Physics for scientists and engineers with modern physics instructor solutions manual 9th solutions

Physics for scientists and engineers with modern physics instructor solutions manual 9th solutions Physics for scientists and engineers with modern physics instructor solutions manual 9th solutions Physics for scientists and engineers with modern physics instructor solutions manual 9th solutions Physics for scientists and engineers with modern physics instructor solutions manual 9th solutions Physics for scientists and engineers with modern physics instructor solutions manual 9th solutions Physics for scientists and engineers with modern physics instructor solutions manual 9th solutions Physics for scientists and engineers with modern physics instructor solutions manual 9th solutions Physics for scientists and engineers with modern physics instructor solutions manual 9th solutions Physics for scientists and engineers with modern physics instructor solutions manual 9th solutions

INSTRUCTOR SOLUTIONS MANUAL Physics and Measurement CHAPTER OUTLINE 1.1 Standards of Length, Mass, and Time 1.2 Matter and Model Building 1.3 Dimensional Analysis 1.4 Conversion of Units 1.5 Estimates and Order-of-Magnitude Calculations 1.6 Significant Figures * An asterisk indicates a question or problem new to this edition ANSWERS TO OBJECTIVE QUESTIONS OQ1.1 The meterstick measurement, (a), and (b) can all be 4.31 cm The meterstick measurement and (c) can both be 4.24 cm Only (d) does not overlap Thus (a), (b), and (c) all agree with the meterstick measurement OQ1.2 Answer (d) Using the relation ⎛ 2.54 cm ⎞ ⎛ m ⎞ = 0.304 m ft = 12 in ⎜ ⎝ in ⎟⎠ ⎜⎝ 100 cm ⎟⎠ we find that ⎛ 0.304 m ⎞ 420 ft ⎜ ⎟⎠ = 132 m ⎝ ft OQ1.3 The answer is yes for (a), (c), and (e) You cannot add or subtract a number of apples and a number of jokes The answer is no for (b) and (d) Consider the gauge of a sausage, kg/2 m, or the volume of a cube, (2 m)3 Thus we have (a) yes; (b) no; (c) yes; (d) no; and (e) yes © 2014 Cengage Learning All Rights Reserved May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part Physics and Measurement OQ1.4 41 € ≈ 41 € (1 L/1.3 €)(1 qt/1 L)(1 gal/4 qt) ≈ (10/1.3) gal ≈ gallons, answer (c) OQ1.6 The number of decimal places in a sum of numbers should be the same as the smallest number of decimal places in the numbers summed 21.4 s 15 s 17.17 s 4.003 s 57.573 s = 58 s, answer (d) OQ1.7 The population is about billion = × 109 Assuming about 100 lb per person = about 50 kg per person (1 kg has the weight of about 2.2 lb), the total mass is about (6 × 109)(50 kg) = × 1011 kg, answer (d) OQ1.8 No: A dimensionally correct equation need not be true Example: chimpanzee = chimpanzee is dimensionally correct Yes: If an equation is not dimensionally correct, it cannot be correct OQ1.9 Mass is measured in kg; acceleration is measured in m/s Force = mass × acceleration, so the units of force are answer (a) kg⋅m/s2 OQ1.10 0.02(1.365) = 0.03 The result is (1.37 ± 0.03) × 10 kg So (d) digits are significant ANSWERS TO CONCEPTUAL QUESTIONS CQ1.1 Density varies with temperature and pressure It would be necessary to measure both mass and volume very accurately in order to use the density of water as a standard CQ1.2 The metric system is considered superior because units larger and smaller than the basic units are simply related by multiples of 10 Examples: km = 103 m, mg = 10–3 g = 10–6 kg, ns = 10–9 s CQ1.3 A unit of time should be based on a reproducible standard so it can be used everywhere The more accuracy required of the standard, the less the standard should change with time The current, very accurate standard is the period of vibration of light emitted by a cesium atom Depending on the accuracy required, other standards could be: the period of light emitted by a different atom, the period of the swing of a pendulum at a certain place on Earth, the period of vibration of a sound wave produced by a string of a specific length, density, and tension, and the time interval from full Moon to full Moon CQ1.4 (a) 0.3 millimeters; (b) 50 microseconds; (c) 7.2 kilograms © 2014 Cengage Learning All Rights Reserved May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part Chapter SOLUTIONS TO END-OF-CHAPTER PROBLEMS Section 1.1 P1.1 (a) Standards of Length, Mass, and Time   Modeling the Earth as a sphere, we find its volume as 4 π r = π ( 6.37 × 106  m ) = 1.08 × 1021  m 3 Its density is then m 5.98 × 1024  kg ρ= = = 5.52 × 103  kg/m 21 V 1.08 × 10  m (b) P1.2 This value is intermediate between the tabulated densities of aluminum and iron Typical rocks have densities around 2000 to 3000 kg/m3 The average density of the Earth is significantly higher, so higher-density material must be down below the surface With V = (base area)(height), V = (π r ) h and ρ = ρ= m , we have V ⎛ 109  mm ⎞ m 1 kg = π r h π ( 19.5 mm )2 ( 39.0 mm ) ⎜⎝ 1 m ⎟⎠ ρ = 2.15 × 10  kg/m P1.3 Let V represent the volume of the model, the same in ρ = Then ρiron = 9.35 kg/V and ρgold = Next, and P1.4 (a) ρgold ρiron mgold = V mgold 9.35 kg ⎛ 19.3 × 103  kg/m ⎞ = ( 9.35 kg ) ⎜ = 22.9 kg 3 ⎝ 7.87 × 10  kg/m ⎟⎠ ρ = m/V and V = ( 4/3 ) π r = ( 4/3 ) π ( d/2 ) = π d /6, where d is the diameter Then ρ = 6m/ π d = (b) mgold m , for both V ( 1.67 × 10−27 kg ) π ( 2.4 × 10 −15 m) = 2.3 × 1017 kg/m 2.3 × 1017 kg/m = 1.0 × 1013 times the density of osmium 3 22.6 × 10 kg/m © 2014 Cengage Learning All Rights Reserved May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part Physics and Measurement P1.5 For either sphere the volume is V = π r and the mass is m = ρV = ρ π r We divide this equation for the larger sphere by the same equation for the smaller: m ρ ( 4/ ) π r3 r3 = = =5 ms ρ ( 4/ ) π rs3 rs3 Then *P1.6 r = rs = ( 4.50 cm ) = 7.69 cm The volume of a spherical shell can be calculated from V = Vo − Vi = π ( r23 − r13 ) From the definition of density, ρ = m = ρV = ρ Section 1.2 P1.7 m , so V ( ) 4π ρ ( r23 − r13 ) π ( r23 − r13 ) = 3 Matter and Model Building   From the figure, we may see that the spacing between diagonal planes is half the distance between diagonally adjacent atoms on a flat plane This diagonal distance may be obtained from the Pythagorean theorem, Ldiag = L2 + L2 Thus, since the atoms are separated by a distance L = 0.200 nm, the diagonal planes are separated by L + L2 = 0.141 nm P1.8 (a) Treat this as a conversion of units using Cu-atom = 1.06 × 10–25 kg, and cm = 10–2 m: kg ⎞ ⎛ 10−2 m ⎞ ⎛ Cu-atom ⎞ ⎛ density = ⎜ 920 ⎟ ⎜ ⎝ m ⎠ ⎝ 1 cm ⎟⎠ ⎜⎝ 1.06 × 10−25  kg ⎟⎠ = 8.42 × 1022 Cu-atom cm © 2014 Cengage Learning All Rights Reserved May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part Chapter (b) Thinking in terms of units, invert answer (a): ⎛ ⎞ (density )−1 = ⎜⎝ 8.42 × 101 cm 22  Cu-atoms ⎟⎠ = 1.19 × 10−23  cm /Cu-atom (c) For a cube of side L, L3 = 1.19 × 10−23  cm → L = 2.28 × 10−8 cm Section 1.3 P1.9 (a) Dimensional Analysis   Write out dimensions for each quantity in the equation vf = vi + ax The variables vf and vi are expressed in units of m/s, so [vf] = [vi] = LT –1 The variable a is expressed in units of m/s2; [a] = LT –2 The variable x is expressed in meters Therefore, [ax] = L2 T –2 Consider the right-hand member (RHM) of equation (a): –1 [RHM] = LT +L2 T –2 Quantities to be added must have the same dimensions Therefore, equation (a) is not dimensionally correct (b) Write out dimensions for each quantity in the equation y = (2 m) cos (kx) For y, [y] = L for m, [2 m] = L and for (kx), [ kx] = ⎡⎣ m –1 x ⎤⎦ = L–1L ( ) Therefore we can think of the quantity kx as an angle in radians, and we can take its cosine The cosine itself will be a pure number with no dimensions For the left-hand member (LHM) and the right-hand member (RHM) of the equation we have [LHM] = [y] = L [RHM] = [2 m][cos (kx)] = L These are the same, so equation (b) is dimensionally correct © 2014 Cengage Learning All Rights Reserved May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part Physics and Measurement P1.10 Circumference has dimensions L, area has dimensions L2, and volume 1/2 has dimensions L Expression (a) has dimensions L(L ) = L , expression (b) has dimensions L, and expression (c) has dimensions L(L ) = L The matches are: (a) and (f), (b) and (d), and (c) and (e) P1.11 (a) Consider dimensions in terms of their mks units For kinetic energy K: ⎡⎛ p ⎞ ⎤ [ p ] kg ⋅ m2 = [ K ] = ⎢⎜ ⎟ ⎥ = s2 ⎣⎝ 2m ⎠ ⎦ kg Solving for [p2] and [p] then gives [ p] = kg ⋅ m2 s2 → [ p] = kg ⋅ m s The units of momentum are kg ⋅ m/s (b) Momentum is to be expressed as the product of force (in N) and some other quantity X Considering dimensions in terms of their mks units, [N ] ⋅ [X ] = [ p ] kg ⋅ m kg ⋅ m ⋅ [X ] = s s [X ] = s Therefore, the units of momentum are N ⋅ s P1.12 ⎡ kg ⋅ m ⎤ [ M ][ L ] We substitute [ kg ] = [M], [ m ] = [L], and [ F ] = ⎢ ⎥ = into ⎣ s ⎦ [T ]2 Newton’s law of universal gravitation to obtain [ M ][L ] = [G ][ M ]2 [T ]2 [L ]2 Solving for [G] then gives [G ] = *P1.13 [L ]3 [ M ][T ]2 = m3 kg ⋅ s The term x has dimensions of L, a has dimensions of LT −2 , and t has dimensions of T Therefore, the equation x = ka mt n has dimensions of L = ( LT −2 ) ( T )n or L1T = LmT n−2m m The powers of L and T must be the same on each side of the equation © 2014 Cengage Learning All Rights Reserved May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part Chapter Therefore, L1 = Lm and m = Likewise, equating terms in T, we see that n – 2m must equal Thus, n = The value of k, a dimensionless constant, cannot be obtained by dimensional analysis P1.14 Summed terms must have the same dimensions (a) [X] = [At3] + [Bt] L = [ A ] T + [ B] T → [ A ] = L/T , and [ B ] = L/T (b) Section 1.4 P1.15 [ dx/dt ] = ⎡⎣ 3At ⎤⎦ + [B] = L/T Conversion of Units   From Table 14.1, the density of lead is 1.13 × 10 kg/m , so we should expect our calculated value to be close to this value The density of water is 1.00 × 103 kg/m3, so we see that lead is about 11 times denser than water, which agrees with our experience that lead sinks Density is defined as ρ = m/V We must convert to SI units in the calculation ( ⎛ 23.94 g ⎞ ⎛ kg ⎞ 100 cm ρ=⎜ ⎝ 2.10 cm ⎟⎠ ⎜⎝ 000 g ⎟⎠ 1 m ( ) ⎛ 23.94 g ⎞ ⎛ kg ⎞ 000 000 cm = ⎜ ⎝ 2.10 cm ⎟⎠ ⎜⎝ 000 g ⎟⎠ 1 m ) = 1.14  × 10  kg/m Observe how we set up the unit conversion fractions to divide out the units of grams and cubic centimeters, and to make the answer come out in kilograms per cubic meter 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 from the tabulated values is possibly due to measurement uncertainty and does not indicate a discrepancy © 2014 Cengage Learning All Rights Reserved May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part Physics and Measurement P1.16 The weight flow rate is ton ⎞ ⎛ 2000 lb ⎞ ⎛ 1 h ⎞ ⎛ 1 min ⎞ ⎛ ⎜⎝ 200 ⎟⎜ ⎟ = 667 lb/s ⎟⎜ ⎟⎜ h ⎠ ⎝ ton ⎠ ⎝ 60 min ⎠ ⎝ 60 s ⎠ P1.17 For a rectangle, Area = Length × Width We use the conversion m = 3.281 ft The area of the lot is then ⎛ 1 m ⎞ ⎛ 1 m ⎞ 125 ft ) ⎜ = 871 m A = LW = ( 75.0 ft ) ⎜ ( ⎟ ⎝ 3.281 ft ⎠ ⎝ 3.281 ft ⎟⎠ P1.18 Apply the following conversion factors: in = 2.54 cm, d = 86 400 s, 100 cm = 1m, and 109 nm = m Then, the rate of hair growth per second is −2 ⎛ ⎞ ( 2.54 cm/in ) ( 10 m/cm ) ( 10 nm/m ) rate = ⎜ in/day ⎟ ⎝ 32 ⎠ 86 400 s/day = 9.19 nm/s This means the proteins are assembled at a rate of many layers of atoms each second! P1.19 The area of the four walls is (3.6 + 3.8 + 3.6 + 3.8) m × (2.5 m) = 37 m Each sheet in the book has area (0.21 m)(0.28 m) = 0.059 m2 The 2 number of sheets required for wallpaper is 37 m /0.059 m = 629 sheets = 629 sheets(2 pages/1 sheet) = 1260 pages The number of pages in Volume are insufficient P1.20 We use the formula for the volume of a pyramid given in the problem and the conversion 43 560 ft2 = acre Then, V = Bh = ⎡⎣( 13.0 acres )( 43 560 ft /acre ) ⎤⎦ × ( 481 ft ) = 9.08 × 107 ft or ANS FIG P1.20 ⎛ 2.83 × 10−2  m ⎞ V = ( 9.08 × 107  ft ) ⎜ ⎟⎠ ⎝ 1 ft    = 2.57 × 106  m © 2014 Cengage Learning All Rights Reserved May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part Chapter P1.21 To find the weight of the pyramid, we use the conversion ton = 000 lbs: Fg = ( 2.50 tons/block ) ( 2.00 × 106  blocks ) ( 000 lb/ton ) = 1.00 × 1010  lbs P1.22 (a) gal ⎛ 30.0 gal ⎞ ⎛ 1 mi ⎞ = 7.14 × 10−2 rate = ⎜ ⎜ ⎟ ⎟ ⎝ 7.00 min ⎠ ⎝ 60 s ⎠ s (b) rate = 7.14 × 10−2 gal ⎛ 231 in ⎞ ⎛ 2.54 cm ⎞ ⎛ 1 m ⎞ ⎜ ⎟ ⎜ ⎟ s ⎜⎝ 1 gal ⎟⎠ ⎝ 1 in ⎠ ⎝ 100 cm ⎠ = 2.70 × 10−4 (c) m3 s To find the time to fill a 1.00-m3 tank, find the rate time/volume: 2.70 × 10−4 *P1.23 m ⎛ 2.70 × 10−4  m ⎞ =⎜ ⎟⎠ s 1 s ⎝ −1 or ⎛ 2.70 × 10−4  m ⎞ ⎜⎝ ⎟⎠ 1 s and so: ⎛ 1 h ⎞ 3.70 × 103  s ⎜ = 1.03 h ⎝ 600 s ⎟⎠ s 1 s ⎛ ⎞ = 3.70 × 103 =⎜ −4 3⎟ ⎝ 2.70 × 10  m ⎠ m It is often useful to remember that the 600-m race at track and field events is approximately mile in length To be precise, there are 609 meters in a mile Thus, acre is equal in area to ⎛ mi ⎞ ⎛ 609 m ⎞ = 4.05 × 103 m ⎟ ⎝ ⎠ ⎝ 640 acres ⎠ mi ( acre ) ⎜ *P1.24 The volume of the interior of the house is the product of its length, width, and height We use the conversion ft = 0.304 m and 100 cm = m V = LWH ⎛ 0.304 m ⎞ ⎛ 0.304 m ⎞ = ( 50.0 ft ) ⎜ × ( 26 ft ) ⎜ ⎟ ⎟⎠ ⎝ ⎠ ⎝ ft ft ⎛ 0.304 m ⎞ × ( 8.0 ft ) ⎜ ⎟⎠ ⎝ ft = 294.5 m = 290 m 3 ⎛ 100 cm ⎞ = ( 294.5 m ) ⎜ = 2.9 × 108 cm ⎝ m ⎟⎠ © 2014 Cengage Learning All Rights Reserved May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part 1224 Particle Physics and Cosmology Let’s check the assumptions If the final particles have any velocity component perpendicular to the initial direction of travel of the photon, then they must be moving with a higher speed after the collision and the incoming photon energy would have to be larger If any one of the particles had a different energy than the other two, then the only way to satisfy both energy and momentum conservation would be for at least two of the particles to have components of velocity perpendicular to the initial direction of motion of the photon, so again the incoming photon energy would have to be larger Therefore, 2.04 MeV represents the minimum energy for the reaction to occur P46.55 We find the number N of neutrinos: 10 46 J = N ( MeV ) = N ( × 1.60 × 10−13 J ) N = 1.0 × 1058 neutrinos The intensity at our location is N ly N 1.0 × 1058 ⎛ ⎞ = = ⎜ ⎟⎠ 2 15 ⎝ A 4π r 9.460 × 10 m 4π ( 1.7 × 10 ly ) = 3.1 × 1014 m −2 The number passing through a body presenting 000 cm2 = 0.50 m2 ⎞ ⎛ 14 14 is then ⎜ 3.1 × 1014 ⎟ ( 0.50 m ) = 1.5 × 10 , or ~ 10 ⎝ m2 ⎠ P46.56 Since the neutrino flux from the Sun reaching the Earth is 0.400 W/m2, the total energy emitted per second by the Sun in neutrinos in all directions is that which would irradiate the surface of a great sphere around it, with the Earth’s orbit as its equator ( 0.400 W/m )( 4π r ) = ( 0.400 W/m ) ⎡⎣ 4π (1.496 × 10 2 11  m ) ⎤ ⎦ = 1.12 × 1023  W In a period of 10 yr, the Sun emits a total energy of ΔE = PΔt E = ( 1.12 × 1023  J/s ) ( 109  yr ) ( 3.156 × 107  s/yr ) = 3.55 × 1039  J carried by neutrinos This energy corresponds to an annihilated mass according to E = mν c = 3.55 × 1039  J or mν = 3.94 × 1022  kg Since the Sun has a mass of 1.989 × 1030 kg, this corresponds to a loss of only about part in × 107 of the Sun’s mass over 109 yr in the form of neutrinos © 2014 Cengage Learning All Rights Reserved May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part Chapter 46 P46.57 1225   In our frame of reference, Hubble’s law is exemplified by v = HR   and v = HR   (a) From the first equation v = HR we may form the equation   − v = −HR This equation expresses Hubble’s law as seen by the observer in the first galaxy cluster, as she looks at us to find our   velocity relative to her (away from her) is − v = H −R ( (b) ) From both equations we may form the equation     v − v = H R − R This equation expresses Hubble’s law as ( ) seen by the observer in the first galaxy cluster, as she looks at cluster two to find the relative velocity of cluster relative to     cluster is v − v = H R − R ( P46.58 ) π − → µ − + ν µ By energy conservation, mπ c = Eµ + Eν = 139.6 MeV [1] Because we assume the antineutrino has no mass, Eν = pν c, and by momentum conservation, pµ = pν ; thus, we can relate the total energies of the muon and antineutrino: ( ) ( ) = ( p c ) + ( m c ) = (E ) + ( m c ) E − E = (m c ) (E + E )(E − E ) = ( m c ) 2 Eµ2 = pµ c + mµ c or and µ ν µ µ 2 2 ν µ 2 2 µ ν ν µ ν 2 µ [2] Substituting [1] into [2], we find that Eµ − Eν (m c ) = (m c ) = (E + E ) m c µ 2 µ µ 2 π ν [3] Subtracting [3] from [1], (E µ ) ( 2Eν = mπ c Eν ) + Eν − Eµ − Eν = mπ c 2 (m c ) − µ µ 2 mπ c 2 mπ c (m c ) − (m c ) = π (m c ) − 2 2mπ c µ 2 ( 139.6 MeV )2 − ( 105.7 MeV )2 = ( 139.6 MeV ) = 29.8 MeV © 2016 Cengage Learning All Rights Reserved May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part 1226 P46.59 Particle Physics and Cosmology Each particle travels in a circle, so each must experience a centripetal force: ∑ F = ma: qvBsin 90° = mv r → mv = qBr The proton and the pion have the same momentum because they have the same magnitude of charge and travel in a circle of the same radius: pp = pπ = p = qBr = ( 1.60 × 10−19 C ) ( 0.250 T ) ( 1.33 m ) = 5.32 × 10−20 kg ⋅ m s so ⎛ MeV ⎞ pc = ( 3.00 × 108 m/s ) ( 5.32 × 10−20 kg ⋅ m s ) ⎜ ⎝ 1.60 × 10−13 J ⎟⎠ = 99.8 MeV Using masses from Table 46.2, we find the total energy of the proton to be Ep = ( pc )2 + ( mpc ) = ( 99.8 MeV ) + ( 938.3 MeV ) 2 = 944 MeV and the total energy of the pion to be Eπ = ( pc )2 + ( mπ c ) = ( 99.8 MeV ) + ( 139.6 MeV ) 2 = 172 MeV The unknown particle was initially at rest; thus, Etotal after = Etotal before = rest energy, and the rest energy of unknown particle is mc = 944 MeV + 172 MeV = 1 116 MeV Mass = 1.12 GeV c From Table 46.2, we see this is a Λ particle P46.60 Each particle travels in a circle, so each must experience a centripetal force: ∑ F = ma: qvBsin 90° = mv r → mv = qBr The particles have the same momentum because they have the same magnitude of charge and travel in a circle of the same radius: p+ = p− = p = eBr → pc = eBrc We find the total energy of the positively charged particle to be E+ , total = ( pc )2 + (E+ )2 = ( qBrc )2 + E+2 © 2014 Cengage Learning All Rights Reserved May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part Chapter 46 1227 and the total energy of the negatively charged particle to be E+ , total = ( pc )2 + (E− )2 = ( qBrc )2 + E−2 The unknown particle was initially at rest; thus, Etotal after = Etotal before = rest energy, and the rest energy of the unknown particle is mc = m= P46.61 ( qBrc )2 + E+2 + ( qBrc )2 + E−2 ( qBrc )2 + E+2 + ( qBrc )2 + E−2 c2 (a) This diagram represents electron–positron annihilation From charge and lepton-number conservation at either vertex, the exchanged particle must be an electron, e− (b) A neutrino collides with a neutron, producing a proton and a muon This is a weak interaction The exchanged particle has charge +e and is a W + ANS FIG P46.61 P46.62 (a) The Feynman diagram in ANS FIG P46.62 shows a neutrino scattering off an electron, and the neutrino and electron not exchange electric charge The neutrino has no electric charge and interacts through the weak interaction (ignoring gravity) The mediator is a Z boson (b) ANS FIG P46.62 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, if the quarks have opposite color charges, no color charge In this case © 2016 Cengage Learning All Rights Reserved May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part 1228 Particle Physics and Cosmology the mediating particle could be a photon or Z boson Depending on the color charges of the d and d quarks, the ephemeral particle could also be a gluon , as suggested in the discussion of Figure 46.13(b) For conservation of both energy and momentum in the collision we would expect two mediating particles; but momentum need not be strictly conserved, according to the uncertainty principle, if the particle travels a sufficiently short distance before producing another matter-antimatter pair of particles, as shown in ANS FIG P46.62(b) P46.63 The expression e −E kB T 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 ∞ ∫e E ∞ ∫e −E kB T −E kB T dE dE ∞ = ∫e E ∞ ∫e −E kB T −E kB T ( −dE ( −dE k BT ) ∞ = k BT ) e −E kB T E e −E kB T ∞ = e −E kB T We require T when this fraction has a value of 0.010 (i.e., 1.00%) P46.64 and E = 1.00 eV = 1.60 × 10−19 J Thus, − 1.60×10−19 J ) ( 1.38×10−23 0.010 0 = e ( or ln ( 0.010 0 ) = − giving T = 2.52 × 103 K ~ 103 K ) J K T 1.60 × 10−19 J 1.16 × 10 K = − , T (1.38 × 10−23 J K )T Σ0 → Λ0 + γ From Table 46.2, mΣ = 192.5 MeV c and mΛ = 1 115.6 MeV c Conservation of energy in the decay requires mΣ c = ( mΛ c + K Λ ) + Eγ or ⎛ p ⎞ mΣ c = ⎜ mΛ c + Λ ⎟ + Eγ 2mΛ ⎠ ⎝ System momentum conservation gives pΛ = pγ , so the last result may be written as ⎛ pγ ⎞ mΣ c = ⎜ mΛ c + + Eγ 2mΛ ⎟⎠ ⎝ © 2014 Cengage Learning All Rights Reserved May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part Chapter 46 or 1229 ⎛ pγ2 c ⎞ mΣ c = ⎜ mΛ c + + Eγ 2mΛ c ⎟⎠ ⎝ Recognizing that pγ c = Eγ , we now have 192.5 MeV = 115.6 MeV + Eγ2 ( 115.6 MeV ) + Eγ Solving this quadratic equation gives Eγ = 74.4 MeV P46.65 p + p → p +π+ + X The protons each have 70.4 MeV of kinetic energy In accord with conservation of momentum for the collision, particle X has zero momentum and thus zero kinetic energy Conservation of system energy then requires ( ) ( mp c + mπ c + mX c = mp c + K p + mp c + K p ) mX c = mp c + 2K p − 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 P46.66 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 system momentum This will be the case when the incident protons have minimum kinetic energy Under these conditions, conservation of energy for the reaction gives ( ) mp c + K p = mp c + mnc + mπ c so the kinetic energy of each of the incident protons is Kp = mnc + mπ c − mp c 2 = 70.4 MeV = ( 939.6 + 139.6 − 938.3) MeV © 2016 Cengage Learning All Rights Reserved May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part 1230 Particle Physics and Cosmology Challenge Problems P46.67 See the discussion of P46.19 in this volume for more details of the mathematical steps used in the following calculations From Table 46.2, mΛ c = 1 115.6 MeV, mp c = 938.3 MeV, and mπ c = 139.6 MeV Since the Λ is at rest, the difference between its rest energy and the rest energies of the proton and the pion is the sum of the kinetic energies of the proton and the pion K p + Kπ = 1 115.6 MeV − 938.3 MeV − 139.6 MeV = 37.7 MeV Now, since p p = pπ = p, applying conservation of relativistic energy to the decay process, we have ⎡ ( 938.3 MeV )2 + p c − 938.3 MeV ⎤ ⎣ ⎦ + ⎡⎣ ( 139.6 MeV )2 + p c − 139.6 MeV ⎤⎦ = 37.7 MeV Solving yields pπ c = p p c = 100.4 MeV Then, Kp = (m c ) p 2 + ( 100.4 MeV )2 − mp c = 5.35 MeV Kπ = ( 139.6 )2 + ( 100.4 MeV )2 − 139.6 = 32.3 MeV P46.68 (a) Let Emin be the minimum total energy of the bombarding particle that is needed to induce the reaction At this energy the product particles all move with the same velocity The product particles are then equivalent to a single particle having mass equal to the total mass of the product particles, moving with the same velocity as each product particle By conservation of energy: Emin + m2 c = ( m3c )2 + ( p3c )2 [1] By conservation of momentum, p3 = p1 , so − ( m1c ) ( p3c )2 = ( p1c )2 = Emin [2] Substitute [2] into [1]: Emin + m2 c = 2 − ( m1c ) ( m3c )2 + Emin © 2014 Cengage Learning All Rights Reserved May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part Chapter 46 1231 Square both sides: 2 Emin + 2Emin m2 c + ( m2 c ) = ( m3 c ) + Emin − ( m1c ) ∴Emin = 2 ( m32 − m12 − m22 ) c 2m2 ∴ K = Emin − m1c m32 − m12 − m22 − 2m1m2 ) c ( = 2m2 ⎡ m32 − ( m1 + m2 ) ⎤⎦ c =⎣ 2m2 Refer to Table 46.2 for the particle masses (b) K = (c) K = [ ( 938.3 )]2 MeV c − [ ( 938.3 )]2 MeV c 2 ( 938.3 MeV c ) = 5.63 GeV ( 497.7 + 1 115.6 )2 MeV c − ( 139.6 + 938.3 )2 MeV c 2 ( 938.3 ) MeV c = 768 MeV (d) K = [ ( 938.3 ) + 135 ]2 MeV c − [ ( 938.3 )]2 MeV c 2 ( 938.3 ) MeV c = 280 MeV P46.69 91.2 × 103 ) − ⎡⎣( 938.3 + 938.3 )2 ⎤⎦ ( = (e) K (a) ΔE = mn − mp − me c MeV c 2 ( 938.3 ) MeV c ( = 4.43 TeV ) From Table 44.2 of masses of isotopes, ΔE = ( 1.008 665 u − 1.007 825 u ) ( 931.5 MeV/u ) = 0.782 MeV (b) Assuming the neutron at rest, momentum conservation for the decay process implies pp = pe Relativistic energy for the system is conserved: (m c ) p 2 + pp2 c + (m c ) e 2 + pe2 c = mn c Since pp = pe = p, we have (m c ) p 2 + p c = mn c − (m c ) e 2 + p2c2 © 2016 Cengage Learning All Rights Reserved May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part 1232 Particle Physics and Cosmology (m c ) p 2 + p c = ( mnc ) − 2mn c 2 (m c ) + p c + (m c ) + p c ) + (m c ) e 2 2 e (m c ) e 2 +p c 2 (m c ) − (m c = n 2 p 2 e 2 2 2 2mn c ( ) ⎡ ( m c )2 − m c 2 + ( m c )2 ⎤ n p e ⎥ − ( m c )2 p2c2 = ⎢ e ⎢ ⎥ 2mn c ⎣ ⎦ Refer to Table 46.2 for the particle masses ⎡ ( 939.6 MeV )2 − ( 938.3 MeV )2 + ( 0.511 MeV )2 ⎤ p c =⎢ ⎥ ( 939.6 MeV ) ⎣ ⎦ 2 − ( 0.511 MeV ) pc = 1.19 MeV From pe c = γ me ve c, we find the speed of the electron: pc γ ve ve = e2 = me c c − ( ve c ) c 2 2 ⎛v ⎞ ⎛v ⎞ ⎛mc ⎞ ⎛v ⎞ ⎡ ⎛mc ⎞ − ⎜ e ⎟ = ⎜ e ⎟ ⎜ e ⎟ → ⎜ e ⎟ ⎢1 + ⎜ e ⎟ ⎝ c⎠ ⎝ c ⎠ ⎝ pe c ⎠ ⎝ c ⎠ ⎢ ⎝ pe c ⎠ ⎣ 2 ⎤ ⎥=1 ⎥⎦ ve 1 = = 2 c + ( 0.511 MeV 1.19 MeV ) + ( me c pe c ) ve = 0.919c To find the speed of the proton, a similar derivation (basically, substituting mp for me), yields vp = c ( + mp c pe c ) = 2.998 × 108 m/s + ( 938.3 MeV 1.19 MeV ) = 3.82 × 105 m/s = 382 km/s (c) The electron is relativistic; the proton is not Our criterion for answers accurate to three significant digits is that the electron is moving at more than one-tenth the speed of light and the proton at less than one-tenth the speed of light © 2014 Cengage Learning All Rights Reserved May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part Chapter 46 P46.70 (a) 1233 At threshold, we consider a photon and a proton colliding headon to produce a proton and a pion at rest, according to p + γ → p + π Energy conservation gives mp c 1− u c 2 + Eγ = mp c + mπ c Momentum conservation gives mp u − u2 c − Eγ c = Combining the equations, we have mp c 1− u c 2 + mp c u = mp c + mπ c 1− u c c ( 938.3 MeV ) ( + u c ) (1 − u c )(1 + u c ) so = 938.3 MeV + 135.0 MeV u = 0.134 c and Eγ = 127 MeV (b) λmaxT = 2.898 mm ⋅ K λmax = (c) 2.898 mm ⋅ K = 1.06 mm 2.73 K hc 1 240 eV ⋅ 10−9 m Eγ = hf = = = 1.17 × 10−3 eV -3 λ 1.06 × 10 m (d) In the primed reference frame, the proton is moving to the right at u′ = 0.134 and the photon is moving to the left with c hf ′ = 1.27 × 108 eV In the unprimed frame, hf = 1.17 × 10−3 eV Using the Doppler effect equation (Equation 39.10), we have for the speed of the primed frame (suppressing units) 1.27 × 108 = 1+ v c 1.17 × 10−3 1− v c v = − 1.71 × 10−22 c Then the speed of the proton is given by u u′ c + v c 0.134 + − 1.71 × 10−22 = = = − 1.30 × 10−22 −22 c + u′v c + 0.134 ( − 1.71 × 10 ) © 2016 Cengage Learning All Rights Reserved May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part 1234 Particle Physics and Cosmology And the energy of the proton is mp c 1− u c 2 = 938.3 MeV − ( − 1.30 × 10−22 ) = 6.19 × 1010 × 938.3 × 106 eV = 5.81 × 1019 eV P46.71 (a) Consider a sphere around us of radius R large compared to the size of galaxy clusters If the matter M inside the sphere has the critical density, then a galaxy of mass m at the surface of the sphere is moving just at escape speed v according to K + U g = 0, GMm mv − = R or The energy of the galaxy-sphere system is conserved, so this equation is true throughout the history of the Universe after the dR Big Bang, where v = Then, dt 2GM ⎛ dR ⎞ ⎜⎝ d t ⎟⎠ = R dR = R −1/2 2GM dt or integrating, R T ∫0 R dR = 2GM ∫0 d t R 3/2 32 R T = 2GM t gives 3/2 R = 2GM T R3 2 R T= = 2GM 2GM R or 2GM =v R From above, T= so 2R 3v Now Hubble’s law says v = HR, so T = (b) T= ( 22 × 10−3 R = HR 3H ⎛ 2.998 × 108 m s ⎞ ⎟⎠ = 9.08 × 10 yr m s ⋅ ly ) ⎜⎝ ly yr = 9.08 billion years © 2014 Cengage Learning All Rights Reserved May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part Chapter 46 P46.72 1235 A photon travels the distance from the Large Magellanic Cloud to us in 170 000 years The hypothetical massive neutrino travels the same distance in 170 000 years plus 10 seconds: c ( 170 000 yr ) = v ( 170 000 yr + 10 s ) 170 000 yr v = c 170 000 yr + 10 s = 1 + {10 s [( 1.7 × 10 yr ) ( 3.156 × 107 s yr )]} = 1 + 1.86 × 10−12 For the neutrino we want to evaluate mc2 in E = γ mc : mc = E v2 = E − = 10 MeV − γ c (1 + 1.86 × 10−12 )2 (1 + 1.86 × 10−12 )2 − (1 + 1.86 × 10−12 )2 = ( 10 MeV ) ( 1.86 × 10−12 ) = ( 10 MeV ) ( 1.93 × 10−6 ) mc ≈ ( 10 MeV ) = 19 eV Then the upper limit on the mass is m= m= P46.73 (a) 19 eV c2 ⎞ 19 eV ⎛ u = 2.1 × 10−8 u 2⎟ ⎜ c ⎝ 931.5 × 10 eV c ⎠ If 2N particles are annihilated, the energy released is 2Nmc2 The E 2Nmc = 2Nmc Since the resulting photon momentum is p = = c c momentum of the system is conserved, the rocket will have momentum 2Nmc directed opposite the photon momentum p = 2Nmc (b) Consider a particle that is annihilated and gives up its rest energy mc2 to another particle which also has initial rest energy mc2 (but no momentum initially) E = p c + ( mc ) © 2016 Cengage Learning All Rights Reserved May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part 1236 Particle Physics and Cosmology Thus, ( 2mc ) = p c + ( mc ) 2 Where p is the momentum the second particle acquires as a result of the annihilation of the first particle Thus ( mc ) = p c + ( mc ) , p = ( mc ) So p = 3mc 2 N N protons and 2 antiprotons) Thus the total momentum acquired by the ejected particles is 3Nmc, and this momentum is imparted to the rocket This process is repeated N times (annihilate p = 3Nmc (c) Method (a) produces greater speed since 2Nmc > 3Nmc © 2014 Cengage Learning All Rights Reserved May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part Chapter 46 1237 ANSWERS TO EVEN-NUMBERED PROBLEMS P46.2 (a) 2.27 × 1023 Hz; (b) 1.32 × 10−15 m P46.4 ~103 Bq P46.6 ~10−18 m P46.8 Baryon number conservation allows the first reaction and forbids the second P46.10 ~10−23 s P46.12 (a) See P46.12(a) for full explanation; (b) Strangeness is not conserved in the second reaction P46.14 (a) ν e ; (b) ν µ ; (c) ν µ ; (d) ν µ , ντ P46.16 ν µ and ν e P46.18 (a) See P46.18(a) for full explanation; (b) Ee = Eγ = 469 MeV, pe = pγ = 469 MeV c ; (c) v = 0.999 999 4c P46.20 The ρ → π + + π − decay must occur via the strong interaction The K 0S → π + + π − decay must occur via the weak interaction P46.22 (a) electron and muon lepton numbers; (b) electron lepton number; (c) charge and strangeness; (d) baryon number; (e) strangeness P46.24 (a) B, charge, Le , and Lτ ; (b) B, charge, Le , Lµ , and Lτ ; (c) S, charge, Le , Lµ , and Lτ ; (d) B, S, charge, Le , Lµ , and Lτ ; (e) B, S, charge, Le , Lµ , and Lτ ; (f) B, S, charge, Le , Lµ , and Lτ P46.26 (a) pΣ+ = 686 MeV c , pπ + = 200 MeV c ; (b) 626 MeV/c; (c) Eπ + = 244 MeV, En = 1.13 GeV; (d) 1.37 GeV; (e) 1.19 GeV/c2; (f) The result in part (e) is within 0.05% of the value ion Table 46.2 P46.28 (a) See table in P46.28(a); (b) See table in P46.28(b) P46.30 (a) Σ + ; (b) π − ; (c) K ; (d) Ξ− P46.32 (a) The reaction has a net of 3u, 0d, and 0s before and after; (b) The reaction has a net of 1u, 1d, and 1s before and after; (c) The reaction must net of 4u, 2d, and 0z before and after; (d) Λ or Σ P46.34 3.34 × 1026 electrons , 9.36 × 1026 up quarks , 8.70 × 1026 down quarks P46.36 mu = 312 MeV c ; md = 314 MeV c © 2016 Cengage Learning All Rights Reserved May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part 1238 P46.38 P46.40 Particle Physics and Cosmology (a) 4.30 × 10−18 m s ; (b) 0.892 nm/s –24 (a) 8.41 × 106 kg ; (b) No It is only the fraction 4.23 × 10 of the mass of the Sun P46.42 1.36 × 1010 yr P46.44 3.15 × 10−6 W m P46.46 ⎛ Z + 2Z ⎞ c ⎛ Z + 2Z ⎞ (a) c ⎜ ; (b) ⎟ H ⎜⎝ Z + 2Z + ⎟⎠ ⎝ Z + 2Z + ⎠ P46.48 P46.50 (a) See P46.48(a) for full explanation; (b) 6.34 × 1061 m /s (a) 1.62 × 10−35 m; (b) 5.39 × 10−44 s P46.52 νµ P46.54 See P46.54 for full explanation P46.56 part in × 107 P46.58 29.8 MeV ( qBrc ) + E+2 + ( qBrc ) + E−2 P46.60 m= P46.62 (a) Z boson; (b) photon or Z boson, gluon P46.64 74.4 MeV P46.66 70.4 MeV P46.68 (a) See P46.68(a) for full explanation; (b) 5.63 GeV; (c) 768 MeV; (d) 280 MeV; (e) 4.43 TeV P46.70 (a) 127 MeV; (b) 1.06 mm; (c) 1.17 × 10–3 eV; (d) 5.81 × 1019 eV; P46.72 19 eV/c2 c2 © 2014 Cengage Learning All Rights Reserved May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part ...1 Physics and Measurement CHAPTER OUTLINE 1.1 Standards of Length, Mass, and Time 1.2 Matter and Model Building 1.3 Dimensional Analysis 1.4 Conversion of Units 1.5 Estimates and Order-of-Magnitude... in part 4 Physics and Measurement P1.5 For either sphere the volume is V = π r and the mass is m = ρV = ρ π r We divide this equation for the larger sphere by the same equation for the smaller:... equation y = (2 m) cos (kx) For y, [y] = L for m, [2 m] = L and for (kx), [ kx] = ⎡⎣ m –1 x ⎤⎦ = L–1L ( ) Therefore we can think of the quantity kx as an angle in radians, and we can take its cosine