge , Le , Lµ , and Lτ e+ + e− → µ + + µ − Baryon number: + → + Le : –1 + → + Charge: +1 – → +1 – Lµ : + → + – Lτ : + → + Strangeness: + → + Conserved quantities are: (f) S, charge , Le , Lµ , and Lτ Σ → Λ0 + γ Baryon number: + → + Le : → + Conserved quantities are: (e) B, charge , Le , Lµ , and Lτ K− + p → Σ0 + n Baryon number: + → + Le : + → + Conserved quantities are: (d) B, charge , Le , and Lτ B, S, charge , Le , Lµ , and Lτ p + n → Λ + Σ− Baryon number: –1 + → –1 + Le : + → + Charge: –1 + → – Lµ : + → + Lτ : + → + Strangeness: + → +1 – Conserved quantities are: B, S, charge , Le , Lµ , and Lτ Chapter 46 Solutions *46.25 (a) K+ + p → ? + p The strong interaction conserves everything +1→ B +1 so Baryon number, +1 + → Q + so Charge, so Lepton numbers, + → L + +1 + → S + so Strangeness, 11 B=0 Q = +1 Le = Lµ = Lτ = S=1 The conclusion is that the particle must be positively charged, a non-baryon, with strangeness of +1 Of particles in Table 46.2, it can only be the K + Thus, this is an elastic scattering process The weak interaction conserves all but strangeness, and ∆S = ±1 (b) Ω− → ? + π − Baryon number, Charge, Lepton numbers, Strangeness, +1 → B + −1 → Q − 0→L+0 −3 → S + so so so so ∆S = 1: B=1 Q=0 Le = Lµ = L τ = S = −2 The particle must be a neutral baryon with strangeness of –2 Thus, it is the Ξ (c) K+ → ? + µ+ + νµ : Baryon number, 0→B+0+0 +1 → Q + + Charge, Lepton Numbers Le , → Le + + Lµ , → Lµ − + Lτ , → Lτ + + 1→ S + + Strangeness: B=0 Q=0 Le = Lµ = Lτ = ∆S = ±1 (for weak interaction): S = so so so so so so The particle must be a neutral meson with strangeness = ⇒ π *46.26 (a) strangeness baryon number charge proton e u 1/3 2e/3 u 1/3 2e/3 d 1/3 –e/3 total e strangeness baryon number charge neutron u 1/3 2e/3 d 1/3 –e/3 d 1/3 –e/3 total (b) *46.27 (a) The number of protons 6.02 × 10 23 molecules 10 protons 26 N p = 1000 g molecule = 3.34 × 10 18.0 g © 2000 by Harcourt, Inc All rights reserved protons 12 Chapter 46 Solutions 6.02 × 10 23 molecules neutrons 26 N n = (1000 g ) molecule = 2.68 × 10 18.0 g and there are So there are for electric neutrality (b) neutrons 3.34 × 1026 electrons The up quarks have number × 3.34 × 1026 + 2.68 × 1026 = 9.36 × 1026 up quarks and there are × 2.68 × 1026 + 3.34 × 1026 = 8.70 × 1026 down quarks Model yourself as 65 kg of water Then you contain 65 × 3.34 × 1026 ~ 1028 electrons 65 × 9.36 × 1026 ~ 1029 up quarks 65 × 8.70 × 1026 ~ 1029 down quarks Only these fundamental particles form your body You have no strangeness, charm, topness or bottomness 46.28 strangeness baryon number charge K0 0 d 1/3 –e/3 s –1/3 e/3 total 0 strangeness baryon number charge Λ0 –1 u 1/3 2e/3 d 1/3 –e/3 s –1 1/3 –e/3 (a) (b) 46.29 total –1 Quark composition of proton = uud and of neutron = udd Thus, if we neglect binding energies, we may write m p = 2m u + m d (1) and m n = m u + 2m d (2) Solving simultaneously, we find 1 [ ] m u = (2m p – m n) = 2(938.3 MeV / c ) − 939.6 MeV / c = 312 MeV/c2 and from either (1) or (2), m d = 314 MeV/c *46.30 In the first reaction, π − + p → K + Λ0 , the quarks in the particles are: ud + uud → ds + uds There is a net of up quark both before and after the reaction, a net of down quarks both Chapter 46 Solutions 13 before and after, and a net of zero strange quarks both before and after Thus, the reaction conserves the net number of each type of quark In the second reaction, π − + p → K + n , the quarks in the particles are: ud + uud → ds + udd In this case, there is a net of up and down quarks before the reaction but a net of up, down, and anti-strange quark after the reaction Thus, the reaction does not conserve the net number of each type of quark 46.31 (a) π − + p → Κ + Λ0 In terms of constituent quarks: ud + uud → ds + uds – + → + 1, or → 1 + → + 1, or → + → –1 + 1, or → up quarks: down quarks: strange quarks: (b) π + + p → Κ+ + Σ+ ud + uud → us + uus ⇒ + → + 2, or → –1 + → + 0, or → 0 + → –1 + 1, or → up quarks: down quarks: strange quarks: (c) Κ − + p → Κ + + Κ + Ω− us + uud → us + ds + sss ⇒ –1 + → + + 0, or → + → + + 0, or → 1 + → –1 – + 3, or → up quarks: down quarks: strange quarks: (d) p + p → K0 + p + π + + ? uud + uud → ds + uud + ud + ? ⇒ The quark combination of ? must be such as to balance the last equation for up, down, and strange quarks up quarks: down quarks: strange quarks: 2+2=0+2+1+? 1+1=1+1–1+? + = –1 + + + ? (has u quark) (has d quark) (has s quark) quark composite = uds = Λ0 or Σ 46.32 Σ0 + p → Σ+ + γ + X dds + uud → uds + + ? The left side has a net 3d, 2u and 1s The right-hand side has 1d, 1u, and 1s leaving 2d and 1u missing The unknown particle is a neutron, udd Baryon and strangeness numbers are conserved © 2000 by Harcourt, Inc All rights reserved 14 Chapter 46 Solutions Compare the given quark states to the entries in Tables 46.4 and 46.5 *46.33 (a) suu = Σ + (b) ud = π − (c) sd = Κ (d) ssd = Ξ − *46.34 (a) (b) ) ( ) ( ) ( ) ( ) ( ) u d d : charge = − 23 e + 31 e + 31 e = This is the antineutron Section 39.4 says *46.35 ( u ud : charge = − 23 e + − 23 e + 31 e = −e This is the antiproton f observer = f source + va c − va c The velocity of approach, va , is the negative of the velocity of mutual recession: va = −v Then, and λ′ = λ 1+ v c 1− v c (1.7 × 10–2 m/s) ly v = HR (Equation 46.7) H= (a) v (2.00 × 106 ly) = 3.4 × 104 m/s λ' = λ (b) v (2.00 × 108 ly) = 3.4 × 106 m/s λ' = 590 + 0.01133 = 597 nm − 0.01133 (c) v (2.00 × 109 ly) = 3.4 × 107 m/s λ' = 590 + 0.1133 = 661 nm − 0.1133 (a) λ' 650 nm = 434 nm = 1.50 = λ + v/c – v/c = 2.24 46.36 46.37 c c 1− v c = λ′ λ + v c v = 0.383c, (b) Equation 46.7, v = HR + v/c − v/c + v/c = 590(1.0001133) = 590.07 nm − v/c 38.3% the speed of light v (0.383)(3.00 × 108) R= H = = 6.76 × 109 light years (1.7 × 10–2) Chapter 46 Solutions 15 Goal Solution A distant quasar is moving away from Earth at such high speed that the blue 434-nm hydrogen line is observed at 650 nm, in the red portion of the spectrum (a) How fast is the quasar receding? You may use the result of Problem 35 (b) Using Hubble's law, determine the distance from Earth to this quasar G: The problem states that the quasar is moving very fast, and since there is a significant red shift of the light, the quasar must be moving away from Earth at a relativistic speed (v > 0.1c) Quasars are very distant astronomical objects, and since our universe is estimated to be about 15 billion years old, we should expect this quasar to be ~10 light-years away O: As suggested, we can use the equation in Problem 35 to find the speed of the quasar from the Doppler red shift, and this speed can then be used to find the distance using Hubble’s law A: (a) 1+ v c λ ′ 650 nm = = 1.498 = λ 434 nm 1− v c Therefore, v = 0.383c or or squared, 1+ v c = 2.243 1− v c 38.3% the speed of light (b) Hubble’s law asserts that the universe is expanding at a constant rate so that the speeds of galaxies are proportional to their distance R from Earth, v = HR so, L: ( ) v (0.383) 3.00 × 10 m / s R= = = 6.76 × 10 ly H 1.70 × 10 -2 m / s ⋅ ly ( ) The speed and distance of this quasar are consistent with our predictions It appears that this quasar is quite far from Earth but not the most distant object in the visible universe *46.38 (a) λ ′n = λ n 1+ v/c = (Z + 1)λ n 1− v/c + v/c – v/c = ( Z + 1) v v + c = ( Z + 1)2 – c ( Z + 1) v ( Z + Z + 2) = Z + Z c Z2 + Z v = c Z + Z + 2 (b) v c Z2 + Z R=H = H Z + Z + 2 © 2000 by Harcourt, Inc All rights reserved 16 Chapter 46 Solutions ( ) The density of the Universe is ρ = 1.20ρc = 1.20 3H π G *46.39 Consider a remote galaxy at distance r The mass interior to the sphere below it is 4π r 3H π r 0.600H r M = ρ = 1.20 = G 8π G both now and in the future when it has slowed to rest from its current speed v = Hr The energy of this galaxy is constant as it moves to apogee distance R: mv 2 − GmM GmM =0− r R so mH r 2 r R so R = 6.00 r − 0.100 = − 0.600 The Universe will expand by a factor of 6.00 *46.40 (a) kBT ≈ 2mp c (b) kBT ≈ 2me c *46.41 (a) Wien’s law: Thus, (b) *46.42 (a) (b) T≈ so T≈ mp c kB = − Gm 0.600H r Gm 0.600H r = − R r G G from its current dimensions (938.3 MeV ) 1.60 × 10 −13 1.38 × 10 −23 J K MeV ( ) (0.511 MeV ) 1.60 × 10 −13 2me c = kB 1.38 × 10 −23 J K MeV ( ) J 13 ~ 10 K J λ maxT = 2.898 × 10 − m ⋅ K λ max = 2.898 × 10 − m ⋅ K 2.898 × 10 − m ⋅ K = = 1.06 × 10 − m = 1.06 mm T 2.73 K This is a microwave hG L= = c3 ~ 1010 K (1.055 × 10 −34 )( J ⋅ s 6.67 × 10 −11 N ⋅ m kg (3.00 × 10 This time is given as T = ms ) )= 1.61 × 10 − 35 m L 1.61 × 10 −35 m = = 5.38 × 10 − 44 s , c 3.00 × 108 m s which is approximately equal to the ultra-hot epoch Chapter 46 Solutions 46.43 (a) ∆E ∆t ≈ h, and ∆E ≈ m= (b) *46.44 (a) 46.45 ∆t ≈ 17 r 1.4 × 10 −15 m = = 4.7 × 10 −24 s c 3.0 × 108 m s h 1.055 × 10 −34 J ⋅ s = = 2.3 × 10 −11 ∆t 4.7 × 10 −24 s MeV J = 1.4 × 10 MeV 1.60 × 10 −13 J ∆E ≈ 1.4 × 10 MeV c ~ 10 MeV c c2 From Table 46.2, mπ c = 139.6 MeV , a pi-meson π – + p → Σ + + π is forbidden by charge conservation (b) µ – → π – + νe is forbidden by energy conservation (c) p → π + + π + + π – is forbidden by baryon number conservation The total energy in neutrinos emitted per second by the Sun is: (0.4)(4π)(1.5 × 1011)2 = 1.1 × 1023 W Over 109 years, the Sun emits 3.6 × 1039 J in neutrinos This represents an annihilated mass mc = 3.6 × 1039 J m = 4.0 × 1022 kg About part in 50,000,000 of the Sun's mass, over 109 years, has been lost to neutrinos © 2000 by Harcourt, Inc All rights reserved 18 Chapter 46 Solutions Goal Solution The energy flux carried by neutrinos from the Sun is estimated to be on the order of 0.4 W/m at Earth's surface Estimate the fractional mass loss of the Sun over 109 years due to the radiation of neutrinos (The mass of the Sun is × 1030 kg The Earth-Sun distance is 1.5 × 1011 m.) G: Our Sun is estimated to have a life span of about 10 billion years, so in this problem, we are examining the radiation of neutrinos over a considerable fraction of the Sun’s life However, the mass carried away by the neutrinos is a very small fraction of the total mass involved in the Sun’s nuclear fusion process, so even over this long time, the mass of the Sun may not change significantly (probably less than 1%) O: The change in mass of the Sun can be found from the energy flux received by the Earth and Einstein’s famous equation, E = mc A: Since the neutrino flux from the Sun reaching the Earth is 0.4 W/m 2, the total energy emitted per second by the Sun in neutrinos in all directions is (0.4 W / m )(4π r ) = (0.4 W / m )(4π )(1.5 × 10 2 11 m ) = 1.13 × 10 23 W In a period of 10 yr, the Sun emits a total energy of (1.13 × 10 23 )( )( ) J / s 10 yr 3.156 × 107 s / yr = 3.57 × 10 39 J in the form of neutrinos This energy corresponds to an annihilated mass of E = mν c = 3.57 × 10 39 J mν = 3.97 × 10 22 kg so Since the Sun has a mass of about × 10 kg, this corresponds to a loss of only about part i n 50 000 000 of the Sun's mass over 10 yr in the form of neutrinos L: It appears that the neutrino flux changes the mass of the Sun by so little that it would be difficult to measure the difference in mass, even over its lifetime! 46.46 p + p → p +π+ + X We suppose the protons each have 70.4 MeV of kinetic energy From conservation of momentum, particle X has zero momentum and thus zero kinetic energy Conservation of energy then requires ( ) ( Mp c + M π c + M X c = Mp c + K p + Mp c + K p ) MX c = Mp c + 2Kp − Mπ c = 938.3 MeV + (70.4 MeV ) − 139.6 MeV = 939.5 MeV X must be a neutral baryon of rest energy 939.5 MeV Thus X is a neutron Chapter 46 Solutions *46.47 19 We find the number N of neutrinos: 1046 J = N(6 MeV) = N(6 × 1.6 × 10–13 J) N = 1.0 × 1058 neutrinos The intensity at our location is ly N 1.0 × 10 58 N 14 = = = 3.1 × 10 / m 2 A 4π r π (1.7 × 10 ly) (3.0 × 10 m / s)(3.16 × 10 s) The number passing through a body presenting 5000 cm2 = 0.50 m2 is then ( ) 3.1 × 1014 0.50 m = 1.5 × 1014 or m *46.48 ~ 1014 E γ + me c = By relativistic energy conservation, By relativistic momentum conservation, c = X= Subtracting (2) from (1), me c = 1= − 3X 1− X X= and − v2 / c2 3me v E γ + me c = 3me c − X2 so (1) (2) − v2 / c2 Eγ Dividing (2) by (1), Solving, 46.49 Eγ 3me c v c − 3me c X − X2 E γ = 4me c = 2.04 MeV m Λc2 = 1115.6 MeV Λ0 → p + π – mπ c = 139.6 MeV m p c = 938.3 MeV The difference between starting mass-energy and final mass-energy is the kinetic energy of the products and p p = pπ = p K p + K π = 37.7 MeV Applying conservation of relativistic energy, (938.3)2 + p c − 938.3 + (139.6)2 + p c − 139.6 = 37.7 MeV Solving the algebra yields p π c = ppc = 100.4 MeV Then, Kp = (m p c 2)2 + (100.4)2 – m pc = 5.35 MeV Kπ = (139.6)2 + (100.4)2 – 139.6 = 32.3 MeV © 2000 by Harcourt, Inc All rights reserved 20 46.50 Chapter 46 Solutions ( pp = 5.32 × 10–20 ) qBr = 1.60 × 10 −19 C (0.250 kg / C ⋅ s)(1.33 m ) Momentum of proton is kg · m s cpp = 1.60 × 10–11 kg m2 = 1.60 × 10–11 J = 99.8 MeV s2 MeV c Therefore, pp = 99.8 The total energy of the proton is Ep = E02 + (cp)2 = (938.3)2 + (99.8)2 = 944 MeV For pion, the momentum qBr is the same (as it must be from conservation of momentum i n a 2-particle decay) pπ = 99.8 MeV c E0π = 139.6 MeV Eπ = E02 + (cp)2 = (139.6)2 + (99.8)2 = 172 MeV Thus, ETotal after = ETotal before = Rest Energy Rest Energy of unknown particle = 944 MeV + 172 MeV = 1116 MeV (This is a Λ0 particle!) Mass = 1116 MeV/c 46.51 Σ → Λ0 + γ From Table 46.2, mΣ = 1192.5 MeV/c and ( mΛ = 1115.6 MeV c ) Conservation of energy requires E0 , Σ = E0 , Λ + K Λ + Eγ , or p2 1192.5 MeV = 1115.6 MeV + Λ + Eγ 2mΛ Momentum conservation gives pΛ = pγ , so the last result may be written as pγ2 1192.5 MeV = 1115.6 MeV + + Eγ 2mΛ or pγ2 c 1192.5 MeV = 1115.6 MeV + + Eγ 2mΛ c Recognizing that mΛ c = 1115.6 MeV we now have 1192.5 MeV = 1115.6 MeV + Solving this quadratic equation, Eγ ≈ 74.4 MeV and pγ c = Eγ , Eγ2 (1115.6 MeV ) + Eγ Chapter 46 Solutions 46.52 21 p + p → p + n +π+ The total momentum is zero before the reaction Thus, all three particles present after the reaction may be at rest and still conserve momentum This will be the case when the incident protons have minimum kinetic energy Under these conditions, conservation of energy gives ( ) mp c + K p = mp c + mnc + mπ c so the kinetic energy of each of the incident protons is Kp = 46.53 mnc + mπ c − mp c 2 Time-dilated lifetime: = (939.6 + 139.6 − 938.3) MeV = T = γ T0 = 0.900 × 10–10 s – v2/c2 = 70.4 MeV 0.900 × 10–10 s – (0.960)2 = 3.214 × 10–10 s distance = (0.960)(3.00 × 108 m/s)(3.214 × 10–10 s) = 9.26 cm 46.54 π − → µ− + νµ From the conservation laws, mπ c = 139.5 MeV = Eµ + Eν and pµ = pν , Eν = pν c Thus, Eµ2 = pµ c or Eµ2 − Eν2 = (105.7 MeV ) Since Eµ + Eν = 139.5 MeV [1] and (Eµ + Eν )(Eµ − Eν ) = (105.7 MeV)2 [2] then Eµ − Eν = Subtracting [3] from [1], 2Eν = 59.4 MeV ( ) [1] + (105.7 MeV ) = ( pν c) + (105.7 MeV ) 2 (105.7 MeV)2 139.5 MeV [2] = 80.1 and © 2000 by Harcourt, Inc All rights reserved [3] Eν = 29.7 MeV 22 Chapter 46 Solutions The expression e −E kBT dE gives the fraction of the photons that have energy between E and E + dE The fraction that have energy between E and infinity is *46.55 ∞ ∞ B −E k B T ∞ B B −E k T dE ∫ e −E k T ( − dE kBT ) e ∫E e E = E∞ = = e −E k T ∞ −E k T −E k T −E k T ∞ dE ∫ e (− dE kBT ) e ∫0 e 0 B B B We require T when this fraction has a value of 0.0100 (i.e., 1.00%) and E = 1.00 eV = 1.60 × 10 −19 J Thus, 0.0100 = e or 46.56 (a) ( − 1.60 × 10 −19 J ) (1.38 × 10 −23 J K )T ln (0.0100) = − ( 1.60 × 10 −19 J ) 1.38 × 10 −23 J K T = − 1.16 × 10 K T giving T = 2.52 × 10 K This diagram represents the annihilation of an electron and an antielectron From charge and leptonnumber conservation at either vertex, the exchanged particle must be an electron, e – (b) This is the tough one A neutrino collides with a neutron, changing it into a proton with release of a muon This is a weak interaction The exchanged particle has charge +1e and is a W + 46.57 (a) (b) (a) The mediator of this weak interaction is a Z0 boson (b) The Feynman diagram shows a down quark and its antiparticle annihilating each other They can produce a particle carrying energy, momentum, and angular momentum, but zero charge, zero baryon number, and, it may be, no color charge In this case the product (a) (b) particle is a photon For conservation of both energy and momentum, we would expect two photons; but momentum need not be strictly conserved, according to the uncertainty principle, if the photon travels a sufficiently short distance before producing another matter-antimatter pair of particles, as shown in Figure P46.57 Depending on the color charges of the d and d quarks, Chapter 46 Solutions 23 the ephemeral particle could also be a gluon , as suggested in the discussion of Figure 46.13(b) © 2000 by Harcourt, Inc All rights reserved ... we note that one million cubic centimeters make one cubic meter Our result is indeed close to the expected value Since the last reported significant digit is not certain, the difference in the... most people, we should suspect that he will not succeed in meeting Emily’s challenge O: Since the bill is released from rest and experiences free fall, we can use the equation y = gt to find the... College Publishers All rights reserved 2 2.5 Chapter Solutions (a) Let d represent the distance between A and B Let t1 be the time for which the walker d has the higher speed in 5.00 m/s = Let