Física para ciencias e ingenieria serway 7ed MS

http://www.elsolucionario.blogspot.com LIBROS UNIVERISTARIOS Y SOLUCIONARIOS DE MUCHOS DE ESTOS LIBROS LOS SOLUCIONARIOS CONTIENEN TODOS LOS EJERCICIOS DEL LIBRO RESUELTOS Y EXPLICADOS DE FORMA CLARA VISITANOS PARA DESARGALOS GRATIS Physics and Measurement CHAPTER OUTLINE 1.1 1.2 1.3 1.4 1.5 1.6 Standards of Length, Mass, and Time Matter and Model-Building Dimensional Analysis Conversion of Units Estimates and Order-ofMagnitude Calculations Significant Figures ANSWERS TO QUESTIONS * An asterisk indicates an item new to this edition Q1.1 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 Q1.2 (a) 0.3 millimeters (b) 50 microseconds (c) 7.2 kilograms *Q1.3 In the base unit we have (a) 0.032 kg (b) 0.015 kg (c) 0.270 kg (d) 0.041 kg (e) 0.27 kg Then the ranking is c=e>d>a>b Q1.4 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 *Q1.5 The answer is yes for (a), (c), and (f ) You cannot add or subtract a number of apples and a number of jokes The answer is no for (b), (d), and (e) Consider the gauge of a sausage, kgր2 m, or the volume of a cube, (2 m)3 Thus we have (a) yes (b) no (c) yes (d) no (e) no (f ) yes *Q1.6 41 € ≈ 41 € (1 Lր1.3 €)(1 qtր1 L)(1 galր4 qt) ≈ (10ր1.3) gal ≈ gallons, answer (c) *Q1.7 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 *Q1.8 0.02(1.365) = 0.03 The result is (1.37 ± 0.03) × 107 kg So (d) digits are significant SOLUTIONS TO PROBLEMS Section 1.1 P1.1 Standards of Length, Mass, and Time 4 Modeling the Earth as a sphere, we find its volume as π r = π ( 6.37 × 10 m ) = 1.08 × 10 21 m 3 m 5.98 × 10 24 kg Its density is then ρ = = = 5.52 × 10 kg m This value is intermediate 21 V 1.08 × 10 m between the tabulated densities of aluminum and iron Typical rocks have densities around 000 to 000 kgրm3 The average density of the Earth is significantly higher, so higher-density material must be down below the surface ISMV1_5103_01.indd 10/27/06 4:33:21 PM P1.2 Chapter With V = ( base area ) ( height ) V = (π r ) h and ρ = ρ= m , we have V ⎛ 10 mm ⎞ m kg = 2 π r h π (19.5 mm ) ( 39.0 mm ) ⎜⎝ m ⎟⎠ ρ = 2.15 × 10 kg m P1.3 and ρgold = *P1.4 m for both Then ρiron = 9.35 kg V V ⎛ 19.3 × 10 kg/m ⎞ = 23.0 kg = 9.35 kg ⎜ ⎝ 7.86 × 10 kg/m ⎟⎠ Let V represent the volume of the model, the same in ρ = mgold V Next, mgold ρgold and mgold = ρiron 9.35 kg ρ = m / V and V = (4 / 3)π r = (4 / 3)π (d / 2)3 = π d / where d is the diameter Then ρ = m / π d = 6(1.67 × 10 −27 kg) = 2.3 × 1017 kg/m π (2.4 × 10 −15 m)3 2.3 × 1017 kg/m /(11.3 × 10 kg/m ) = it is 20 × 1012 times the density of lead P1.5 4 π r and the mass is m = ρV = ρ π r We divide 3 this equation for the larger sphere by the same equation for the smaller: For either sphere the volume is V = mᐉ ρ 4π rᐉ3 rᐉ3 = = = ms ρ 4π rs3 rs3 Then rᐉ = rs = 4.50 cm (1.71) = 7.69 cm Section 1.2 P1.6 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 Section 1.3 Dimensional Analysis P1.7 (a) This is incorrect since the units of [ ax ] are m s , while the units of [ v ] are m s (b) −1 This is correct since the units of [ y ] are m, and cos ( kx ) is dimensionless if [ k ] is in m P1.8 (a) Circumference has dimensions of L (b) Volume has dimensions of L3 (c) Area has dimensions of L2 Expression (i) has dimension L ( L2 ) 1/ = L2 , so this must be area (c) Expression (ii) has dimension L, so it is (a) Expression (iii) has dimension L ( L2 ) = L3, so it is (b) Thus, (a) = ii; (b) = iii;(c) = i ISMV1_5103_01.indd 10/27/06 12:27:01 PM Physics and Measurement P1.9 Inserting the proper units for everything except G, ⎡ kg m ⎤ = G [ kg ] ⎢⎣ s ⎥⎦ [ m ]2 Multiply both sides by [ m ]2 and divide by [ kg ] ; the units of G are Section 1.4 P1.10 m3 kg ⋅ s Conversion of Units Apply the following conversion factors: in = 2.54 cm, d = 86 400 s, 100 cm = m, and 10 nm = m −2 ⎛ in day⎞ ( 2.54 cm in ) (10 m cm ) (10 nm m ) = 9.19 nm s ⎝ 32 ⎠ 86 400 s day This means the proteins are assembled at a rate of many layers of atoms each second! P1.11 Conceptualize: We must calculate the area and convert units Since a meter is about feet, we should expect the area to be about A ≈ 30 m 50 m = 1500 m ( )( ) Categorize: We model the lot as a perfect rectangle to use Area = Length × Width Use the conversion:1 m = 3.281 ft 1m ⎞ 1m ⎞ Analyze: A = LW = (100 ft ) ⎛ = 390 m = 1.39 × 10 m (150 ft ) ⎛ ⎝ 3.281 ft ⎠ ⎝ 3.281 ft ⎠ Finalize: Our calculated result agrees reasonably well with our initial estimate and has the proper units of m Unit conversion is a common technique that is applied to many problems P1.12 (a) V = ( 40.0 m ) ( 20.0 m ) (12.0 m ) = 9.60 × 10 m V = 9.60 × 10 m ( 3.28 ft m ) = 3.39 × 10 ft 3 (b) The mass of the air is m = ρairV = (1.20 kg m ) ( 9.60 × 10 m ) = 1.15 × 10 kg The student must look up weight in the index to find Fg = mg = (1.15 × 10 kg ) ( 9.80 m s ) = 1.13 × 10 N Converting to pounds, Fg = (1.13 × 10 N ) (1 lb 4.45 N ) = 2.54 × 10 lb *P1.13 The area of the four walls is (3.6 + 3.8 + 3.6 + 3.8)m (2.5 m) = 37 m2 Each sheet in the book has area (0.21 m) (0.28 m) = 0.059 m2 The number of sheets required for wallpaper is 37 m2ր0.059 m2 = 629 sheets = 629 sheets(2 pagesր1 sheet) = 1260 pages The pages from volume one are inadequate, but the full version has enough pages ISMV1_5103_01.indd 10/27/06 4:18:09 PM P1.14 Chapter (a) Seven minutes is 420 seconds, so the rate is r= (b) 30.0 gal = 7.14 × 10 −2 gal s 420 s Converting gallons first to liters, then to m 3, ⎛ 3.786 L ⎞ ⎛ 10 −3 m ⎞ r = ( 7.14 × 10 −2 gal s ) ⎜ ⎝ gal ⎟⎠ ⎜⎝ L ⎟⎠ r = 2.70 × 10 −4 m s (c) At that rate, to fill a 1-m tank would take ⎛ m3 ⎞⎛ 1h ⎞ = 1.03 h t=⎜ −4 ⎝ 2.70 × 10 m s ⎟⎠ ⎜⎝ 600 ⎟⎠ P1.15 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 number This density value tells us that lead is about 11 times denser than water, which agrees with our experience that lead sinks Density is defined as mass per volume, in ρ = m We must convert to SI units in the calculation V 23.94 g ⎛ kg ⎞ ⎛ 100 cm ⎞ 23.94 g ⎛ kg ⎞ ⎛ 000 000 cm ⎞ = ρ= = 1.14 × 10 kg m ⎜ ⎟ ⎟ ⎟ ⎟ ⎜ 3⎜ 3⎜ 1000 g m 1000 g ⎠⎝ ⎠⎝ ⎠ 2.10 cm ⎝ 2.10 cm ⎝ 1m ⎠ At one step in the calculation, we note that one million cubic centimeters make one cubic meter Our result is indeed close to the expected value Since the last reported significant digit is not certain, the difference in the two values is probably due to measurement uncertainty and should not be a concern One important common-sense check on density values is that objects which sink in water must have a density greater than g cm 3, and objects that float must be less dense than water P1.16 The weight flow rate is 200 P1.17 (a) (b) ton ⎛ 000 lb ⎞ ⎛ h ⎞ ⎛ ⎞ = 667 lb s h ⎝ ton ⎠ ⎝ 60 ⎠ ⎝ 60 s ⎠ ⎛ × 1012 $ ⎞ ⎛ h ⎞ ⎛ day ⎞ ⎛ yr ⎞ ⎜⎝ 000 $ s ⎟⎠ ⎜⎝ 600 s ⎟⎠ ⎝ 24 h ⎠ ⎜⎝ 365 days ⎟⎠ = 250 years The circumference of the Earth at the equator is 2π ( 6.378 × 10 m ) = 4.01 × 10 m The length of one dollar bill is 0.155 m so that the length of trillion bills is 1.24 × 1012 m Thus, the trillion dollars would encircle the Earth 1.24 × 1012 m = 3.09 × 10 times 4.01 × 10 m P1.18 ⎡(13.0 acres ) ( 43 560 ft acre ) ⎤⎦ Bh = ⎣ ( 481 ft ) 3 = 9.08 × 10 ft , V= h B or ⎛ 2.83 × 10 −2 m ⎞ V = ( 9.08 × 10 ft ) ⎜ ⎟⎠ ⎝ ft FIG P1.18 = 2.57 × 10 m ISMV1_5103_01.indd 10/27/06 12:27:03 PM Physics and Measurement P1.19 Fg = ( 2.50 tons block ) ( 2.00 × 10 blocks ) ( 000 lb ton ) = 1.00 × 1010 lbs P1.20 (a) ⎛d ⎞ 300 ft ⎞ = 6.79 × 10 −3 ft , or dnucleus, scale = dnucleus, real ⎜ atom, scale ⎟ = ( 2.40 × 10 −15 m ) ⎛ ⎝ 1.06 × 10 −10 m ⎠ ⎝ datom, real ⎠ dnucleus, scale = ( 6.79 × 10 −3 ft ) ( 304.8 mm ft ) = 2.07 mm (b) 3 ⎛ datom ⎞ ⎛ 1.06 × 10 −10 m ⎞ Vatom / ⎛ ratom ⎞ 4π ratom = = = =⎜ ⎜⎝ 2.40 × 10 −15 m ⎟⎠ ⎜⎝ d ⎟ Vnucleus 4π rnucleus / ⎝ rnucleus ⎟⎠ nucleus ⎠ = 8.62 × 1013 times as large V 3.78 × 10 −3 m = = 1.51 ì 10 m ( or 151 m ) 25.0 m A P1.21 V = At so t = P1.22 (a) ⎛ ( 6.37 × 10 m ) (100 cm m ) ⎞ ⎛r ⎞ AEarth 4π rEarth = = ⎜ Earth ⎟ = ⎜ ⎟ = 13.4 1.74 × 108 cm AMoon 4π rMoon ⎝ rMoon ⎠ ⎠ ⎝ (b) ⎛ ( 6.37 × 10 m ) (100 cm m ) ⎞ VEarth / ⎛ rEarth ⎞ 4π rEarth = = =⎜ ⎜ ⎟ = 49.1 1.74 × 108 cm VMoon 4π rMoon / ⎝ rMoon ⎟⎠ ⎝ ⎠ 2 3 P1.23 To balance, mFe = mAl or ρFeVFe = ρAlVAl ⎛ 4⎞ ρFe ⎜ ⎟ π rFe = ρAl ⎛ ⎞ π rAl ⎝ ⎝ 3⎠ 3⎠ ⎛ρ ⎞ rAl = rFe ⎜ Fe ⎟ ⎝ ρAl ⎠ P1.24 1/ 7.86 ⎞ = ( 2.00 cm ) ⎛ ⎝ 2.70 ⎠ 1/ = 2.86 cm The mass of each sphere is mAl = ρAlVAl = 4π ρAl rAl 3 mFe = ρFeVFe = 4π ρFe rFe and Setting these masses equal, 4π ρAl rAl 4π ρFe rFe and r = r ρFe = Al Fe ρAl 3 The resulting expression shows that the radius of the aluminum sphere is directly proportional to the radius of the balancing iron sphere The sphere of lower density has larger radius The ρ fraction Fe is the factor of change between the densities, a number greater than Its cube root ρAl is a number much closer to The relatively small change in radius implies a change in volume sufficient to compensate for the change in density ISMV1_5103_01.indd 10/27/06 12:27:04 PM Chapter Section 1.5 P1.25 Estimates and Order-of-Magnitude Calculations Model the room as a rectangular solid with dimensions m by m by m, and each ping-pong ball as a sphere of diameter 0.038 m The volume of the room is × × = 48 m 3, while the volume of one ball is 4π ⎛ 0.038 m ⎞ = 2.87 × 10 −5 m 3 ⎝ ⎠ 48 Therefore, one can fit about ~ 10 ping-pong balls in the room 2.87 × 10 −5 As an aside, the actual number is smaller than this because there will be a lot of space in the room that cannot be covered by balls In fact, even in the best arrangement, the so-called “best packing fraction” is π = 0.74 so that at least 26% of the space will be empty Therefore, the above estimate reduces to 1.67 × 10 × 0.740 ~ 10 P1.26 A reasonable guess for the diameter of a tire might be 2.5 ft, with a circumference of about ft Thus, the tire would make 50 000 mi 280 ft mi rev ft = × 10 rev ~ 10 rev P1.27 Assume the tub measures 1.3 m by 0.5 m by 0.3 m One-half of its volume is then )( ( )( ) V = ( 0.5) (1.3 m ) ( 0.5 m ) ( 0.3 m ) = 0.10 m The mass of this volume of water is mwater = ρwaterV = (1 000 kg m ) ( 0.10 m ) = 100 kg ~ 10 kg Pennies are now mostly zinc, but consider copper pennies filling 50% of the volume of the tub The mass of copper required is mcopper = ρcopperV = (8 920 kg m ) ( 0.10 m ) = 892 kg ~ 10 kg *P1.28 The time required for the task is bad yr ⎞ ⎛ s ⎞ h ⎞ ⎛ working day ⎞ ⎛ = 58 yr 10 $ ⎜ ⎟ ⎛ ⎜ ⎠ ⎝ $ ⎠ ⎝ 3600 s ⎠ ⎝ ⎝ 300 working days ⎟⎠ 16 h Since you are already around 20 years old, you would have a miserable life and likely die before accomplishing the task You have better things to Say no P1.29 Assume: Total population = 10 7; one out of every 100 people has a piano; one tuner can serve about 000 pianos (about per day for 250 weekdays, assuming each piano is tuned once per year) Therefore, ⎛ tuner ⎞ ⎛ piano ⎞ # tuners ~ ⎜ (10 people ) = 100 tuners ⎝ 000 pianos ⎟⎠ ⎜⎝ 100 people ⎟⎠ Section 1.6 P1.30 Significant Figures METHOD ONE We treat the best value with its uncertainty as a binomial ( 21.3 ± 0.2 ) cm ( 9.8 ± 0.1) cm, A = [ 21.3 ( 9.8 ) ± 21.3 ( 0.1) ± 0.2 ( 9.8 ) ± ( 0.2 ) ( 0.1)] cm The first term gives the best value of the area The cross terms add together to give the uncertainty and the fourth term is negligible A = 209 cm ± cm METHOD TWO We add the fractional uncertainties in the data 0.2 0.1 ⎞ A = ( 21.3 cm ) ( 9.8 cm ) ± ⎛ + = 209 cm ± 2% = 209 cm ± cm ⎝ 21.3 9.8 ⎠ ISMV1_5103_01.indd 10/28/06 2:41:27 AM Physics and Measurement P1.31 P1.32 (a) (b) (c) (d) r = ( 6.50 ± 0.20 ) cm = ( 6.50 ± 0.20 ) × 10 −2 m m = (1.85 ± 0.02 ) kg ρ= ( m )π r3 also, δ ρ = δ m + 3δ r ρ m r In other words, the percentages of uncertainty are cumulative Therefore, δ ρ 0.02 ( 0.20 ) = + = 0.103, ρ 1.85 6.50 ρ= 1.85 ( ) π ( 6.5 × 10 −2 m) = 1.61 × 10 kg m and ρ ± δ ρ = (1.61 ± 0.17 ) × 10 kg m = (1.6 ± 0.2 ) × 10 kg m P1.33 P1.34 (a) 756.?? 37.2? 0.83 + 2.5? 796 / / = 797 (b) 0.003 ( s.f.) × 356.3 ( s.f.) = 1.140 16 = ( s.f.) 1.1 (c) 5.620 ( s.f.) × π ( >4 s.f.) = 17.656= ( s.f.) 17.66 We work to nine significant digits: ⎛ 365.242 199 d ⎞ ⎛ 24 h ⎞ ⎛ 60 ⎞ ⎛ 60 s ⎞ yr = yr ⎜ ⎟⎠ ⎝ d ⎠ ⎝ h ⎠ ⎝ ⎠ = 31 556 926.0 s ⎝ yr *P1.35 The tax amount is $1.36 − $1.25 = $0.11 The tax rate is $0.11ր$1.25 = 0.088 = 8.80% *P1.36 (a) We read from the graph a vertical separation of 0.3 spaces = 0.015 g (b) Horizontally, 0.6 spaces = 30 cm2 (c) Because the graph line goes through the origin, the same percentage describes the vertical and the horizontal scatter: 30 cm2ր380 cm2 = 8% (d) Choose a grid point on the line far from the origin: slope = 0.31 g ր600 cm2 = 0.000 52 gրcm2 = (0.000 52 gրcm2)(10 000 cm2ր1 m2) = 5.2 g/m2 ISMV1_5103_01.indd (e) For any and all shapes cut from this copy paper, the mass of the cutout is proportional to its area The proportionality constant is 5.2 g/m2 ± 8%, where the uncertainty is estimated (f ) This result should be expected if the paper has thickness and density that are uniform within the experimental uncertainty The slope is the areal density of the paper, its mass per unit area 10/28/06 2:41:45 AM Chapter *P1.37 15 players = 15 players (1 shift ր1.667 player) = shifts *P1.38 Let o represent the number of ordinary cars and s the number of trucks We have o = s + 0.947s = 1.947s, and o = s + 18 We eliminate o by substitution: s + 18 = 1.947s 0.947s = 18 and s = 18ր0.947 = 19 *P1.39 Let s represent the number of sparrows and m the number of more interesting birds We have sրm = 2.25 and s + m = 91 We eliminate m by substitution: m = sր2.25 s + sր2.25 = 91 1.444s = 91 s = 91ր1.444 = 63 *P1.40 For those who are not familiar with solving equations numerically, we provide a detailed solution It goes beyond proving that the suggested answer works The equation x − x + x − 70 = is quartic, so we not attempt to solve it with algebra To find how many real solutions the equation has and to estimate them, we graph the expression: x −3 −2 −1 y = x − x + x − 70 158 −24 −70 −70 −66 −52 26 270 We see that the equation y = has two roots, one around x = −2.2 and the other near x = +2.7 To home in on the first of these solutions we compute in sequence: When x = −2.2, y = −2.20 The root must be between x = −2.2 and x = −3 When x = −2.3, y = 11.0 The root is between x = −2.2 and x = −2.3 When x = −2.23, y = 1.58 The root is between x = −2.20 and x = −2.23 When x = −2.22, y = 0.301 The root is between x = −2.20 and −2.22 When x = −2.215, y = −0.331 The root is between x = −2.215 and −2.22 We could next try x = −2.218, but we already know to three-digit precision that the root is x = −2.22 *P1.41 y x FIG P1.40 We require sin θ = −3 cos θ , or sin θ = −3 , or tan θ = −3 tan θ cos θ For tan −1 ( −3) = arc tan ( −3) , your calculator may return −71.6°, but this angle is not between 0° and 360° as the problem requires The tangent function is negative in the second quadrant (between 90° and 180°) and in the fourth quadrant (from 270° to 360°) The solutions to the equation are then 360° − 71.6° = 288° and 180° − 71.6° = 108° 360° θ FIG P1.41 ISMV1_5103_01.indd 10/27/06 12:27:07 PM Physics and Measurement *P1.42 We draw the radius to the initial point and the radius to the final point The angle θ between these two radii has its sides perpendicular, right side to right side and left side to left side, to the 35° angle between the original R and final tangential directions of travel A most useful theorem from geometry then identifies these angles as equal: θ = 35° The whole θ circumference of a 360° circle of the same radius is 2π R By proportion, then 2π R = 840 m 360° 35° 360° 840 m 840 m R= = = 1.38 × 10 m 0.611 2π 35° i N 35.0° f W E S FIG P1.42 We could equally well say that the measure of the angle in radians is 2π radians ⎞ 840 m θ = 35° = 35° ⎛ = 0.611 rad = ⎝ 360 ° ⎠ R Solving yields R = 1.38 km *P1.43 Mass is proportional to cube of length: m = kᐉ3 mƒ րmi = (ᐉf րᐉi)3 Length changes by 15.8%: ᐉf = ᐉi + 0.158 ᐉi = 1.158 ᐉi Mass increase: mf = mi + 17.3 kg mf = 1.1583 = 1.553 Eliminate by substitution: m f − 17.3 kg mf = 1.553 mf − 26.9 kg *P1.44 26.9 kg = 0.553 mf mf = 26.9 kg ր0.553 = 48 kg We use substitution, as the most generally applicable method for solving simultaneous equations We substitute p = 3q into each of the other two equations to eliminate p: ⎧3qr = qs ⎪ 2 ⎨1 ⎪⎩ 3qr + qs = qt ⎧3r = s 3r + ( 3r ) = t These simplify to ⎨ 2 We substitute to eliminate s: We solve for the ⎩3r + s = t 12r = t combination t : r t2 = 12 r2 t = either 3.46 or − 3.46 r *P1.45 Solve the given equation for ∆t: (a) Making d three times larger with d in the bottom of the fraction makes ∆t nine times smaller (b) ∆t is inversely proportional to the square of d (c) Plot ∆t on the vertical axis and 1/d on the horizontal axis (d) ISMV1_5103_01.indd ∆t = 4QLրkπd 2(Th − Tc) = [4QLրkπ (Th − Tc)] [1ր d 2] From the last version of the equation, the slope is 4QL/kπ(Th − Tc) Note that this quantity is constant as both ∆t and d vary 10/28/06 2:42:03 AM 604 P46.29 Chapter 46 (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: ( m c ) + ( p c) 2 Emin + m2 c = (1) By conservation of momentum: p3 = p1 ∴ ( p3 c ) = ( p1c ) = Emin − ( m1c ) 2 Emin + m2 c = Substitute (2) in (1): (2) (m c ) 2 + Emin − ( m1c ) Square both sides: 2 Emin + Emin m2 c + ( m2 c ) = ( m3 c ) + Emin − ( m1c ) ∴ Emin = (m 2 − m12 − m22 ) c 2 m2 ∴ K = Emin − m1c (m = − m12 − m22 − m1 m2 ) c 2 m2 ⎡m − ( m + m )2 ⎤ c 2 ⎣ ⎦ = 2m2 Refer to Table 46.2 for the particle masses ⎡⎣4 ( 938.3)⎤⎦ MeV2 c − ⎡⎣2 ( 938.3)⎤⎦ MeV2 c = = 5.63 GeV ( 938.3 MeV c ) (b) K (c) K = ( 497.7 + 1115.6) 2 (e) 2 ( 938.3) MeV c = 768 MeV K ⎡⎣2 ( 938.3) + 135⎤⎦ MeV2 c − ⎡⎣2 ( 938.3)⎤⎦ MeV2 c = = 280 MeV ( 938.3) MeV c K ⎡ 91.2 × 10 − ⎡ 938.3 + 938.3 ⎤ MeV2 c ⎤ )⎦ ) ⎣( ⎥⎦ ⎣⎢( = = 4.43 TeV ( 938.3) MeV c 2 (d) MeV2 c − (139.6 + 938.3) MeV2 c 2 Particle Physics and Cosmology Section 46.7 Finding Patterns in the Particles Section 46.8 Quarks Section 46.9 Multicolored Quarks Section 46.10 The Standard Model P46.30 (a) 605 The number of protons ⎛ 6.02 × 10 23 molecules ⎞ ⎛ 10 protons ⎞ 26 N p = 000 g ⎜ ⎟ = 3.34 × 10 protons ⎟⎜ ⎝ ⎠ molecule 18.0 g ⎝ ⎠ and there are ⎛ 6.02 × 10 23 molecules ⎞ ⎛ neutrons ⎞ 26 Nn = 000 g ⎜ ⎟ = 2.68 × 10 neutrons ⎟⎜ 18.0 g ⎝ ⎠ ⎝ molecule ⎠ So there are for electric neutrality (b) 3.34 × 10 26 electrons The up quarks have number ( 3.34 × 10 26 ) + 2.68 × 10 26 = 9.36 × 10 26 up quarks and there are ( 2.68 × 10 26 ) + 3.34 × 10 26 = 8.70 × 10 26 down quarkss Model yourself as 65 kg of water Then you contain: 65 ( 3.34 × 10 26 ) ~10 28 electrons 65 ( 9.36 × 10 26 ) ~10 29 up quarks 65 ( 8.70 × 10 26 ) ~10 29 down quarks Only these fundamental particles form your body You have no strangeness, charm, topness, or bottomness P46.31 (a) proton strangeness baryon number charge (b) strangeness baryon number charge e u 1/3 2e/3 u 1/3 2e/3 d 1/3 –e/3 total e neutron u d d total 0 1/3 2e/3 1/3 –e/3 1/3 –e/3 606 P46.32 Chapter 46 Quark composition of proton = uud and of neutron = udd Thus, if we neglect binding energies, we may write and mp = mu + md (1) mn = mu + md (2) Solving simultaneously, mu = we find 1 mp − mn) = ⎡⎣2 ( 938 MeV c ) − 939.6 meV c ⎦⎤ = 312 MeV c ( 3 and from either (1) or (2), md = 314 MeV c P46.33 (a) strangeness baryon number charge (b) strangeness baryon number charge P46.34 K0 d s total 0 1/3 –e/3 –1/3 e/3 0 Λ0 u d s total –1 0 1/3 2e/3 1/3 –e/3 –1 1/3 –e/3 –1 In the first reaction, π − + p → K + Λ , the quarks in the particles are ud + uud → ds + uds There is a net of up quark both before and after the reaction, a net of down quarks both before and after, and a net of zero strange quarks both before and after Thus, the reaction conserves the net number of each type of quark In the second reaction, π − + p → K + n, the quarks in the particles are ud + uud → ds + udd In this case, there is a net of up and down quarks before the reaction but a net of up, down, and anti-strange quark after the reaction Thus, the reaction does not conserve the net number of each type of quark P46.35 (a) π − + p → K0 + Λ0 In terms of constituent quarks: up quarks: down quarks: strange quarks: (b) (c) π + + p → K + + Σ+ ud + uud → ds + uds −1 + → + 1, + → + 1, + → −1 + 1, or 1→1 2→2 0→0 or or du + uud → us + uus up quarks: + → + 2, or 3→ down quarks: −1 + → + 0, or 0→0 strange quarks: + → −1 + 1, or 0→0 or or or 1→1 1→1 1→1 K − + p → K + + K + Ω− up quarks: down quarks: strange quarks: continued on next page us + uud → us + ds + sss −1 + → + + 0, + → + + 0, + → −1 − + 3, Particle Physics and Cosmology (d) p + p → K0 + p + π + + ? 607 uud + uud → ds + uud + ud + ? The quark combination of ? must be such as to balance the last equation for up, down, and strange quarks up quarks: + = + +1+ ? (has u quark) down quarks: 1+1 = 1+1 −1+ ? (has d quark) + = −1 + + + ? (has s quark) strange quarks: quark composition = uds = Λ or Σ P46.36 Σ0 + p → Σ+ + γ + X dds + uud → uds + + ? The left side has a net 3d, 2u, and 1s The right-hand side has 1d, 1u, and 1s leaving 2d and 1u missing The unknown particle is a neutron, udd Baryon and strangeness numbers are conserved P46.37 P46.38 Compare the given quark states to the entries in Tables 46.4 and 46.5: (a) suu = Σ+ (b) ud = π − (c) sd = K (d) ssd = Ξ− (a) uud: ⎛ ⎞ ⎛ ⎞ ⎛1 ⎞ charge = ⎜− e ⎟ + ⎜− e ⎟ + ⎜ e ⎟ = −e This is the antiproton ⎝ ⎠ ⎝ ⎠ ⎝3 ⎠ (b) udd: ⎛ ⎞ ⎛1 ⎞ ⎛1 ⎞ charge = ⎜− e ⎟ + ⎜ e ⎟ + ⎜ e ⎟ = This is the antineutron ⎝ ⎠ ⎝3 ⎠ ⎝3 ⎠ Section 46.11 The Cosmic Connection *P46.39 We let r in Hubble’s law represent any distance (a) ⎞⎛ ⎛ ly ⎞ ⎛ ⎞ m yr c 1.85 m ⎜ ⎟⎜ ⎟ ⎟⎜ s ⋅ ly ⎝ c ⋅ yr ⎠ ⎝ × 10 m s ⎠ ⎝ 3.156 × 10 s ⎠ m 1.85 m = 1.80 × 10 −18 1.85 m = 3.32 × 10 −18 m s = 17 × 10 −3 s ⋅ 9.47 × 1015 m s v = Hr = 17 × 10 −3 This is unobservably small (b) v = Hr = 1.80 × 10 −18 3.84 × 10 m = 6.90 × 10 −10 m s s again too small to measure 608 P46.40 Chapter 46 Section 39.4 says fobserver = fsource + va c − va c The velocity of approach, va , is the negative of the velocity of mutual recession: va = −v c c 1− v c = λ′ λ 1+ v c Then, P46.41 (a) λ′ = λ 1+ v c 1− v c λ′ = λ and 510 nm = 434 nm 1+ v c = 1.381 1− v c v v + = 1.381 − 1.381 c c v = 0.160 or c 2.38 (b) v = HR : R= (a) λn′ = λn 1+ v c 1− v c 1+ v c 1− v c 1.18 = P46.42 1+ v = 0.381 c v = 0.160 c = 4.80 × 10 m s v 4.80 × 10 m s = = 2.82 × 10 ly H 17 × 10 −3 m s ⋅ ly 1+ v c = ( Z + 1) 1− v c 1+ v c = ( Z + 1) λn 1− v c ⎛v⎞ 2 ⎜ ⎟ ( Z + Z + 2) = Z + Z ⎝c⎠ ⎛v⎞ v 2 = ( Z + 1) − ⎜ ⎟ ( Z + 1) ⎝c⎠ c ⎛ Z + 2Z ⎞ v = c⎜ ⎝ Z + Z + ⎟⎠ (b) P46.43 R= v = HR v c ⎛ Z + 2Z ⎞ = H H ⎜⎝ Z + Z + ⎟⎠ (1.7 × 10 H= m s) −2 ly 1+ v c = 590 (1.000 113 3) = 590.07 nm 1− v c (a) v ( 2.00 × 10 ly ) = 3.4 × 10 m s λ′ = λ (b) v ( 2.00 × 10 ly ) = 3.4 × 10 m s λ ′ = 590 + 0.011 33 = 597 nm − 0.011 33 (c) v ( 2.00 × 10 ly ) = 3.4 × 10 m s λ ′ = 590 + 0.113 = 661 nm − 0.113 Particle Physics and Cosmology *P46.44 (a) (b) 609 What we can see is limited by the finite age of the Universe and by the finite speed of light We can see out only to a look-back time equal to a bit less than the age of the Universe Every year on your birthday the Universe also gets a year older, and light now in transit from still more distant objects arrives at Earth So the radius of the visible Universe expands at the speed of light, which is dr/dt = c = ly/yr The volume of the visible section of the Universe is (4/3)p r where r = 13.7 billion light-years The rate of volume increase is ⎛ × 10 ms 3.156 × 10 s ⎞ dV d 43 π r dr = = π 3r = π r c = π ⎜13.7 × 10 ly ⎟ × 10 ly dt dt dt ⎝ ⎠ m s = 6.34 × 1061 m3/s *P46.45 (a) The volume of the sphere bounded by the Earth’s orbit is 4 π r = π (1.496 × 1011 m ) = 1.40 × 10 34 m 3 m = ρV = × 10 −28 kg m 1.40 × 10 34 m = 8.41 × 10 kg V= (b) By Gauss’s law, the dark matter would create a gravitational field acting on the Earth to accelerate it toward the Sun It would shorten the duration of the year in the same way that 8.41 × 10 kg of extra material in the Sun would This has the fractional effect of 8.41 × 10 kg = 4.23 × 10 −24 of the mass of the Sun It is immeasurably small 1.99 × 10 30 kg P46.46 (a) Wien’s law: 2.898 × 10 −3 m ⋅ K 2.898 × 10 −3 m ⋅ K = = 1.06 × 10 −3 m = 1.06 mm T 2.73 K This is a microwave Thus, (b) P46.47 λmaxT = 2.898 × 10 −3 m ⋅ K λmax = We assume that the fireball of the Big Bang is a black body I = eσ T = (1) ( 5.67 × 10 −8 W m ⋅ K ) ( 2.73 K ) = 3.15 × 10 −6 W m As a bonus, we can find the current power of direct radiation from the Big Bang in the portion of the Universe observable to us If it is fourteen billion years old, the fireball is a perfect sphere of radius fourteen billion light years, centered at the point halfway between your eyes: ⎛ × 10 m s ⎞ ⎟ ( 3.156 × 10 s yr ) ⎝ ly yr ⎠ P = IA = I (4 π r ) = ( 3.15 × 10−6 W m ) ( π ) (14 × 10 ly ) ⎜ P = × 10 47 W 610 P46.48 Chapter 46 The density of the Universe is ⎛ 3H ⎞ ρ = 1.20 ρ c = 1.20 ⎜ ⎟ ⎝ 8π G ⎠ Consider a remote galaxy at distance r The mass interior to the sphere below it is ⎛ H ⎞ ⎛ ⎞ 0.600 H r M = ρ ⎛ π r ⎞ = 1.20 ⎜ πr = ⎝3 ⎠ ⎠ ⎝ 8π G ⎟⎠ ⎝ G both now and in the future when it has slowed to rest from its current speed v = Hr The energy of this galaxy-sphere system is constant as the galaxy moves to apogee distance R: GmM GmM = 0− mv − r R so Gm ⎛ 0.600 H r ⎞ Gm ⎛ 0.600 H r ⎞ mH r − = − ⎟⎠ ⎟⎠ r ⎜⎝ G R ⎜⎝ G r so R = 6.00 r R The Universe will expand by a factor of 6.00 from its current dimensions −0.100 = −0.600 P46.49 (a) kBT ≈ m p c so (b) (a) (b) m p c2 kB = ( 938.3 MeV) ⎛ 1.60 × 10 −13 J ⎞ 13 ⎜ ⎟ ~ 10 K (1.38 × 10−23 J K ) ⎝ MeV ⎠ kBT ≈ me c so P46.50 T≈ T≈ ( 0.511 MeV) ⎛ 1.60 × 10 −13 J ⎞ me c 10 = ⎜ ⎟ ~ 10 K kB (1.38 × 10−23 J K ) ⎝ MeV ⎠ The Hubble constant is defined in v = HR The distance R between any two far-separated objects opens at constant speed according to R = v t Then the time t since the Big Bang is found from v = H vt = Ht t= H ⎛ × 10 m s ⎞ 1 10 = ⎜ ⎟ = 1.76 × 10 yr = 17.6 billion years H 17 × 10 −3 m s ⋅ ly ⎝ ly yr ⎠ Particle Physics and Cosmology P46.51 (a) 611 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 GMm mv − =0 R The energy of the galaxy-sphere system is conserved, so this equation is true throughout the dR history of the Universe after the Big Bang, where v = Then dt R T ⎛ dR ⎞ GM dR R dR = GM ∫ d t = R−1/2 GM ⎜ ⎟ = ∫ 0 dt R ⎝dt ⎠ K + Ug = R 3/ 32 R = GM t 3/ R = GM T T 0 T= 2GM =v R From above, (b) P46.52 R R3 2 = GM 2GM R 2R 3v so T= Now Hubble’s law says v = HR So T= T= (17 × 10 −3 R = HR 3H ⎛ × 10 m s ⎞ 10 ⎜ ⎟ = 1.18 × 10 yr = 11.8 billion years ly yr m s ⋅ ly ) ⎝ ⎠ In our frame of reference, Hubble’s law is exemplified by v1 = H R1 and v = H R From these we may form the equations −v1 = − H R1 and v − v1 = H R − R1 These equations express Hubble’s law as seen by the observer in the first galaxy cluster, as she looks at us to ( ( ) ) ( ) find −v1 = H −R1 and as she looks at cluster two to find v − v1 = H R − R1 Section 46.12 *P46.53 (a) (b) (c) Problems and Perspectives L= G = c3 (1.055 × 10 − 34 J ⋅ s) ( 6.67 × 10 −11 N ⋅ m kg ) ( 3.00 × 10 m s) = 1.61 × 10 −35 m L 1.61 × 10 − 35 m = = 5.38 × 10 −44 s , c 3.00 × 10 m s which is approximately equal to the duration of the ultra-hot epoch The Planck time is given as T = Yes The uncertainty principle foils any attempt at making observations of things when the age of the Universe was less than the Planck time The opaque fireball of the Big Bang, measured as the cosmic microwave background radiation, prevents us from receiving visible light from things before the Universe was a few hundred thousand years old Walls of more profound fire hide all information from still earlier times 612 Chapter 46 Additional Problems P46.54 We find the number N of neutrinos: 10 46 J = N ( MeV) = N ( × 1.60 × 10 −13 J ) N = 1.0 × 10 58 neutrinos The intensity at our location is ⎛ ⎞ 1.0 × 10 58 ly N N −2 14 ⎜ ⎟ = = 2 ⎜ 3.00 × 10 m s 3.16 × 10 s ⎟ = 3.1 × 10 m A 4π r )( )⎠ π (1.7 × 10 ly ) ⎝ ( The number passing through a body presenting 000 cm = 0.50 m ⎛ ⎞ 14 0.50 m ) = 1.5 × 1014 ⎜ 3.1 × 10 ⎟( ⎝ ⎠ m is then ~ 1014 or P46.55 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) v 170 000 yr 1 = = = c 170 000 yr + 10 s + 10 s ⎡(1.7 × 10 yr ) ( 3.156 × 10 s yr ) ⎤ + 1.86 × 10 −12 ⎣ ⎦ { } For the neutrino we want to evaluate mc in E = γ mc 2: v2 E mc = = E − = 10 MeV − = 10 MeV γ c (1 + 1.86 × 10−12 ) mc ≈ 10 MeV (1.86 × 10 −12 ) (1 + 1.86 × 10 ) − (1 + 1.86 × 10 ) −12 −12 = 10 MeV (1.93 × 10 −6 ) = 19 eV Then the upper limit on the mass is m= P46.56 P46.57 19 eV c2 or m= ⎞ 19 eV ⎛ u = 2.1 × 10 −8 u ⎜ ⎟ c ⎝ 931.5 × 10 eV c ⎠ (a) π − + p → Σ+ + π is forbidden by charge conservation (b) µ− → π − + νe is forbidden by energy conservation (c) p → π+ + π+ + π− is forbidden by baryon number conservation The total energy in neutrinos emitted per second by the Sun is: ( 0.4 ) ⎡⎣4π (1.5 × 1011 ) ⎤⎦ W = 1.1 × 10 23 W Over 10 years, the Sun emits 3.6 × 10 39 J in neutrinos This represents an annihilated mass m c = 3.6 × 10 39 J m = 4.0 × 10 22 kg About part in 50 000 000 of the Sun’s mass, over 10 years, has been lost to neutrinos Particle Physics and Cosmology P46.58 613 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 ( ) ( M p c2 + M π c2 + M X c2 = M p c2 + K p + M p c2 + K p ) M X c = M p c + K p − M π 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 P46.59 (a) If 2N particles are annihilated, the energy released is Nmc The resulting photon E Nmc momentum is p = = = Nmc Since the momentum of the system is conserved, c c the rocket will have momentum 2Nmc directed opposite the photon momentum p = Nmc (b) Consider a particle that is annihilated and gives up its rest energy mc to another particle which also has initial rest energy mc (but no momentum initially) E = p c + ( mc ) 2 2 Thus ( mc ) = p c + ( mc ) 2 where p is the momentum the second particle acquires as a result of the annihilation of the 2 first particle Thus ( mc ) = p c + ( mc ) , p = ( mc ) So p = 3mc N N protons and antiprotons) Thus the total 2 momentum acquired by the ejected particles is 3Nmc, and this momentum is imparted to the rocket This process is repeated N times (annihilate p = Nmc (c) P46.60 Method (a) produces greater speed since Nmc > 3Nmc By relativistic energy conservation in the reaction, Eγ + me c = By relativistic momentum conservation for the system, Eγ Dividing (2) by (1), X= Subtracting (2) from (1), me c = Solving, = − 3X 1− X and X = c so Eγ = me c = 2.04 MeV = 3me c − v2 c2 3me v (2) − v2 c2 Eγ Eγ + me c = 3me c 1− X (1) v c − 3me c X 1− X 614 P46.61 Chapter 46 mΛ c = 1115.6 MeV Λ0 → p + π − m p c = 938.3 MeV mπ c = 139.6 MeV The difference between starting rest energy and final rest energy is the kinetic energy of the products K p + K π = 37.7 MeV and p p = pπ = p Applying conservation of relativistic energy to the decay process, we have ⎡ ⎣⎢ (938.3) 2 ⎤ ⎡ ⎤ + p c − 938.3⎥ + ⎢ (139.6 ) + p c − 139.6⎥ = 37.7 MeV ⎦ ⎣ ⎦ Solving the algebra yields pπ c = p p c = 100.4 MeV Then, P46.62 Kp = (m c ) Kπ = (139.6 )2 + (100.4 )2 − 139.6 = 32.3 MeV 2 p + (100.4 ) − m p c = 5.35 MeV 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 ( ) m p c + K p = m p c + mn c + mπ c so the kinetic energy of each of the incident protons is Kp = P46.63 mn c + mπ c − m p c 2 = (939.6 + 139.6 − 938.3) MeV = 70.4 MeV Σ0 → Λ0 + γ From Table 46.2, mΣ = 1192.5 MeV c and mΛ = 1115.6 MeV c Conservation of energy in the decay requires ( ) E0, Σ = Eo, Λ + K Λ + Eγ or ⎛ p ⎞ 1192.5 MeV = ⎜1115.6 MeV + Λ ⎟ + Eγ mΛ ⎠ ⎝ System momentum conservation gives pΛ = pγ , so the last result may be written as ⎛ p2 ⎞ 1192.5 MeV = ⎜⎜1115.6 MeV + γ ⎟⎟ + Eγ mΛ ⎠ ⎝ or ⎛ p c2 ⎞ 1192.5 MeV = ⎜⎜1115.6 MeV + γ ⎟⎟ + Eγ mΛ c ⎠ ⎝ Recognizing that mΛ c = 1115.6 MeV we now have 1192.5 MeV = 1115.6 MeV + Solving this quadratic equation gives Eγ = 74.4 MeV and pγ c = Eγ Eγ 2 (1115.6 MeV) + Eγ Particle Physics and Cosmology P46.64 615 The momentum of the proton is qBr = (1.60 × 10 −19 C) ( 0.250 kg C ⋅ s)(1.33 m ) p p = 5.32 × 10 −20 kg ⋅ m s cp p = 1.60 × 10 −11 kg ⋅ m s = 1.60 × 10 −11 J = 99.8 MeV p p = 99.8 MeV c Therefore E p = E02 + ( cp ) = The total energy of the proton is (938.3) + (99.8) 2 = 944 MeV For the pion, the momentum qBr is the same (as it must be from conservation of momentum in a 2-particle decay) pπ = 99.8 MeV c E0π = 139.6 MeV Eπ = E02 + ( cp ) = (139.6 )2 + ( 99.8)2 = 172 MeV Thus Etotal after = Etotal before = Rest energy Rest energy of unknown particle = 944 MeV + 172 MeV = 1116 MeV (This is a Λ particle!) Mass = 1116 MeV c P46.65 π − → µ− + νµ : From the conservation laws for the decay, mπ c = 139.6 MeV = Eµ + Eν P46.66 ( ) [1] and pµ = pν , Eν = pν c : Eµ2 = pµ c + (105.7 MeV ) = ( pν c ) + (105.7 MeV ) or Eµ2 − Eν2 = (105.7 MeV) Since Eµ + Eν = 139.6 MeV and (E then Eµ − Eν = Subtracting [3] from [1], Eν = 59.6 MeV µ 2 )( 2 [2] [1] ) + Eν Eµ − Eν = (105.7 MeV ) (105.7 MeV )2 139.6 MeV = 80.0 and [3] Eν = 29.8 MeV The expression e− E kBT dE gives the fraction of the photons that have energy between E and E + dE The fraction that have energy between E and infinity is ∞ ∞ ∫ e− E kBT dE E ∞ ∫e − E kBT dE = ∫e E ∞ ∫e − E kBT (−dE kBT ) = − E kBT (−dE kBT ) e− E kBT e ∞ E − E kBT ∞ = e− E kBT 0 We require T when this fraction has a value of 0.0100 (i.e., 1.00%) and E = 1.00 eV = 1.60 × 10 −19 J Thus, − 1.60 ×10 −19 J ) (1.38×10 −23 J K )T 0.010 = e ( or ln ( 0.010 ) = − [2] 1.60 × 10 −19 J 1.16 × 10 K = − giving T = 2.52 × 10 K T (1.38 × 10 −23 J K ) T 616 P46.67 Chapter 46 (a) This diagram represents the annihilation of an electron and an antielectron From charge and lepton-number conservation at either vertex, the exchanged particle must be an electron, e − (b) This is the tough one A neutrino collides with a neutron, changing it into a proton with release of a muon This is a weak interaction The exchanged particle has charge +e and is a W + FIG P46.67 P46.68 (a) The mediator of this weak interaction is a Z boson (b) The Feynman diagram shows a down quark and its antiparticle annihilating each other They can produce a particle carrying energy, momentum, and angular momentum, but zero charge, zero baryon number, and, it may be, no color charge In this case the product particle is a photon FIG P46.68 For conservation of both energy and momentum in the collision we would expect two photons; but momentum need not be strictly conserved, according to the uncertainty principle, if the photon travels a sufficiently short distance before producing another matter-antimatter pair of particles, as shown in Figure P46.68 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) P46.69 (a) At threshold, we consider a photon and a proton colliding head-on to produce a proton and a pion at rest, according to p + γ → p + π Energy conservation gives mp c 2 1− u c + Eγ = mp c + mπ c mp u Momentum conservation gives 1− u c 2 − Eγ c = Combining the equations, we have mp c 1− u c 2 + mp c u = mp c + mπ c c 1− u c 938.3 MeV (1 + u c ) (1 − u c) (1 + u c) (b) so u = 0.134 c and Eγ = 127 MeV λmaxT = 2.898 mm ⋅ K λmax = continued on next page 2.898 mm ⋅ K = 1.06 mm 2.73 K = 938.3 MeV + 135.0 MeV Particle Physics and Cosmology (c) (d) 617 hc 240 eV ⋅ 10 −9 m = = 1.17 × 10 −3 eV −3 λ 1.06 × 10 m u′ In the primed reference frame, the proton is moving to the right at = 0.134 and c the photon is moving to the left with hf ′ = 1.27 × 10 eV In the unprimed frame, hf = 1.17 × 10 −3 eV Using the Doppler effect equation from Section 39.4, we have for the speed of the primed frame Eγ = hf = 1.27 × 10 = 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 c + u ′v c + 0.134 (1 − 1.71 × 10 −22 ) And the energy of the proton is m p c2 1− u c 2 = 938.3 MeV − (1 − 1.30 × 10 ) −22 = 6.19 × 1010 × 938.3 × 10 eV = 5.81 × 1019 eV ANSWERS TO EVEN PROBLEMS P46.2 ~103 Bq P46.4 2.27 × 10 23 Hz; 1.32 fm P46.6 ν µ and ν e P46.8 (b) The range is inversely proportional to the mass of the field particle (c) Our rule describes the electromagnetic, weak, and gravitational interactions For the electromagnetic and gravitational interactions, we take the limiting form of the rule with infinite range and zero mass for the field particle For the weak interaction, 98.7 eV ⋅ nm/90 GeV ≈ 10–18 m = 10–3 fm, in agreement with the tabulated information (d) ~10–16 m P46.10 ~10 −23 s P46.12 νµ P46.14 (b) The second violates strangeness conservation P46.16 The second violates conservation of baryon number P46.18 0.828c P46.20 (a) ν e P46.22 See the solution P46.24 (a) not allowed; violates conservation of baryon number (b) strong interaction interaction (d) weak interaction (e) electromagnetic interaction P46.26 (a) K+ (b) Ξ0 (b) ν µ (c) ν µ (c) p (d) ν µ + ν τ (c) weak 618 P46.28 Chapter 46 (a) 686 MeV 200 MeV and c c (b) 627 MeV c (c) 244 MeV, 1130 MeV, 370 MeV (d) 1190 MeV c 2, 0.500 c P46.30 (a) 3.34 × 1026 e−, 9.36 × 1026 u, 8.70 × 1026 d (b) ~1028 e−, ~1029 u, ~1029 d I have zero strangeness, charm, topness, and bottomness P46.32 mu = 312 MeV/c2 P46.34 See the solution P46.36 a neutron, udd P46.38 (a) –e, antiproton b) 0, antineutron P46.40 See the solution P46.42 ⎛ Z + 2Z ⎞ (a) v = c ⎜ ⎝ Z + Z + ⎟⎠ md = 314 MeV/c2 (b) c ⎛ Z + 2Z ⎞ ⎜ ⎟ H ⎝ Z + 2Z + ⎠ P46.44 (a) What we can see is limited by the finite age of the Universe and by the finite speed of light We can see out only to a look-back time equal to a bit less than the age of the Universe Every year on your birthday the Universe also gets a year older, and light now in transit from still more distant objects arrives at Earth So the radius of the visible Universe expands at the speed of light, which is ly/yr (b) 6.34 × 1061 m3/s P46.46 (a) 1.06 mm P46.48 6.00 P46.50 (a) See the solution P46.52 See the solution P46.54 ~1014 P46.56 (a) charge (b) energy P46.58 neutron P46.60 2.04 MeV P46.62 70.4 MeV P46.64 116 MeV/c2 P46.66 2.52 × 103 K P46.68 (a) Z0 boson (b) microwave (b) 17.6 Gyr (c) baryon number (b) gluon or photon ... in One Dimension P2.3 (a) Let d represent the distance between A and B Let t be the time for which the walker has d the higher speed in 5.00 m s = Let t represent the longer time for the return... again at the end of the time interval Q2.3 Yes Yes If the speed of the object varies at all over the interval, the instantaneous velocity will sometimes be greater than the average velocity and... g ( j) We should expect agreement in parts b-c-d, because those parts are about a uniform sphere of density 4.7 g/cm3 We should expect agreement in parts e- f-g, because those parts are about