Tipler p a fisica para la ciencia y la tecnología vol 1 vol 2 (solucionario)

www.elsolucionario.org Chapter Systems of Measurement Conceptual Problems *1 • Determine the Concept The fundamental physical quantities in the SI system include mass, length, and time Force, being the product of mass and acceleration, is not a fundamental quantity (c) is correct • Picture the Problem We can express and simplify the ratio of m/s to m/s2 to determine the final units m s = m ⋅ s = s and (d ) is correct m m ⋅s s Express and simplify the ratio of m/s to m/s2: • Determine the Concept Consulting Table 1-1 we note that the prefix giga means 109 (c ) is correct • Determine the Concept Consulting Table 1-1 we note that the prefix mega means 106 (d ) is correct *5 • Determine the Concept Consulting Table 1-1 we note that the prefix pico means 10−12 (a ) is correct • Determine the Concept Counting from left to right and ignoring zeros to the left of the first nonzero digit, the last significant figure is the first digit that is in doubt Applying this criterion, the three zeros after the decimal point are not significant figures, but the last zero is significant Hence, there are four significant figures in this number (c) is correct Chapter • Determine the Concept Counting from left to right, the last significant figure is the first digit that is in doubt Applying this criterion, there are six significant figures in this number (e) is correct • Determine the Concept The advantage is that the length measure is always with you The disadvantage is that arm lengths are not uniform; if you wish to purchase a board of ″two arm lengths″ it may be longer or shorter than you wish, or else you may have to physically go to the lumberyard to use your own arm as a measure of length • (a) True You cannot add ″apples to oranges″ or a length (distance traveled) to a volume (liters of milk) (b) False The distance traveled is the product of speed (length/time) multiplied by the time of travel (time) (c) True Multiplying by any conversion factor is equivalent to multiplying by Doing so does not change the value of a quantity; it changes its units www.elsolucionario.org Estimation and Approximation *10 •• Picture the Problem Because θ is small, we can approximate it by θ ≈ D/rm provided that it is in radian measure We can solve this relationship for the diameter of the moon Express the moon’s diameter D in terms of the angle it subtends at the earth θ and the earth-moon distance rm: D = θ rm Find θ in radians: θ = 0.524° × Substitute and evaluate D: D = (0.00915 rad )(384 Mm ) 2π rad = 0.00915 rad 360° = 3.51 × 106 m Systems of Measurement *11 •• Picture the Problem We’ll assume that the sun is made up entirely of hydrogen Then we can relate the mass of the sun to the number of hydrogen atoms and the mass of each Express the mass of the sun MS as the product of the number of hydrogen atoms NH and the mass of each atom MH: M S = NHM H Solve for NH: NH = MS MH Substitute numerical values and evaluate NH: NH = 1.99 × 1030 kg = 1.19 × 1057 1.67 × 10 −27 kg 12 •• Picture the Problem Let P represent the population of the United States, r the rate of consumption and N the number of aluminum cans used annually The population of the United States is roughly 3×108 people Let’s assume that, on average, each person drinks one can of soft drink every day The mass of a soft-drink can is approximately 1.8 ×10−2 kg (a) Express the number of cans N used annually in terms of the daily rate of consumption of soft drinks r and the population P: N = rP∆t Substitute numerical values and approximate N: ⎛ 1can ⎞ ⎟⎟ × 108 people N = ⎜⎜ ⋅ person d ⎝ ⎠ ⎛ d⎞ × (1 y )⎜⎜ 365.24 ⎟⎟ y⎠ ⎝ ( ≈ 1011 cans (b) Express the total mass of aluminum used per year for soft drink cans M as a function of the number of cans consumed and the mass m per can: M = Nm ) www.elsolucionario.org Chapter Substitute numerical values and evaluate M: (c) Express the value of the aluminum as the product of M and the value at recycling centers: ( )( M = 1011 cans/y 1.8 × 10−2 kg/can ) ≈ × 109 kg/y Value = ($1 / kg )M ( = ($1 / kg ) × 109 kg/y ) = $2 × 10 / y = billion dollars/y 13 •• Picture the Problem We can estimate the number of words in Encyclopedia Britannica by counting the number of volumes, estimating the average number of pages per volume, estimating the number of words per page, and finding the product of these measurements and estimates Doing so in Encyclopedia Britannica leads to an estimate of approximately 200 million for the number of words If we assume an average word length of five letters, then our estimate of the number of letters in Encyclopedia Britannica becomes 109 (a) Relate the area available for one letter s2 and the number of letters N to be written on the pinhead to the area of the pinhead: Solve for s to obtain: Substitute numerical values and evaluate s: Ns = π d where d is the diameter of the pinhead s= πd 4N ⎡ s= (b) Express the number of atoms per letter n in terms of s and the atomic spacing in a metal datomic: n= Substitute numerical values and evaluate n: n= ⎣ cm ⎞⎤ ⎛ ⎟ in ⎠⎥⎦ ⎝ ≈ 10−8 m 10 π ⎢(161 in )⎜ 2.54 ( ) s d atomic 10 −8 m ≈ 20 atoms × 10 −10 atoms/m *14 •• Picture the Problem The population of the United States is roughly × 108 people Assuming that the average family has four people, with an average of two cars per Systems of Measurement family, there are about 1.5 × 108 cars in the United States If we double that number to include trucks, cabs, etc., we have × 108 vehicles Let’s assume that each vehicle uses, on average, about 12 gallons of gasoline per week (a) Find the daily consumption of gasoline G: Assuming a price per gallon P = $1.50, find the daily cost C of gasoline: (b) Relate the number of barrels N of crude oil required annually to the yearly consumption of gasoline Y and the number of gallons of gasoline n that can be made from one barrel of crude oil: Substitute numerical values and estimate N: ( ) G = 3×108 vehicles (2 gal/d ) = ×108 gal/d ( ) C = GP = × 108 gal/d ($1.50 / gal) = $9 × 108 / d ≈ $1 billion dollars/d N= Y G∆t = n n (6 ×10 N= ) gal/d (365.24 d/y ) 19.4 gal/barrel ≈ 1010 barrels/y 15 •• Picture the Problem We’ll assume a population of 300 million (fairly accurate as of September, 2002) and a life expectancy of 76 y We’ll also assume that a diaper has a volume of about half a liter In (c) we’ll assume the disposal site is a rectangular hole in the ground and use the formula for the volume of such an opening to estimate the surface area required (a) Express the total number N of disposable diapers used in the United States per year in terms of the number of children n in diapers and the number of diapers D used by each child in 2.5 y: N = nD Use the daily consumption, the number of days in a year, and the estimated length of time a child is in diapers to estimate the number of diapers D required per child: D= diapers 365.24 d × × 2.5 y d y ≈ × 103 diapers/child Chapter Use the assumed life expectancy to estimate the number of children n in diapers: ⎛ y ⎞ ⎟⎟ 300 × 10 children n = ⎜⎜ ⎝ 76 y ⎠ ≈ 10 children Substitute to obtain: N = 107 children ( ( ( ) × × 10 diapers/child ) ) ≈ × 1010 diapers (b) Express the required landfill volume V in terms of the volume of diapers to be buried: Substitute numerical values and evaluate V: (c) Express the required volume in terms of the volume of a rectangular parallelepiped: V = NVone diaper ( ) V = × 1010 diapers (0.5 L/diaper ) ≈ 1.5 × 107 m V = Ah www.elsolucionario.org V 1.5 × 10 m A= = = 1.5 × 10 m Solve and evaluate h: h Use a conversion factor to express this area in square miles: 10 m A = 1.5 × 106 m × mi 2.590 km ≈ 0.6 mi 16 ••• Picture the Problem The number of bits that can be stored on the disk can be found from the product of the capacity of the disk and the number of bits per byte In part (b) we’ll need to estimate (i) the number of bits required for the alphabet, (ii) the average number of letters per word, (iii) an average number of words per line, (iv) an average number of lines per page, and (v) a book length in pages (a) Express the number of bits Nbits as a function of the number of bits per byte and the capacity of the hard disk Nbytes: N bits = N bytes (8 bits/byte) = (2 × 109 bytes)(8 bits/byte) = 1.60 × 1010 bits www.elsolucionario.org Systems of Measurement (b) Assume an average of letters/word and bits/character to estimate the number of bytes required per word: bits characters bits ×8 = 64 character word word bytes =8 word words bytes bytes ×8 = 4800 page word page Assume 10 words/line and 60 lines/page: 600 Assume a book length of 300 pages and approximate the number bytes required: 300pages × 4800 Divide the number of bytes per disk by our estimated number of bytes required per book to obtain an estimate of the number of books the 2-gigabyte hard disk can hold: N books = bytes = 1.44 × 106 bytes page × 109 bytes 1.44 ì 106 bytes/book 1400 books *17 ãã Picture the Problem Assume that, on average, four cars go through each toll station per minute Let R represent the yearly revenue from the tolls We can estimate the yearly revenue from the number of lanes N, the number of cars per minute n, and the $6 toll per car C R = NnC = 14 lanes × h d $6 cars × 60 × 24 × 365.24 × = $177M h d y car Units 18 • Picture the Problem We can use the metric prefixes listed in Table 1-1 and the abbreviations on page EP-1 to express each of these quantities (a) (c) 1,000,000 watts = 10 watts × 10 −6 meter = µm = MW (d) (b) −3 0.002gram = × 10 g = mg 30,000 seconds = 30 × 103 s = 30 ks Chapter 19 • Picture the Problem We can use the definitions of the metric prefixes listed in Table 1-1 to express each of these quantities without prefixes (c) (a) 40 µW = 40 × 10 W = 0.000040 W MW = × 106 W = 3,000,000 W (b) (d) −6 −9 ns = × 10 s = 0.000000004 s 25 km = 25 × 103 m = 25,000 m *20 • Picture the Problem We can use the definitions of the metric prefixes listed in Table 1-1 to express each of these quantities without abbreviations (a) 10 −12 boo = picoboo (e) 106 phone = 1megaphone (b) 10 low = gigalow (f) 10 −9 goat = nanogoat (c) 10 −6 phone = microphone (g) 1012 bull = terabull (d) 10 −18 boy = attoboy 21 •• Picture the Problem We can determine the SI units of each term on the right-hand side of the equations from the units of the physical quantity on the left-hand side (a) Because x is in meters, C1 and C2t must be in meters: C1 is in m; C2 is in m/s (b) Because x is in meters, ½C1t2 must be in meters: C1 is in m/s (c) Because v2 is in m2/s2, 2C1x must be in m2/s2: C1 is in m/s (d) The argument of trigonometric function must be dimensionless; i.e without units Therefore, because x C1 is in m; C2 is in s −1 Systems of Measurement is in meters: (e) The argument of an exponential function must be dimensionless; i.e without units Therefore, because v is in m/s: C1 is in m/s; C2 is in s −1 22 •• Picture the Problem We can determine the US customary units of each term on the right-hand side of the equations from the units of the physical quantity on the left-hand side (a) Because x is in feet, C1 and C2t must be in feet: C1 is in ft; C2 is in ft/s (b) Because x is in feet, ½C1t2 must be in feet: C1 is in ft/s (c) Because v2 is in ft2/s2, 2C1x must be in ft2/s2: C1 is in ft/s (d) The argument of trigonometric function must be dimensionless; i.e without units Therefore, because x is in feet: C1 is in ft; C2 is in s −1 (e) The argument of an exponential function must be dimensionless; i.e without units Therefore, because v is in ft/s: C1 is in ft/s; C2 is in s −1 Conversion of Units 23 • Picture the Problem We can use the formula for the circumference of a circle to find the radius of the earth and the conversion factor mi = 1.61 km to convert distances in meters into distances in miles (a) The Pole-Equator distance is one-fourth of the circumference: c = × 107 m www.elsolucionario.org Elementary Particles and the Beginning of the Universe 1419 Because Le = before and after the Lepton number: 0→0+0+0+0=0 decay, the lepton number is conserved The decay satisfies all conservation laws and is allowed Quarks 21 • Picture the Problem For each quark combination we can determine the baryon number B, the charge Q, and the strangeness S and then use Table 41-1 to find a match and complete the following table (a) (b) (c) (d) (e) (f) Combination uud udd uus dds uss dss B 1 1 1 Q +1 +1 −1 −1 S 0 −1 −1 −2 −2 hadron p+ n Σ+ Σ− Ξ0 Ξ− 22 • Picture the Problem For each quark combination we can determine the baryon number B, the charge Q, and the strangeness S and then use Table 41-1 to find a match and complete the following table Combination (a) (b) (c) (d) ud ud us us B Q +1 S 0 0 −1 +1 −1 +1 −1 hadron π+ π− K+ K− 23 • Determine the Concept From Table 41-2 we see that to satisfy the conditions of charge = +2 and zero strangeness, charm, topness, and bottomness, the quark combination must be uuu 24 • Picture the Problem Because K + and K0 are mesons, they consist of a quark and an antiquark We can use Tables 41-1 and 41-2 to find combinations of quarks with the correct values for electric charge, baryon number, and strangeness for these particles 1420 Chapter 41 (a) For K + we need: Q = +1 B=0 S = +1 A combination of quarks with these properties is us (b) For K0 we need: Q=0 B=0 S = +1 A combination of quarks with these properties is ds 25 • Determine the Concept Because D + and D− are mesons, they consist of a quark and an antiquark (a) B = 0, so we must look for a combination of quark and antiquark Because it has charm of +1, one of the quarks must be c Because the charge is +e, the antiquark must be d The possible combination for D+ is cd (b) Because D− is the antiparticle of D+, the quark combination is c d 26 • Picture the Problem We can use Tables 41-1 and 41-2 to find combinations of quarks with the correct values for electric charge, baryon number, and strangeness for these particles Because K – and K are mesons, they consist of a quark and an antiquark (a) For K – we need: Q = −1 B=0 S = −1 A combination of quarks with these properties is u s (b) For K we need: Q=0 B=0 S = −1 A combination of quarks with these properties is d s Elementary Particles and the Beginning of the Universe 1421 Remarks: An alternative solution could take advantage of our results in Problem 20 for the antiparticles K + and K *27 •• Picture the Problem Because Λ0, p – , and Σ– are baryons, they are made up of three quarks We can use Tables 41-1 and 41-2 to find combinations of quarks with the correct values for electric charge, baryon number, and strangeness for these particles (a) For Λ0 we need: Q=0 B = +1 S = −1 The quark combination that satisfies these conditions is uds (b) For p – we need: Q = −1 B = −1 S = +1 The quark combination that satisfies these conditions is u u d www.elsolucionario.org Q = −1 (c) For Σ– we need: B = +1 S = −1 The quark combination that satisfies these conditions is dds 28 •• Picture the Problem Because n , Ξ0 , and Σ+ are baryons, they are made up of three quarks We can use Tables 41-1 and 41-2 to find combinations of quarks with the correct values for electric charge, baryon number, and strangeness for these particles (a) For n we need: Q=0 B = −1 S=0 The quark combination that satisfies these conditions is u d d (b) For Ξ0 we need: Q=0 B = +1 S = −2 www.elsolucionario.org 1422 Chapter 41 The quark combination that satisfies these conditions is uss (c) For Σ+ we need: Q = +1 B = +1 S = −1 The quark combination that satisfies these conditions is uus 29 •• Picture the Problem Because Ω– and Ξ– are baryons, they are made up of three quarks We can use Tables 41-1 and 41-2 to find combinations of quarks with the correct values for electric charge, baryon number, and strangeness for these particles (a) For Ω– we need: Q = −1 B = +1 S = −3 The quark combination that satisfies these conditions is sss (b) For Ξ– we need: Q = −1 B = +1 S = −2 The quark combination that satisfies these conditions is ssd 30 •• Picture the Problem We can use Table 41-2 to identify the properties of the particles made up of the given quarks (a) For ddd: Q = −1 B = +1 S= (b) For uc : Q= B= S= charm = − Elementary Particles and the Beginning of the Universe 1423 Q = +1 (c) For ub : B= S= bottomness = − (d) For s s s : Q = +1 B = −1 S = +3 The Evolution of the Universe *31 • Picture the Problem We can use Hubble’s law to find the distance from the earth to this galaxy v = Hr The recessional velocity of galaxy is related to its distance by Hubble’s law: Solve for r: r= v H ( 0.025) c (0.025) (3 × 105 km/s ) = r= Substitute numerical values and evaluate r: 23 km/s 106 c ⋅ y 23 km/s 106 c ⋅ y = 3.26 ì 108 c y 32 ã Picture the Problem We can use Hubble’s law to find the speed of the galaxy v = Hr The recessional velocity of galaxy is related to its distance by Hubble’s law: Substitute numerical values and evaluate v: ⎛ 23 km/s ⎞ c ⎛ ⎞ ⎟⎟ 12 × 109 c ⋅ y ⎜ v = ⎜⎜ ⎟ = 0.920c ⎝ 3.00 × 10 km/s ⎠ ⎝ 10 c ⋅ y ⎠ ( ) 1424 Chapter 41 33 •• Picture the Problem We can substitute for f ′ and f0, using v = fλ, in Equation 39-16b to + v/c − v/c show that the relativistic wavelength shift is λ' = λ0 From Equation 39-16b: f' = f Express f ′ and f0 in terms of λ′ and λ0: f' = Substitute for f ′ and f0 to obtain: c λ' Solve for λ′: = − v/c + v/c c c and f = λ' λ0 c λ0 λ' = λ0 − v/c + v/c 1+ v c 1− v c *34 •• Picture the Problem Using Hubble’s law, we can rewrite the equation from Problem 31 as λ' = λ0 + Hr/c − Hr/c From Problem 33 we have: Use Hubble’s law to relate v to r: Substitute for v to obtain: λ' = λ0 + v/c − v/c v = Hr λ' = λ0 + Hr/c − Hr/c (a) For r = 5×106 c⋅y: ⎛ 23 km/s ⎞ (5 × 10 c ⋅ y ) ⎟ + ⎜⎜ 10 c ⋅ y ⎟⎠ (3 × 10 km/s ) ⎝ λ' = 656.3 nm = 656.6 nm ⎛ 23 km/s ⎞ (5 × 10 c ⋅ y ) ⎟⎟ − ⎜⎜ ⎝ 10 c ⋅ y ⎠ (3 × 10 km/s ) (b) For r = 50 ×106 c⋅y: www.elsolucionario.org Elementary Particles and the Beginning of the Universe 1425 ⎛ 23 km/s ⎞ (50 × 106 c ⋅ y ) ⎟ + ⎜⎜ 10 c ⋅ y ⎟⎠ (3 × 105 km/s ) ⎝ λ' = 656.3 nm = 658.8 nm ⎛ 23 km/s ⎞ (50 × 106 c ⋅ y ) ⎟⎟ − ⎜⎜ ⎝ 10 c ⋅ y ⎠ (3 × 10 km/s ) (c) For r = 500 ×106 c⋅y: ⎛ 23 km/s ⎞ ⎟ + ⎜⎜ 10 c ⋅ y ⎟⎠ ⎝ λ' = 656.3 nm ⎛ 23 km/s ⎞ ⎟⎟ − ⎜⎜ ⎝ 10 c ⋅ y ⎠ (500 ×10 c ⋅ y ) (3 ×10 km/s) = (500 ×10 c ⋅ y) (3 ×10 km/s) 682.0 nm (5 ×10 c ⋅ y) (3 ×10 km/s) = (5 ×10 c ⋅ y ) (3 ×10 km/s) 983.0 nm 6 (d) For r = 5×109 c⋅y: ⎛ 23 km/s ⎞ ⎟ + ⎜⎜ 10 c ⋅ y ⎟⎠ ⎝ λ' = 656.3 nm ⎛ 23 km/s ⎞ ⎟⎟ − ⎜⎜ ⎝ 10 c ⋅ y ⎠ 9 General Problems www.elsolucionario.org 35 • Determine the Concept (a) It must be a meson, and it must consist of a quark and its antiquark (b) The π0 is its own antiparticle (c) The Ξ is a baryon; it cannot be its own antiparticle; the antiparticle is the Ξ0 = u s s 36 •• Picture the Problem Examination of the decay products will reveal whether all the final products are stable A decay process is allowed if energy, charge, baryon number, and lepton number are conserved (a) Yes, the final products are stable (b) Add the reactions to obtain: 1426 Chapter 41 Ξ → p + + e − + ν e +ν µ + ν µ + 2γ (c) Charge conservation: → e+ + e− = Charge is conserved Baryon number: 1→1+0=1 Baryon number is conserved Lepton number: 0→0+1−1+1−1=0 Lepton number is conserved Strangeness: −2 → Strangeness is not conserved However, the reaction is allowed via the weak interaction because in the first two decays ∆S = +1 (d) No; the rest masses of the decay products would be greater than the rest mass of the Ξ , violating energy conservation *37 •• Picture the Problem The π o particle is composed of two quarks, uu Hence, the reaction π o → γ + γ is equivalent to uu → γ + γ (a) The u and u annihilate resulting in the photons (b) Two or more photons are required to conserve linear momentum 38 •• Picture the Problem A decay process is allowed if energy, charge, baryon number, and lepton number are conserved (a) Energy conservation: Because mΛ > mp + mπ , energy conservation is not violated Charge conservation: → −1 = Because the total charge is before and after the decay, charge conservation is not violated Elementary Particles and the Beginning of the Universe 1427 Baryon number: 1→1+0=1 Because there is no change in baryon number, baryon number is conserved Lepton number: 0→0+0=0 Because lepton number is on both sides, lepton number is conserved The decay satisfied all conservation laws and is allowed (b) Energy conservation: Because mΣ < mn + m p , energy is not conserved Charge conservation: −1 → − = −1 Because the total charge does not change, charge is conserved Baryon number: +1 → − = Because B changes from + to 0, Lepton number: 0→0+0 =0 Because L is on both sides, baryon number is not conserved lepton number is conserved Because the decay violates both energy conservation and baryon number, it is not allowed (c) Energy conservation: Energy is conserved Charge conservation: −1 → − + + = −1 Because the total charge does not Baryon number: 0→0+0+0 =0 Lepton number: 1→1−1+1 =1 change, charge is conserved Because B does not change, baryon number is conserved Because L does not change, lepton number is conserved The decay satisfied all conservation laws and is allowed www.elsolucionario.org 1428 Chapter 41 Remarks: The decay in Part (c) is the decay process for the muon (see Example 41-2) *39 ãã Picture the Problem We can systematically determine Q, B, S, and s for each reaction and then use these values to identify the unknown particles (a) For the strong reaction: p + π − → Σo + ? Charge number: +1 − = + Q ⇒ Q = Baryon number: +1 + = +1 + B ⇒ B = Strangeness: + = −1 + S ⇒ S = +1 Spin: + 12 + = + 12 + s ⇒ s = These properties indicate that the particle is the kaon, K (b) For the strong reaction: p + p → π + + n + K+ + ? Charge number: +1 + = +1 + + + Q ⇒ Q = Baryon number: +1 + = + + + B ⇒ B = +1 Strangeness: + = + + + S ⇒ S = −1 Spin: + 12 + = + 12 + + s ⇒ s = + 12 These properties indicate that the particle is either the Σ or the Λ0 baryon (c) For the strong reaction: p + K − → Ξ− + ? Charge number: +1 − = −1 + Q ⇒ Q = +1 Baryon number: +1 + = +1 + B ⇒ B = Strangeness: − = −2 + S ⇒ S = −1 Spin: + 12 + = + 12 + s ⇒ s = These properties indicate that the particle is the kaon, K + 40 •• Picture the Problem We can systematically determine Q, B, S, and s for the reaction and then use these values to identify the unknown particle The Q value for the reaction is given by Q = −(∆m ) c and the expression for the threshold energy for the reaction is given in the problem statement Elementary Particles and the Beginning of the Universe 1429 (a) For the strong reaction: p + p → Λo + K o + p + ? Charge number: +1 + = + + + Q ⇒ Q = +1 baryon number: +1 + =+1 + + + B ⇒ B = strangeness: + = −1 + + + S ⇒ S = spin: + 12 + = + 12 + + +s ⇒s=0 These properties indicate that the unknown particle is a pion, π + p + p → Λo + K o + p + π + (b) The reaction is: The Q-value for the reaction is: [ ] Q = (mp + mp ) − (M Λo + M Ko + M p + M π + ) c Use Table 41-1 to find the mass-energy values: Q = [(938.3 + 938.3) − (1116 + 497.7 + 938.3 + 139.6 )]MeV = − 815 MeV Because Q < 0, the reaction is endothermic www.elsolucionario.org (c) The threshold energy for this reaction is: K th = − Q (mp + mp + M Λo + M Ko + M p + M π + ) 2mp Using Table 41-1 to find the mass-energy values, substitute numerical values and evaluate Kth: K th = − − 815 MeV (938.3 + 938.3 + 1116 + 497.7 + 938.3 + 139.6)MeV 2(938.3 MeV ) = 1984 MeV = 1.984 GeV 41 •• Picture the Problem We can solve the equation derived in Problem 31 for the recessional velocity of the galaxy and then use Hubble’s equation to estimate the distance to the galaxy (a) From Problem 31 we have: λ' = λ0 + v/c − v/c 1430 Chapter 41 Solve for the radicand: + v/c ⎛ λ' ⎞ =⎜ ⎟ − v/c ⎜⎝ λ0 ⎟⎠ Substitute numerical values for λ′ and λ0: + v/c ⎛ 1458 nm ⎞ =⎜ ⎟ = 4.935 − v/c ⎝ 656.3 nm ⎠ Simplify to obtain: v ⎛ v⎞ 4.953⎜1 − ⎟ = + c ⎝ c⎠ 2 and v 5.953 = 3.953 c Solve for v: v = 0.664c = 0.664(3 ×108 m/s ) = 1.99 ×108 m/s = 1.99 ×10 km/s (b) From the Hubble equation we have: r= Substitute numerical values and evaluate r: r= v H 1.99 × 105 km/s = 8.65 × 109 c ⋅ y 23 km/s 106 c ⋅ y 42 ••• Picture the Problem We can use the masses of the parent and daughters to find the total kinetic energy of the decay products under the assumption that the Λ0 is initially at rest Application of conservation of energy and the definition of kinetic energy will yield the ratio of the kinetic energy of the pion to the kinetic energy of the proton Finally, we can use our results in (a) and (b) to find the kinetic energies of the proton and the pion for this decay (a) The total kinetic energy of the decay products is given by: K tot = (mΛ − mp − mπ )c Substitute numerical values (see Table 41-1) and evaluate Ktot: MeV MeV MeV ⎞ ⎛ K tot = ⎜1116 − 938.3 − 139.6 ⎟c = 38.1 MeV c c c ⎠ ⎝ (b) The ratio of the kinetic energies is given by: Kπ 12 mπ vπ2 mπ vπ2 = = Kp mp vp2 mp vp www.elsolucionario.org Elementary Particles and the Beginning of the Universe 1431 vπ mp = vp mπ Use conservation of momentum (nonrelativistic) to obtain: mπ vπ = mp v p ⇒ Substitute for the ratio of the speeds to obtain: Kπ mπ ⎛ mp ⎜ = K p mp ⎜⎝ mπ Substitute numerical values and evaluate the ratio of the kinetic energies: MeV 938.3 Kπ c = 6.72 = K p 139.6 MeV c2 (c) Express the total kinetic energy in terms of Kπ and Kp: K p + Kπ = K tot Use our results in (a) and (b) to obtain: K p + 6.72 K p = 38.1 MeV Solve for Kp: K p = 4.94 MeV Substitute in equation (1) to obtain: Kπ = K tot − K p = 33.2 MeV m ⎞ ⎟⎟ = p mπ ⎠ (1) *43 ••• Picture the Problem The total kinetic energy of the decay products is the rest energy of the Σ0 particle We can find the momentum of the photon from its energy and use the conservation of momentum to calculate the kinetic energy of the Λ0 (a) The total kinetic energy of the decay products is given by: K tot = (mΣ ) c Substitute numerical values (see Table 41-1) and evaluate Ktot: MeV ⎞ ⎛ K tot = ⎜1193 ⎟ c = 1193 MeV c ⎠ ⎝ (b) The momentum of the photon is given by: pγ = Substitute numerical values and evaluate pγ: Eγ E − mΛ c = c c MeV ⎞ ⎛ 1193 MeV − ⎜1116 ⎟ c c ⎠ ⎝ pγ = c MeV = 77.0 c 1432 Chapter 41 (c) The kinetic energy of the Λ0 is given by: pΛ2 KΛ = 2mΛ or, because pΛ = pγ , KΛ = pγ2 2mΛ Substitute numerical values and evaluate KΛ: MeV ⎞ ⎛ ⎜ 77.0 ⎟ c ⎠ ⎝ = 2.66 MeV KΛ = MeV ⎞ ⎛ 2⎜1116 ⎟ c ⎠ ⎝ (d) A better estimate of the energy of the photon is: Eγ = E − mΛ c − K Λ Substitute numerical values and evaluate Eγ: MeV ⎞ ⎛ Eγ = 1193 MeV − ⎜1116 ⎟c − 2.66 MeV = 74.3 MeV c ⎠ ⎝ The improved estimate of the momentum of the photon is: pγ = Eλ 74.3 MeV MeV = = 74.3 c c c 44 ••• Picture the Problem The solution strategy is outlined in the problem statement (a) Express ∆t = t2 − t1 in terms of u2 and u1: Noting that u1u2 ≈ c2, we have: ∆t = t2 − t1 = ∆t ≈ x x x(u1 − u2 ) − = u2 u1 u1u2 x∆u c2 (1) where ∆u = u1 − u2 (b) From Equation 39-25 we have: ⎛ ⎛ mc ⎞ ⎞ ⎛ mc ⎞ u ⎟⎟ = ⎜1 − ⎜⎜ ⎟⎟ ⎟ = − ⎜⎜ ⎜ c E E ⎝ ⎠ ⎠ ⎟⎠ ⎝ ⎝ Expand binomially to obtain: u ⎛ mc ⎞ ⎟ = − ⎜⎜ c ⎝ E ⎟⎠ 12 Elementary Particles and the Beginning of the Universe 1433 (c) Express u1 − u2 in terms of E1, E2, and mc2: u1 − u2 = = (mc ) ⎛⎜⎜ E1 − ( ) (E − E22 2 ⎝ c mc 2 2 ⎞ ⎟ E12 ⎟⎠ 2 2 ) 2E E Substitute numerical values and evaluate ∆u: [ ] ⎛ eV ⎞ 2 c⎜ 20 c ⎟ (20 MeV ) − (5 MeV ) c ⎠ ∆u = ⎝ = 7.50 × 10 −12 c 2 2(20 MeV ) (5 MeV ) (1.7 ×10 )( c ⋅ y 7.5 × 10−12 c c2 31.56 Ms = 1.275 × 10−6 y × y Use equation (1) to evaluate ∆t: ∆t ≈ ) = 40.2 s (d) Using mc2 = 40 eV for the rest energy of a neutrino: www.elsolucionario.org ⎛ eV ⎞ c 40 c [(20 MeV ) − (5 MeV ) ] ⎜ ∆u = ⎝ 2 ⎟ c ⎠ 2 2(20 MeV ) (5 MeV ) Use equation (1) to evaluate ∆t: = 3.00 × 10 −11 c (1.7 ×10 )( c ⋅ y × 10−11 c c2 31.56 Ms = 5.1× 10−6 y × y ∆t ≈ ) = 161s Remarks: Note that the spread in the arrival time for neutrinos from a supernova can be used to estimate the mass of a neutrino ... ) 1. 22 m ( 49 ) = − 12 3 g Remarks: The final velocity we obtained in part (a) , approximately 12 1 mph, is about the same as the terminal velocity for an "average" man This solution is probably... constant maximum velocity: ∆x 12 = vmax ∆t 12 + 12 a1 2 (∆t 12 ) Substitute for these displacements and solve for a: 10 0 m = 12 a( 3 s ) + a( 3 s )(6.79 s ) 2 www.elsolucionario.org = (a? ??t )∆t 01 12 = a( 3... N= Y G∆t = n n (6 ? ?10 N= ) gal/d (365 .24 d /y ) 19 .4 gal/barrel ≈ 10 10 barrels /y 15 •• Picture the Problem We’ll assume a population of 300 million (fairly accurate as of September, 20 02) and a