Solutions manual for fundamentals of physic 9th by halliday

Solutions manual for fundamentals of physic 9th by halliday Solutions manual for fundamentals of physic 9th by halliday Solutions manual for fundamentals of physic 9th by halliday Solutions manual for fundamentals of physic 9th by halliday Solutions manual for fundamentals of physic 9th by halliday Solutions manual for fundamentals of physic 9th by halliday Solutions manual for fundamentals of physic 9th by halliday Solutions manual for fundamentals of physic 9th by halliday Solutions manual for fundamentals of physic 9th by halliday

SOLUTION MANUAL FOR c2011 VOLUME PART 1 Measurement Motion Along a Straight Line Vectors Motion in Two and Three Dimensions Force and Motion — I Force and Motion — II Kinetic Energy and Work Potential Energy and Conservation of Energy Center of Mass and Linear Momentum 10 Rotation 11 Rolling, Torque, and Angular Momentum PART 12 Equilibrium and Elasticity 13 Gravitation 14 Fluids 15 Oscillations 16 Waves — I 17 Waves — II 18 Temperature, Heat, and the First Law of Thermodynamics 19 The Kinetic Theory of Gases 20 Entropy and the Second Law of Thermodynamics VOLUME PART 21 Electric Charge 22 Electric Fields 23 Gauss’ Law 24 Electric Potential 25 Capacitance 26 Current and Resistance 27 Circuits 28 Magnetic Fields 29 Magnetic Fields Due to Currents 30 Induction and Inductance 31 Electromagnetic Oscillations and Alternating Current 32 Maxwell’s Equations; Magnetism of Matter PART 33 Electromagnetic Waves 34 Images 35 Interference 36 Diffraction 37 Relativity PART 38 Photons and Matter Waves 39 More About Matter Waves 40 All About Atoms 41 Conduction of Electricity in Solids 42 Nuclear Physics 43 Energy from the Nucleus 44 Quarks, Leptons, and the Big Bang   Chapter 1 Various geometric formulas are given in Appendix E (a) Expressing the radius of the Earth as R = ( 6.37 × 106 m )(10−3 km m ) = 6.37 × 103 km, its circumference is s = 2π R = 2π (6.37 × 103 km) = 4.00 × 104 km (b) The surface area of Earth is A = 4π R = 4π ( 6.37 × 103 km ) = 5.10 × 108 km (c) The volume of Earth is V = π 4π R = 6.37 × 103 km 3 ( ) = 1.08 × 1012 km3 The conversion factors are: gry = 1/10 line , line = 1/12 inch and point = 1/72 inch The factors imply that gry = (1/10)(1/12)(72 points) = 0.60 point Thus, gry2 = (0.60 point)2 = 0.36 point2, which means that 0.50 gry = 0.18 point The metric prefixes (micro, pico, nano, …) are given for ready reference on the inside front cover of the textbook (see also Table 1–2) (a) Since km = × 103 m and m = × 106 μm, ( )( ) 1km = 103 m = 103 m 106 μ m m = 109 μ m The given measurement is 1.0 km (two significant figures), which implies our result should be written as 1.0 × 109 μm (b) We calculate the number of microns in centimeter Since cm = 10−2 m, ( )( ) 1cm = 10−2 m = 10−2 m 106 μ m m = 104 μ m We conclude that the fraction of one centimeter equal to 1.0 μm is 1.0 × 10−4 (c) Since yd = (3 ft)(0.3048 m/ft) = 0.9144 m, CHAPTER ( ) 1.0 yd = ( 0.91m ) 106 μ m m = 9.1 × 105 μ m (a) Using the conversion factors inch = 2.54 cm exactly and picas = inch, we obtain ⎛ inch ⎞ ⎛ picas ⎞ 0.80 cm = ( 0.80 cm ) ⎜ ⎟⎜ ⎟ ≈ 1.9 picas ⎝ 2.54 cm ⎠ ⎝ inch ⎠ (b) With 12 points = pica, we have ⎛ inch ⎞ ⎛ picas ⎞ ⎛ 12 points ⎞ 0.80 cm = ( 0.80 cm ) ⎜ ⎟⎜ ⎟⎜ ⎟ ≈ 23 points ⎝ 2.54 cm ⎠ ⎝ inch ⎠ ⎝ pica ⎠ Given that furlong = 201.168 m , rod = 5.0292 m and chain = 20.117 m , we find the relevant conversion factors to be rod 1.0 furlong = 201.168 m = (201.168 m ) = 40 rods, 5.0292 m and chain 1.0 furlong = 201.168 m = (201.168 m ) = 10 chains 20.117 m Note the cancellation of m (meters), the unwanted unit Using the given conversion factors, we find (a) the distance d in rods to be d = 4.0 furlongs = ( 4.0 furlongs ) 40 rods = 160 rods, furlong (b) and that distance in chains to be d = 4.0 furlongs = ( 4.0 furlongs ) 10 chains = 40 chains furlong We make use of Table 1-6 (a) We look at the first (“cahiz”) column: fanega is equivalent to what amount of cahiz? We note from the already completed part of the table that cahiz equals a dozen fanega Thus, fanega = 12 cahiz, or 8.33 × 10−2 cahiz Similarly, “1 cahiz = 48 cuartilla” (in the already completed part) implies that cuartilla = 48 cahiz, or 2.08 × 10−2 cahiz Continuing in this way, the remaining entries in the first column are 6.94 × 10−3 and 3.47 ×10−3 (b) In the second (“fanega”) column, we find 0.250, 8.33 × 10−2, and 4.17 × 10−2 for the last three entries (c) In the third (“cuartilla”) column, we obtain 0.333 and 0.167 for the last two entries (d) Finally, in the fourth (“almude”) column, we get = 0.500 for the last entry (e) Since the conversion table indicates that almude is equivalent to medios, our amount of 7.00 almudes must be equal to 14.0 medios (f) Using the value (1 almude = 6.94 × 10−3 cahiz) found in part (a), we conclude that 7.00 almudes is equivalent to 4.86 × 10−2 cahiz (g) Since each decimeter is 0.1 meter, then 55.501 cubic decimeters is equal to 0.055501 7.00 7.00 m3 or 55501 cm3 Thus, 7.00 almudes = 12 fanega = 12 (55501 cm3) = 3.24 × 104 cm3 We use the conversion factors found in Appendix D acre ⋅ ft = (43,560 ft ) ⋅ ft = 43,560 ft Since in = (1/6) ft, the volume of water that fell during the storm is V = (26 km )(1/6 ft) = (26 km )(3281ft/km) (1/6 ft) = 4.66 × 107 ft Thus, 4.66 × 107 ft × 103 acre ⋅ ft V = = 11 4.3560 × 10 ft acre ⋅ ft From Fig 1-4, we see that 212 S is equivalent to 258 W and 212 – 32 = 180 S is equivalent to 216 – 60 = 156 Z The information allows us to convert S to W or Z (a) In units of W, we have ⎛ 258 W ⎞ 50.0 S = ( 50.0 S) ⎜ ⎟ = 60.8 W ⎝ 212 S ⎠ (b) In units of Z, we have ⎛ 156 Z ⎞ 50.0 S = ( 50.0 S) ⎜ ⎟ = 43.3 Z ⎝ 180 S ⎠ The volume of ice is given by the product of the semicircular surface area and the thickness The area of the semicircle is A = πr2/2, where r is the radius Therefore, the volume is CHAPTER V = π r z where z is the ice thickness Since there are 103 m in km and 102 cm in m, we have ⎛ 103 m ⎞ r = ( 2000 km ) ⎜ ⎟ ⎝ 1km ⎠ ⎛ 102 cm ⎞ ⎜ ⎟ = 2000 × 10 cm ⎝ 1m ⎠ In these units, the thickness becomes ⎛ 102 cm ⎞ z = 3000 m = ( 3000 m ) ⎜ ⎟ = 3000 × 10 cm 1m ⎝ ⎠ which yields V = π 2000 × 105 cm ( ) ( 3000 × 10 2 ) cm = 1.9 × 1022 cm3 10 Since a change of longitude equal to 360° corresponds to a 24 hour change, then one expects to change longitude by 360° / 24 = 15° before resetting one's watch by 1.0 h 11 (a) Presuming that a French decimal day is equivalent to a regular day, then the ratio of weeks is simply 10/7 or (to significant figures) 1.43 (b) In a regular day, there are 86400 seconds, but in the French system described in the problem, there would be 105 seconds The ratio is therefore 0.864 12 A day is equivalent to 86400 seconds and a meter is equivalent to a million micrometers, so 3.7 m 106 μ m m = 31 μ m s 14 day 86400 s day b b gc gb h g 13 The time on any of these clocks is a straight-line function of that on another, with slopes ≠ and y-intercepts ≠ From the data in the figure we deduce tC = 594 tB + , 7 tB = 33 662 tA − 40 These are used in obtaining the following results (a) We find t B′ − t B = when t'A − tA = 600 s 33 ( t ′A − t A ) = 495 s 40 (b) We obtain t C′ − t C = b g b g 2 495 = 141 s t B′ − t B = 7 (c) Clock B reads tB = (33/40)(400) − (662/5) ≈ 198 s when clock A reads tA = 400 s (d) From tC = 15 = (2/7)tB + (594/7), we get tB ≈ −245 s 14 The metric prefixes (micro (μ), pico, nano, …) are given for ready reference on the inside front cover of the textbook (also Table 1–2) ⎛ 100 y ⎞ ⎛ 365 day ⎞ ⎛ 24 h ⎞ ⎛ 60 ⎞ (a) μ century = (10−6 century ) ⎜ ⎟⎜ ⎟⎜ ⎟⎜ ⎟ = 52.6 century y day h ⎝ ⎠⎝ ⎠⎝ ⎠⎝ ⎠ (b) The percent difference is therefore 52.6 − 50 = 4.9% 52.6 15 A week is days, each of which has 24 hours, and an hour is equivalent to 3600 seconds Thus, two weeks (a fortnight) is 1209600 s By definition of the micro prefix, this is roughly 1.21 × 1012 μs 16 We denote the pulsar rotation rate f (for frequency) f = rotation 1.55780644887275 × 10−3 s (a) Multiplying f by the time-interval t = 7.00 days (which is equivalent to 604800 s, if we ignore significant figure considerations for a moment), we obtain the number of rotations: ⎛ ⎞ rotation N =⎜ ⎟ ( 604800 s ) = 388238218.4 −3 × 1.55780644887275 10 s ⎝ ⎠ which should now be rounded to 3.88 × 108 rotations since the time-interval was specified in the problem to three significant figures (b) We note that the problem specifies the exact number of pulsar revolutions (one million) In this case, our unknown is t, and an equation similar to the one we set up in part (a) takes the form N = ft, or ⎛ rotation × 106 = ⎜ −3 ⎝ 1.55780644887275 × 10 ⎞ ⎟t s⎠ CHAPTER which yields the result t = 1557.80644887275 s (though students who this calculation on their calculator might not obtain those last several digits) (c) Careful reading of the problem shows that the time-uncertainty per revolution is ± ×10− 17 s We therefore expect that as a result of one million revolutions, the uncertainty should be ( ± × 10−17 )(1× 106 )= ± ×10− 11 s 17 None of the clocks advance by exactly 24 h in a 24-h period but this is not the most important criterion for judging their quality for measuring time intervals What is important is that the clock advance by the same amount in each 24-h period The clock reading can then easily be adjusted to give the correct interval If the clock reading jumps around from one 24-h period to another, it cannot be corrected since it would impossible to tell what the correction should be The following gives the corrections (in seconds) that must be applied to the reading on each clock for each 24-h period The entries were determined by subtracting the clock reading at the end of the interval from the clock reading at the beginning CLOCK A B C D E Sun -Mon −16 −3 −58 +67 +70 Mon -Tues −16 +5 −58 +67 +55 Tues -Wed −15 −10 −58 +67 +2 Wed -Thurs −17 +5 −58 +67 +20 Thurs -Fri −15 +6 −58 +67 +10 Fri -Sat −15 −7 −58 +67 +10 Clocks C and D are both good timekeepers in the sense that each is consistent in its daily drift (relative to WWF time); thus, C and D are easily made “perfect” with simple and predictable corrections The correction for clock C is less than the correction for clock D, so we judge clock C to be the best and clock D to be the next best The correction that must be applied to clock A is in the range from 15 s to 17s For clock B it is the range from -5 s to +10 s, for clock E it is in the range from -70 s to -2 s After C and D, A has the smallest range of correction, B has the next smallest range, and E has the greatest range From best to worst, the ranking of the clocks is C, D, A, B, E 18 The last day of the 20 centuries is longer than the first day by ( 20 century ) ( 0.001 s century ) = 0.02 s The average day during the 20 centuries is (0 + 0.02)/2 = 0.01 s longer than the first day Since the increase occurs uniformly, the cumulative effect T is T = ( average increase in length of a day )( number of days ) ⎛ 0.01 s ⎞ ⎛ 365.25 day ⎞ =⎜ ⎟⎜ ⎟ ( 2000 y ) y ⎝ day ⎠ ⎝ ⎠ = 7305 s or roughly two hours 19 When the Sun first disappears while lying down, your line of sight to the top of the Sun is tangent to the Earth’s surface at point A shown in the figure As you stand, elevating your eyes by a height h, the line of sight to the Sun is tangent to the Earth’s surface at point B Let d be the distance from point B to your eyes From the Pythagorean theorem, we have d + r = (r + h) = r + 2rh + h or d = 2rh + h , where r is the radius of the Earth Since r  h , the second term can be dropped, leading to d ≈ 2rh Now the angle between the two radii to the two tangent points A and B is θ, which is also the angle through which the Sun moves about Earth during the time interval t = 11.1 s The value of θ can be obtained by using θ 360 ° This yields θ= = t 24 h (360°)(11.1 s) = 0.04625° (24 h)(60 min/h)(60 s/min) Using d = r tan θ , we have d = r tan θ = 2rh , or r= 2h tan θ Using the above value for θ and h = 1.7 m, we have r = 5.2 ×106 m 1626 CHAPTER 44 which implies that the difference between v and c is ⎛ ⎛ ⎛ ⎞ c − v = c ⎜⎜1 − − ⎟⎟ ≈ c ⎜1 − ⎜ − + γ ⎠ ⎝ ⎝ 2γ ⎝ ⎞⎞ ⎟⎟ ⎠⎠ where we use the binomial expansion (see Appendix E) in the last step Therefore, ⎛ ⎞ ⎛ ⎞ c − v ≈ c ⎜ ⎟ = (299792458 m s) ⎜ = 0.0266 m s ≈ 2.7 cm s ⎟ ⎝ 2γ ⎠ ⎝ 2(75000) ⎠ 10 From Eq 37-52, the Lorentz factor is γ = 1+ 80 MeV K =1+ = 1.59 mc 135 MeV Solving Eq 37-8 for the speed, we find γ= 1− ( v c ) ⇒ v = c 1− γ2 which yields v = 0.778c or v = 2.33 × 108 m/s Now, in the reference frame of the laboratory, the lifetime of the pion is not the given τ value but is “dilated.” Using Eq 37-9, the time in the lab is c h t = γτ = (159 ) 8.3 × 10−17 s = 13 × 10−16 s Finally, using Eq 37-10, we find the distance in the lab to be x = vt = ( 2.33 × 108 m s ) (1.3 ×10−16 s ) = 3.1× 10−8 m 11 (a) The conservation laws considered so far are associated with energy, momentum, angular momentum, charge, baryon number, and the three lepton numbers The rest energy of the muon is 105.7 MeV, the rest energy of the electron is 0.511 MeV, and the rest energy of the neutrino is zero Thus, the total rest energy before the decay is greater than the total rest energy after The excess energy can be carried away as the kinetic energies of the decay products and energy can be conserved Momentum is conserved if the electron and neutrino move away from the decay in opposite directions with equal magnitudes of momenta Since the orbital angular momentum is zero, we consider only spin angular momentum All the particles have spin = / The total angular momentum after the decay must be either = (if the spins are aligned) or zero (if the spins are antialigned) Since the spin before the decay is / 2, angular momentum cannot be 1627 conserved The muon has charge –e, the electron has charge –e, and the neutrino has charge zero, so the total charge before the decay is –e and the total charge after is –e Charge is conserved All particles have baryon number zero, so baryon number is conserved The muon lepton number of the muon is +1, the muon lepton number of the muon neutrino is +1, and the muon lepton number of the electron is Muon lepton number is conserved The electron lepton numbers of the muon and muon neutrino are and the electron lepton number of the electron is +1 Electron lepton number is not conserved The laws of conservation of angular momentum and electron lepton number are not obeyed and this decay does not occur (b) We analyze the decay μ − → e + +ν e + ν μ in the same way We find that charge and the muon lepton number Lμ are not conserved (c) For the process μ + → π + +ν μ , we find that energy cannot be conserved because the mass of muon is less than the mass of a pion Also, muon lepton number Lμ is not conserved 12 (a) Noting that there are two positive pions created (so, in effect, its decay products are doubled), then we count up the electrons, positrons, and neutrinos: 2e + + e − + 5v + 4v (b) The final products are all leptons, so the baryon number of A2+ is zero Both the pion and rho meson have integer-valued spins, so A2+ is a boson (c) A2+ is also a meson (d) As stated in (b), the baryon number of A2+ is zero 13 The formula for Tz as it is usually written to include strange baryons is Tz = q – (S + B)/2 Also, we interpret the symbol q in the Tz formula in terms of elementary charge units; this is how q is listed in Table 44-3 In terms of charge q as we have used it in previous chapters, the formula is q Tz = − ( B + S ) e For instance, Tz = + 12 for the proton (and the neutral Xi) and Tz = − 12 for the neutron (and the negative Xi) The baryon number B is +1 for all the particles in Fig 44-4(a) Rather than use a sloping axis as in Fig 44-4 (there it is done for the q values), one reproduces (if one uses the “corrected” formula for Tz mentioned above) exactly the same pattern using regular rectangular axes (Tz values along the horizontal axis and Y values along the vertical) with the neutral lambda and sigma particles situated at the origin 14 (a) From Eq 37-50, 1628 CHAPTER 44 Q = −Δmc = (mΣ+ + mK + − mπ+ − m p )c = 1189.4MeV + 493.7MeV − 139.6MeV − 938.3MeV = 605MeV (b) Similarly, Q = − Δmc = ( mΛ0 + mπ − mK − − m p )c = 1115.6 MeV + 135.0 MeV − 493.7 MeV − 938.3 MeV = −181 MeV 15 (a) The lambda has a rest energy of 1115.6 MeV, the proton has a rest energy of 938.3 MeV, and the kaon has a rest energy of 493.7 MeV The rest energy before the decay is less than the total rest energy after, so energy cannot be conserved Momentum can be conserved The lambda and proton each have spin = / and the kaon has spin zero, so angular momentum can be conserved The lambda has charge zero, the proton has charge +e, and the kaon has charge –e, so charge is conserved The lambda and proton each have baryon number +1, and the kaon has baryon number zero, so baryon number is conserved The lambda and kaon each have strangeness –1 and the proton has strangeness zero, so strangeness is conserved Only energy cannot be conserved (b) The omega has a rest energy of 1680 MeV, the sigma has a rest energy of 1197.3 MeV, and the pion has a rest energy of 135 MeV The rest energy before the decay is greater than the total rest energy after, so energy can be conserved Momentum can be conserved The omega and sigma each have spin = / and the pion has spin zero, so angular momentum can be conserved The omega has charge –e, the sigma has charge –e, and the pion has charge zero, so charge is conserved The omega and sigma have baryon number +1 and the pion has baryon number 0, so baryon number is conserved The omega has strangeness –3, the sigma has strangeness –1, and the pion has strangeness zero, so strangeness is not conserved (c) The kaon and proton can bring kinetic energy to the reaction, so energy can be conserved even though the total rest energy after the collision is greater than the total rest energy before Momentum can be conserved The proton and lambda each have spin = and the kaon and pion each have spin zero, so angular momentum can be conserved The kaon has charge –e, the proton has charge +e, the lambda has charge zero, and the pion has charge +e, so charge is not conserved The proton and lambda each have baryon number +1, and the kaon and pion each have baryon number zero; baryon number is conserved The kaon has strangeness –1, the proton and pion each have strangeness zero, and the lambda has strangeness –1, so strangeness is conserved Only charge is not conserved 16 To examine the conservation laws associated with the proposed reaction p + p → Λ + Σ + + e− , we make use of particle properties found in Tables 44-3 and 44-4 1629 (a) With q(p) = +1, q(p) = −1, q(Λ ) = 0, q (Σ + ) = +1, and q(e− ) = −1 , we have + (−1) = + + (−1) Thus, the process conserves charge (b) With B (p) = +1, B(p) = −1, B(Λ ) = 1, B(Σ + ) = +1, and B(e − ) = , we have + (−1) ≠ + + Thus, the process does not conserve baryon number (c) With Le (p) = Le (p) = 0, Le (Λ ) = Le (Σ + ) = 0, and Le (e − ) = , + ≠ + + , so the process does not conserve electron lepton number we have (d) All the particles on either side of the reaction equation are fermions with s = 1/ Therefore, (1/ 2) + (1/ 2) ≠ (1/ 2) + (1/ 2) + (1/ 2) and the process does not conserve spin angular momentum S (Λ ) = 1, S (Σ + ) = +1, and S (e − ) = , we have (e) With S (p) = S (p) = 0, + ≠ + + , so the process does not conserve strangeness (f) The process does conserve muon lepton number since all the particles involved have muon lepton number of zero 17 To examine the conservation laws associated with the proposed decay process Ξ − → π − + n + K − + p , we make use of particle properties found in Tables 44-3 and 44-4 (a) With q (Ξ − ) = −1 , q(π − ) = −1, q(n) = 0, q (K − ) = −1, and q(p) = +1, we have −1 = −1 + + (−1) + Thus, the process conserves charge (b) Since B (Ξ − ) = +1 , B (π − ) = 0, B (n) = +1, B(K − ) = 0, and B (p) = +1, we have +1 ≠ + + + = Thus, the process does not conserve baryon number (c) Ξ − , n and p are fermions with s = 1/ , while π − and K − are mesons with spin zero Therefore, +1/ ≠ + (1/ 2) + + (1/ 2) and the process does not conserve spin angular momentum (d) Since S (Ξ − ) = −2 , S (π − ) = 0, S (n) = 0, S (K − ) = −1, and S (p) = 0, we have −2 ≠ + + (−1) + 0, so the process does not conserve strangeness 18 (a) Referring to Tables 44-3 and 44-4, we find that the strangeness of K0 is +1, while it is zero for both π+ and π– Consequently, strangeness is not conserved in this decay; K → π + + π − does not proceed via the strong interaction (b) The strangeness of each side is –1, which implies that the decay is governed by the strong interaction 1630 CHAPTER 44 (c) The strangeness or Λ0 is –1 while that of p + π– is zero, so the decay is not via the strong interaction (d) The strangeness of each side is –1; it proceeds via the strong interaction 19 For purposes of deducing the properties of the antineutron, one may cancel a proton from each side of the reaction and write the equivalent reaction as π+ → p + n Particle properties can be found in Tables 44-3 and 44-4 The pion and proton each have charge +e, so the antineutron must be neutral The pion has baryon number zero (it is a meson) and the proton has baryon number +1, so the baryon number of the antineutron must be –1 The pion and the proton each have strangeness zero, so the strangeness of the antineutron must also be zero In summary, for the antineutron, (a) q = 0, (b) B = –1, (c) and S = 20 (a) From Eq 37-50, Q = − Δmc = ( mΛ0 − m p − mπ − )c = 1115.6 MeV − 938.3 MeV − 139.6 MeV = 37.7 MeV (b) We use the formula obtained in Problem 44-6 (where it should be emphasized that E is used to mean the rest energy, not the total energy): Kp = c EΛ − E p EΛ h −E 2 π a1115.6 MeV − 938.3 MeVf − a139.6 MeVf = 2a1115.6 MeVf 2 = 5.35 MeV (c) By conservation of energy, K π − = Q − K p = 37.7 MeV − 5.35 MeV = 32.4 MeV 21 (a) As far as the conservation laws are concerned, we may cancel a proton from each side of the reaction equation and write the reaction as p → Λ0 + x Since the proton and the lambda each have a spin angular momentum of = , the spin angular momentum of x must be either zero or = Since the proton has charge +e and the lambda is neutral, x must have charge +e Since the proton and the lambda each have a baryon number of +1, the 1631 baryon number of x is zero Since the strangeness of the proton is zero and the strangeness of the lambda is –1, the strangeness of x is +1 We take the unknown particle to be a spin zero meson with a charge of +e and a strangeness of +1 Look at Table 44-4 to identify it as a K+ particle (b) Similar analysis tells us that x is a spin - 12 antibaryon (B = –1) with charge and strangeness both zero Inspection of Table 44-3 reveals that it is an antineutron (c) Here x is a spin-0 (or spin-1) meson with charge zero and strangeness +1 According to Table 44-4, it could be a K particle 22 Conservation of energy (see Eq 37-47) leads to K f = − Δmc + Ki = ( mΣ − − mπ − − mn )c + Ki = 1197.3 MeV − 139.6 MeV − 939.6 MeV + 220 MeV = 338 MeV 23 (a) Looking at the first three lines of Table 44-5, since the particle is a baryon, we determine that it must consist of three quarks To obtain a strangeness of –2, two of them must be s quarks Each of these has a charge of –e/3, so the sum of their charges is –2e/3 To obtain a total charge of e, the charge on the third quark must be 5e/3 There is no quark with this charge, so the particle cannot be constructed In fact, such a particle has never been observed (b) Again the particle consists of three quarks (and no antiquarks) To obtain a strangeness of zero, none of them may be s quarks We must find a combination of three u and d quarks with a total charge of 2e The only such combination consists of three u quarks 24 If we were to use regular rectangular axes, then this would appear as a right triangle Using the sloping q axis as the problem suggests, it is similar to an “upside down” equilateral triangle as we show below 1632 CHAPTER 44 The leftmost slanted line is for the –1 charge, and the rightmost slanted line is for the +2 charge 25 (a) We indicate the antiparticle nature of each quark with a “bar” over it Thus, u u d represents an antiproton (b) Similarly, u d d represents an antineutron b g 26 (a) The combination ddu has a total charge of − 13 − 13 + 23 = , and a total strangeness of zero From Table 44-3, we find it to be a neutron (n) (b) For the combination uus, we have Q = + 23 + 23 − 13 = and S = + – = –1 This is the Σ+ particle (c) For the quark composition ssd, we have Q = − 13 − 13 − 13 = −1 and S = – – + = – This is a Ξ − 27 The meson K is made up of a quark and an anti-quark, with net charge zero and strangeness S = −1 The quark with S = −1 is s By charge neutrality condition, the antiquark must be d Therefore, the constituents of K are s and d 28 (a) Using Table 44-3, we find q = and S = –1 for this particle (also, B = 1, since that is true for all particles in that table) From Table 44-5, we see it must therefore contain a strange quark (which has charge –1/3), so the other two quarks must have charges to add to zero Assuming the others are among the lighter quarks (none of them being an antiquark, since B = 1), then the quark composition is sud (b) The reasoning is very similar to that of part (a) The main difference is that this particle must have two strange quarks Its quark combination turns out to be uss b g 29 (a) The combination ssu has a total charge of − 13 − 13 + 23 = , and a total strangeness of – From Table 44-3, we find it to be the Ξ particle (b) The combination dds has a total charge of ( − 13 − 13 − 13 ) = −1 , and a total strangeness of –1 From Table 44-3, we find it to be the Σ − particle 30 From γ = + K/mc2 (see Eq 37-52) and v = β c = c − γ −2 (see Eq 37-8), we get F H v = c 1− 1+ (a) Therefore, for the Σ*0 particle, K mc I K −2 1633 ⎛ 1000MeV ⎞ v = (2.9979 × 10 m s) − ⎜1 + ⎟ ⎝ 1385MeV ⎠ −2 = 2.4406 × 108 m s For Σ0, ⎛ 1000 MeV ⎞ v′ = (2.9979 ×10 m s) − ⎜1 + ⎟ ⎝ 1192.5 MeV ⎠ −2 = 2.5157 × 108 m s Thus Σ0 moves faster than Σ*0 (b) The speed difference is Δv = v′ − v = (2.5157 − 2.4406)(108 m s) = 7.51×106 m s 31 First, we find the speed of the receding galaxy from Eq 37-31: − ( f f ) − (λ0 λ ) = β= + ( f f ) + (λ0 λ ) = − (590.0 nm 602.0 nm) = 0.02013 + (590.0 nm 602.0 nm) where we use f = c/λ and f0 = c/λ0 Then from Eq 44-19, ( ) v β c ( 0.02013 ) 2.998 ×10 m s r= = = = 2.77 × 108 ly H H 0.0218 m s ⋅ ly 32 Since λ = λ0 1+ β = 2λ0 1− β ⇒ 1+ β =2, 1− β the speed of the receding galaxy is v = β c = 3c / Therefore, the distance to the galaxy when the light was emitted is v β c (3 / 5)c (0.60)(2.998 ×108 m/s) r= = = = = 8.3× 109 ly H H H 0.0218 m s ⋅ ly 33 We apply Eq 37-36 for the Doppler shift in wavelength: Δλ λ = v c CHAPTER 44 1634 where v is the recessional speed of the galaxy We use Hubble’s law to find the recessional speed: v = Hr, where r is the distance to the galaxy and H is the Hubble constant 21.8 × 10−3 sm⋅ly Thus, ( ) ( v = 21.8 ×10−3 m/s ⋅ ly )( 2.40 ×10 ) ly = 5.23×106 m/s and ⎛ 5.23×106 m s ⎞ v Δλ = λ = ⎜ ⎟ (656.3 nm) = 11.4 nm c ⎝ 3.00 ×10 m s ⎠ Since the galaxy is receding, the observed wavelength is longer than the wavelength in the rest frame of the galaxy Its value is 656.3 nm + 11.4 nm = 667.7 nm ≈ 668 nm 34 (a) Using Hubble’s law given in Eq 44-19, the speed of recession of the object is ( ) v = Hr = ( 0.0218 m/s ⋅ ly ) 1.5 ×104 ly = 327 m/s Therefore, the extra distance of separation one year from now would be d = vt = (327 m/s)(365 d)(86400 s/d) = 1.0 × 1010 m (b) The speed of the object is v = 327 m/s ≈ 3.3 × 102 m/s 35 Letting v = Hr = c, we obtain r= c 3.0 ×108 m s = = 1.376 × 1010 ly ≈ 1.4 ×1010 ly H 0.0218 m s ⋅ ly 36 (a) Letting v(r ) = Hr ≤ ve = 2G M r , we get M r ≥ H 2G Thus, ρ= M M 3H = ≥ 4π r 4π r 8π G (b) The density being expressed in H-atoms/m3 is equivalent to expressing it in terms of ρ0 = mH/m3 = 1.67 × 10–27 kg/m3 Thus, 1635 ( )( ( 0.0218 m s ⋅ ly ) 1.00 ly 9.460 × 1015 m H atoms m3 3H H atoms m = ρ= 8πG ρ0 8π 6.67×10−11 m3 kg ⋅ s 1.67 × 10−27 kg m3 ( ) ( )( = 5.7 H atoms m3 37 (a) From f = c/λ and Eq 37-31, we get λ0 = λ 1− β 1− β = (λ0 + Δλ ) 1+ β 1+ β Dividing both sides by λ0 leads to = (1 + z ) 1− β 1+ β where z = Δλ / λ0 We solve for β: β= z2 + 2z (1 + z ) − = (1 + z )2 + z + z + (b) Now z = 4.43, so b4.43g + 2b4.43g = 0.934 β= b4.43g + 2b4.43g + 2 (c) From Eq 44-19, ( ) v β c ( 0.934 ) 3.0 × 10 m s = = = 1.28 ×1010 ly r= H H 0.0218 m s ⋅ ly 38 Using Eq 39-33, the energy of the emitted photon is ⎛1 ⎞ E = E3 − E2 = −(13.6 eV) ⎜ − ⎟ = 1.89 eV ⎝3 ⎠ and its wavelength is λ0 = hc 1240 eV ⋅ nm = = 6.56 × 10−7 m E 1.89 eV Given that the detected wavelength is λ = 3.00 ×10−3 m , we find ) ) 1636 CHAPTER 44 λ 3.00 ×10−3 m = = 4.57 ×103 −7 λ0 6.56 ×10 m 39 (a) From Eq 41-29, we know that N N1 = e − ΔE kT We solve for ΔE: ΔE = kT ln N1 ⎛ − 0.25 ⎞ = ( 8.62 × 10−5 eV K ) ( 2.7K ) ln ⎜ ⎟ N2 ⎝ 0.25 ⎠ = 2.56 × 10−4 eV ≈ 0.26 meV (b) Using hc = 1240eV ⋅ nm, we get λ= hc 1240eV ⋅ nm = = 4.84 × 106 nm ≈ 4.8mm ΔE 2.56 ×10−4 eV 40 From Fgrav = GMm r = mv r we find M ∞ v Thus, the mass of the Sun would be ⎛v ⎞ ⎛ 47.9 km M s′ = ⎜ Mercury ⎟ M s = ⎜ ⎝ 4.74 km ⎝ vPluto ⎠ s⎞ ⎟ M s = 102 M s s⎠ 41 (a) The gravitational force on Earth is only due to the mass M within Earth’s orbit If r is the radius of the orbit, R is the radius of the new Sun, and MS is the mass of the Sun, then FrI M =G J H RK F 150 × 10 m I =G × 10 kgh = 3.27 × 10 H 5.90 × 10 mJK c199 11 Ms 30 12 25 kg The gravitational force on Earth is given by GMm r , where m is the mass of Earth and G is the universal gravitational constant Since the centripetal acceleration is given by v2/r, where v is the speed of Earth, GMm r = mv r and GM = v= r c6.67 × 10 −11 hc m3 s2 ⋅ kg 3.27 × 1025 kg 150 × 10 m 11 (b) The ratio is 1.21×102 m s = 0.00405 2.98 ×104 m s (c) The period of revolution is h = 121 × 10 m s 1637 11 2πr 2π (1.50 × 10 m ) = = 7.82 ×109 s = 247 y v 1.21×10 m s T= Note: An alternative way to calculate the speed ratio and the periods is as follows Since v ∼ M , the ratio of the speeds can be obtained as v = v0 M ⎛r⎞ =⎜ ⎟ MS ⎝ R ⎠ 3/ ⎛ 1.50 × 1011 m ⎞ =⎜ ⎟ 12 ⎝ 5.90 × 10 m ⎠ 3/ = 0.00405 In addition, since T ∼ 1/ v ∼ 1/ M , we have T = T0 MS ⎛R⎞ = T0 ⎜ ⎟ M ⎝r⎠ 3/ ⎛ 5.90 × 1012 m ⎞ = (1 y) ⎜ ⎟ 11 ⎝ 1.50 × 10 m ⎠ 3/ = 247 y 42 (a) The mass of the portion of the galaxy within the radius r from its center is given by M ′ = r R M Thus, from GM ′m r = mv r (where m is the mass of the star) we get b g FG IJ H K GM ′ GM r v= = r r R =r GM R3 (b) In the case where M' = M, we have T= r πr πr = πr = v GM GM 43 (a) For the universal microwave background, Wien’s law leads to T= 2898 μ m ⋅ K λmax = 2898 mm ⋅ K = 2.6 K 1.1mm (b) At “decoupling” (when the universe became approximately “transparent”), λmax = 2898 μ m ⋅ K 2898 μ m ⋅ K = = 0.976 μ m = 976 nm T 2970K 44 (a) We substitute λ = (2898 μm·K)/T into the expression: E = hc / λ = (1240 eV·nm)/λ First, we convert units: 1638 CHAPTER 44 2898 μm·K = 2.898 × 106 nm·K and 1240 eV·nm = 1.240 × 10–3 MeV·nm Thus, × 10 c1240 E= h = c4.28 × 10 2.898 × 10 nm ⋅ K −3 MeV ⋅ nm T −10 h MeV K T (b) The minimum energy required to create an electron-positron pair is twice the rest energy of an electron, or 2(0.511 MeV) = 1.022 MeV Hence, T= E 4.28 × 10 −10 MeV K = 1022 MeV = 2.39 × 109 K −10 4.28 × 10 MeV K 45 Since only the strange quark (s) has nonzero strangeness, in order to obtain S = –1 we need to combine s with some non-strange anti-quark (which would have the negative of the quantum numbers listed in Table 44-5) The difficulty is that the charge of the strange quark is –1/3, which means that (to obtain a total charge of +1) the anti-quark would have to have a charge of + 43 Clearly, there are no such anti-quarks in our list Thus, a meson with S = –1 and q = +1 cannot be formed with the quarks/anti-quarks of Table 44-5 Similarly, one can show that, since no quark has q = − 43 , there cannot be a meson with S = +1 and q = –1 46 Assuming the line passes through the origin, its slope is 0.40c/(5.3 × 109 ly) Then, 1 5.3 × 109 ly 5.3 × 109 y T= = = = ≈ 13 × 109 y H slope 0.40 c 0.40 47 The energy released would be twice the rest energy of Earth, or E = 2mc2 = 2(5.98 × 1024 kg)(2.998 × 108 m/s)2 = 1.08 × 1042 J The mass of Earth can be found in Appendix C As in the case of annihilation between an electron and a positron, the total energy of the Earth and the anti-Earth after the annihilation would appear as electromagnetic radiation 48 We note from track 1, and the quantum numbers of the original particle (A), that positively charged particles move in counterclockwise curved paths, and — by inference — negatively charged ones move along clockwise arcs This immediately shows that tracks 1, 2, 4, 6, and belong to positively charged particles, and tracks 5, and belong to negatively charged ones Looking at the fictitious particles in the table (and noting that each appears in the cloud chamber once [or not at all]), we see that this observation (about charged particle motion) greatly narrows the possibilities: tracks 2, 4, 6, 7, ↔ particles C , F , H , J tracks 5,8,9 ↔ particles D, E , G 1639 This tells us, too, that the particle that does not appear at all is either B or I (since only one neutral particle “appears”) By charge conservation, tracks 2, and are made by particles with a single unit of positive charge (note that track is made by one with a single unit of negative charge), which implies (by elimination) that track is made by particle H This is confirmed by examining charge conservation at the end-point of track Having exhausted the charge-related information, we turn now to the fictitious quantum numbers Consider the vertex where tracks 2, 3, and meet (the Whimsy number is listed here as a subscript): tracks 2, ↔ particles C2 , F0 , J −6 tracks ↔ particle B4 or I The requirement that the Whimsy quantum number of the particle making track must equal the sum of the Whimsy values for the particles making tracks and places a powerful constraint (see the subscripts above) A fairly quick trial and error procedure leads to the assignments: particle F makes track 4, and particles J and I make tracks and 3, respectively Particle B, then, is irrelevant to this set of events By elimination, the particle making track (the only positively charged particle not yet assigned) must be C At the vertex defined by A → F + C + track _ , b g where the charge of that particle is indicated by the subscript, we see that Cuteness number conservation requires that the particle making track has Cuteness = –1, so this must be particle G We have only one decision remaining: tracks 8,9, ↔ particles D, E Re-reading the problem, one finds that the particle making track must be particle D since it is the one with seriousness = Consequently, the particle making track must be E Thus, we have the following: (a) Particle A is for track (b) Particle J is for track (c) Particle I is for track (d) Particle F is for track (e) Particle G is for track (f) Particle C is for track 1640 CHAPTER 44 (g) Particle H is for track (h) Particle D is for track (i) Particle E is for track 49 (a) We use the relativistic relationship between speed and momentum: mv p = γmv = b g 1− v c , which we solve for the speed v: v = 1− c ( pc / mc2 ) + For an antiproton mc2 = 938.3 MeV and pc = 1.19 GeV = 1190 MeV, so v = c 1− b1190 MeV 938.3 MeVg + = 0.785 c (b) For the negative pion mc2 = 193.6 MeV, and pc is the same Therefore, v = c 1− b1190 MeV 193.6 MeVg + = 0.993c (c) Since the speed of the antiprotons is about 0.78c but not over 0.79c, an antiproton will trigger C2 (d) Since the speed of the negative pions exceeds 0.79c, a negative pion will trigger C1 (e) We use Δt = d/v, where d = 12 m For an antiproton Δt = = 5.1× 10−8 s = 51ns 0.785 ( 2.998 × 10 m s ) (f) For a negative pion Δt = 12 m = 4.0 × 10−8 s = 40 ns 0.993 2.998 × 10 m s c h ... Thus, the mass of a cubic meter of water is 1000 kg (b) We divide the mass of the water by the time taken to drain it The mass is found from M = ρV (the product of the volume of water and its... …) are given for ready reference on the inside front cover of the textbook (see also Table 1–2) The surface area A of each grain of sand of radius r = 50 μm = 50 × 10−6 m is given by A = 4π(50... equal, we can solve for d Thus, d = 25 m The volume of a small part of the mud over a patch of area of 4.0 m2 is (4.0)d = 100 m3 Since each cubic meter corresponds to a mass of 1900 kg (stated