Alan giambattista, betty richardson, robert c richardson instructors solution manuals to physics mcgrawhill (2010)

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Des c a r gaL i br osUni v er s i t a r i osGr a t i senPDF Chapter INTRODUCTION Conceptual Questions Knowledge of physics is important for a full understanding of many scientific disciplines, such as chemistry, biology, and geology Furthermore, much of our current technology can only be understood with knowledge of the underlying laws of physics In the search for more efficient and environmentally safe sources of energy, for example, physics is essential Also, many study physics for the sense of fulfillment that comes with learning about the world we inhabit Without precise definitions of words for scientific use, unambiguous communication of findings and ideas would be impossible Even when simplified models not exactly match real conditions, they can still provide insight into the features of a physical system Often a problem would become too complicated if one attempted to match the real conditions exactly, and an approximation can yield a result that is close enough to the exact one to still be useful (a) (b) Scientific notation eliminates the need to write many zeros in very large or small numbers Also, the appropriate number of significant digits is unambiguous when written this way In scientific notation the decimal point is placed after the first (leftmost) numeral The number of digits written equals the number of significant figures Not all of the significant digits are precisely known The least significant digit (rightmost) is an estimate and is less precisely known than the others It is important to list the correct number of significant figures so that we can indicate how precisely a quantity is known and not mislead the reader by writing digits that are not at all known to be correct The kilogram, meter, and second are three of the base units used in the SI system 10 The SI system uses a well-defined set of internationally agreed upon standard units and makes measurements in terms of these units and their powers of ten The U.S Customary system contains units that are primarily of historical origin and are not based upon powers of ten As a result of this international acceptance and the ease of manipulation that comes from dealing with powers of ten, scientists around the world prefer to use the SI system 11 Fathoms, kilometers, miles, and inches are units with dimensions of length Grams and kilograms are units with dimensions of mass Years, months, and seconds are units with dimensions of time 12 The first step toward successfully solving almost any physics problem is to thoroughly read the question and obtain a precise understanding of the scenario The second step is to visualize the problem, often making a quick sketch to outline the details of the situation and the known parameters 13 Trends in a set of data are often the most interesting aspect of the outcome of an experiment Such trends are more apparent when data is plotted graphically rather than listed in numerical tables 14 The statement gives a numerical value for the speed of sound in air, but fails to indicate the units used for the measurement Without units, the reader cannot relate the speed to one given in familiar units such as km/s 165 Chapter 1: Introduction Physics 15 After solving a problem, it is a good idea to check that the solution is reasonable and makes intuitive sense It may also be useful to explore other possible methods of solution as a check on the validity of the first Problems Strategy The new fence will be 100% + 37% = 137% of the height of the old fence Solution Find the height of the new fence 1.37 × 1.8 m = 2.5 m Strategy There are 60 s 60 24 h × × = 86, 400 seconds in one day and 24 hours in one day 1h 1d Solution Find the ratio of the number of seconds in a day to the number of hours in a day 86, 400 24 × 3600 = = 3600 24 24 Strategy Relate the surface area S to the radius r using S = 4π r Solution Find the ratio of the new radius to the old S1 = 4π r12 and S2 = 4π r22 = 1.160S1 = 1.160(4π r12 ) 4π r22 = 1.160(4π r12 ) r22 = 1.160r12 ⎛ r2 ⎞ ⎜⎜ ⎟⎟ = 1.160 ⎝ r1 ⎠ r2 = 1.160 = 1.077 r1 The radius of the balloon increases by 7.7% Strategy Relate the surface area S to the radius r using S = 4π r Solution Find the ratio of the new radius to the old S1 = 4π r12 and S2 = 4π r22 = 2.0 S1 = 2.0(4π r12 ) 4π r22 = 2.0(4π r12 ) r22 = 2.0r12 ⎛ r2 ⎞ ⎜⎜ ⎟⎟ = 2.0 ⎝ r1 ⎠ r2 = 2.0 = 1.4 r1 The radius of the balloon increases by a factor of 1.4 166 Physics Chapter 1: Introduction Strategy The surface area S and the volume V are given by S = 6s and V = s3 , respectively Solution Find the ratio of the surface area to the volume S 6s = = V s s Strategy To find the factor Samantha’s height increased, divide her new height by her old height Subtract from this value and multiply by 100 to find the percent increase Solution Find the factor 1.65 m = 1.10 1.50 m Find the percentage 1.10 − = 0.10, so the percent increase is 10 % Strategy Recall that area has dimensions of length squared Solution Find the ratio of the area of the park as represented on the map to the area of the actual park map length map area = = 10−4 , so = (10−4 ) = 10−8 actual length 10, 000 actual area Strategy Let X be the original value of the index Solution Find the net percentage change in the index for the two days (first day change) × (second day change) = [ X × (1 + 0.0500)] × (1 − 0.0500) = 0.9975 X The net percentage change is (0.9975 − 1) × 100% = −0.25%, or down 0.25% Strategy Use a proportion Solution Find Jupiter’s orbital period T R3 T ∝ R3 , so J = J = 5.193 Thus, TJ = 5.193/2 TE = 11.8 yr TE2 RE3 10 Strategy The area of the circular garden is given by A = π r Let the original and final areas be A1 = π r12 and A2 = π r22 , respectively Solution Calculate the percentage increase of the area of the garden plot π r − π r12 r2 − r2 1.252 r12 − r12 ∆A 1.252 − × 100% = × 100% = × 100% = × 100% = × 100% = 56% A π r12 r12 r12 11 Strategy The area of the poster is given by A = w Let the original and final areas be A1 = 1w1 and A2 = w2 , respectively Solution Calculate the percentage reduction of the area A2 = w2 = (0.800 )(0.800w1 ) = 0.640 1w1 = 0.640 A1 A1 − A2 A − 0.640 A1 × 100% = × 100% = 36.0% A1 A1 167 free online eBooks and Solutions Manual can be found at www.elsolucionario.org Chapter 1: Introduction Physics 12 Strategy The volume of the rectangular room is given by V = wh Let the original and final volumes be V1 = 1w1h1 and V2 = w2 h2 , respectively Solution Find the factor by which the volume of the room increased V2 w h (1.50 )(2.00 w1 )(1.20h1 ) = 2 = = 3.60 V1 1w1h1 1w1h1 13 (a) Strategy Rewrite the numbers so that the power of 10 is the same for each Then add and give the answer with the number of significant figures determined by the less precise of the two numbers Solution Perform the operation with the appropriate number of significant figures 3.783 × 106 kg + 1.25 × 108 kg = 0.03783 × 108 kg + 1.25 × 108 kg = 1.29 × 108 kg (b) Strategy Find the quotient and give the answer with the number of significant figures determined by the number with the fewest significant figures Solution Perform the operation with the appropriate number of significant figures (3.783 ì 106 m) ữ (3.0 ì 102 s) = 1.3 × 108 m s 14 (a) Strategy Move the decimal point eight places to the left and multiply by 108 Solution Write the number in scientific notation 290,000,000 people = 2.9 × 108 people (b) Strategy Move the decimal point 15 places to the right and multiply by 10−15 Solution Write the number in scientific notation 0.000 000 000 000 003 m = 3.8 × 10−15 m 15 (a) Strategy Rewrite the numbers so that the power of 10 is the same for each Then subtract and give the answer with the number of significant figures determined by the less precise of the two numbers Solution Perform the calculation using an appropriate number of significant figures 3.68 × 107 g − 4.759 × 105 g = 3.68 × 107 g − 0.04759 × 107 g = 3.63 × 107 g (b) Strategy Find the quotient and give the answer with the number of significant figures determined by the number with the fewest significant figures Solution Perform the calculation using an appropriate number of significant figures 6.497 × 104 m = 1.273 × 102 m 5.1037 × 102 m 168 www.elsolucionario.org Physics Chapter 1: Introduction 16 (a) Strategy Rewrite the numbers so that the power of 10 is the same for each Then add and give the answer with the number of significant figures determined by the less precise of the two numbers Solution Write your answer using the appropriate number of significant figures 6.85 × 10−5 m + 2.7 × 10−7 m = 6.85 × 10−5 m + 0.027 × 10−5 m = 6.88 × 10−5 m (b) Strategy Add and give the answer with the number of significant figures determined by the less precise of the two numbers Solution Write your answer using the appropriate number of significant figures 702.35 km + 1897.648 km = 2600.00 km (c) Strategy Multiply and give the answer with the number of significant figures determined by the number with the fewest significant figures Solution Write your answer using the appropriate number of significant figures 5.0 m × 4.3 m = 22 m (d) Strategy Find the quotient and give the answer with the number of significant figures determined by the number with the fewest significant figures Solution Write your answer using the appropriate number of significant figures ( 0.04 π ) cm = 0.01 cm (e) Strategy Find the quotient and give the answer with the number of significant figures determined by the number with the fewest significant figures Solution Write your answer using the appropriate number of significant figures ( 0.040 π ) m = 0.013 m 17 Strategy Multiply and give the answer in scientific notation with the number of significant figures determined by the number with the fewest significant figures Solution Solve the problem (3.2 m) × (4.0 × 10−3 m) × (1.3 × 10−8 m) = 1.7 × 10−10 m3 18 Strategy Follow the rules for identifying significant figures Solution (a) All three digits are significant, so 7.68 g has significant figures (b) The first zero is not significant, since it is used only to place the decimal point The digits and are significant, as is the final zero, so 0.420 kg has significant figures (c) The first two zeros are not significant, since they are used only to place the decimal point The digits and are significant, so 0.073 m has significant figures (d) All three digits are significant, so 7.68 × 105 g has significant figures 169 www.elsolucionario.org Chapter 1: Introduction Physics (e) The zero is significant, since it comes after the decimal point The digits and are significant as well, so 4.20 × 103 kg has significant figures (f) Both and are significant, so 7.3 × 10−2 m has significant figures (g) Both and are significant The two zeros are significant as well, since they come after the decimal point, so 2.300 × 104 s has significant figures 19 Strategy Divide and give the answer with the number of significant figures determined by the number with the fewest significant figures Solution Solve the problem 3.21 m 3.21 m = = 459 m s 7.00 ms 7.00 × 10−3 s 20 Strategy Convert each length to meters Then, rewrite the numbers so that the power of 10 is the same for each Finally, add and give the answer with the number of significant figures determined by the less precise of the two numbers Solution Solve the problem 3.08 × 10−1 km + 2.00 × 103 cm = 3.08 × 102 m + 2.00 × 101 m = 3.08 × 102 m + 0.200 × 102 m = 3.28 × 102 m 21 Strategy There are approximately 39.37 inches per meter Solution Find the thickness of the cell membrane in inches 7.0 × 10−9 m × 39.37 inches m = 2.8 × 10−7 inches 22 (a) Strategy There are approximately 3.785 liters per gallon and 128 ounces per gallon Solution Find the number of fluid ounces in the bottle 128 fl oz gal 1L × × 355 mL × = 12.0 fluid ounces gal 3.785 L 103 mL (b) Strategy From part (a), we have 355 mL = 12.0 fluid ounces Solution Find the number of milliliters in the drink 355 mL 16.0 fl oz × = 473 mL 12.0 fl oz 23 Strategy There are approximately 3.281 feet per meter Solution Convert to meters and identify the order of magnitude (a) 1595.5 ft × (b) 6016 ft × 1m = 4.863 × 102 m ; the order of magnitude is 102 3.281 ft 1m = 1.834 × 103 m ; the order of magnitude is 103 3.281 ft 170 www.elsolucionario.org Physics Chapter 1: Introduction 24 Strategy There are 3600 seconds in one hour and 1000 m in one kilometer Solution Convert 1.00 kilometers per hour to meters per second 1.00 km 1h 1000 m × × = 0.278 m s 1h 3600 s km 25 (a) Strategy There are 60 seconds in one minute, 5280 feet in one mile, and 3.28 feet in one meter Solution Express 0.32 miles per minute in meters per second 0.32 mi 5280 ft 1m × × × = 8.6 m s 60 s mi 3.28 ft (b) Strategy There are 60 minutes in one hour Solution Express 0.32 miles per minute in miles per hour 0.32 mi 60 × = 19 mi h 1h 26 Strategy There are 0.6214 miles in kilometer Solution Find the length of the marathon race in miles 0.6214 mi = 26.22 mi 42.195 km × km 27 Strategy Calculate the change in the exchange rate and divide it by the original price to find the drop Solution Find the actual drop in the value of the dollar over the first year 1.27 − 1.45 −0.18 = = −0.12 1.45 1.45 The actual drop is 0.12 or 12% 28 Strategy There are 1000 watts in one kilowatt and 100 centimeters in one meter Solution Convert 1.4 kW m to W cm 1.4 kW 1000 W ⎛ m ⎞ × ×⎜ ⎟ = 0.14 W cm kW ⎝ 100 cm ⎠ m2 29 Strategy There are 1000 grams in one kilogram and 100 centimeters in one meter Solution Find the density of mercury in units of g cm3 1.36 × 104 kg 1000 g ⎛ m ⎞ × ×⎜ ⎟ = 13.6 g cm kg ⎝ 100 cm ⎠ m3 30 Strategy The distance traveled d is equal to the rate of travel r times the time of travel t There are 1000 milliseconds in one second Solution Find the distance the molecule would move 459 m 1s d = rt = × 7.00 ms × = 3.21 m 1s 1000 ms 171 www.elsolucionario.org Chapter 1: Introduction Physics 31 Strategy There are 1000 meters in a kilometer and 1,000,000 millimeters in a kilometer Solution Find the product and express the answer in km3 with the appropriate number of significant figures km km (3.2 km) × (4.0 m) × (13 × 10−3 mm) × × = 1.7 × 10−10 km3 1000 m 1,000,000 mm 32 (a) Strategy There are 12 inches in one foot and 2.54 centimeters in one inch Solution Find the number of square centimeters in one square foot 2 ⎛ 12 in ⎞ ⎛ 2.54 cm ⎞ ft × ⎜ ⎟ ×⎜ ⎟ = 929 cm ft in ⎝ ⎠ ⎝ ⎠ (b) Strategy There are 100 centimeters in one meter Solution Find the number of square centimeters in one square meter ⎛ 100 cm ⎞ m2 × ⎜ ⎟ = 1× 10 cm ⎝ 1m ⎠ (c) Strategy Divide one square meter by one square foot Estimate the quotient Solution Find the approximate number of square feet in one square meter m 10, 000 cm = ≈ 11 ft 929 cm 33 (a) Strategy There are 12 inches in one foot, 2.54 centimeters in one inch, and 60 seconds in one minute Solution Express the snail’s speed in feet per second 5.0 cm 1min in ft × × × = 2.7 × 10−3 ft s 60 s 2.54 cm 12 in (b) Strategy There are 5280 feet in one mile, 12 inches in one foot, 2.54 centimeters in one inch, and 60 minutes in one hour Solution Express the snail’s speed in miles per hour 5.0 cm 60 in ft mi × × × × = 1.9 × 10−3 mi h 1h 2.54 cm 12 in 5280 ft 34 Strategy A micrometer is 10−6 m and a millimeter is 10−3 m; therefore, a micrometer is 10−6 10−3 = 10−3 mm Solution Find the area in square millimeters 103 mm 150 àm ì = 1.5 ì 104 mm àm ⎟⎟ ⎝ ⎠ 35 Strategy Replace each quantity in U = mgh with its SI base units Solution Find the combination of SI base units that are equivalent to joules U = mgh ⇒ J = kg × m s × m = kg ⋅ m ⋅ s −2 172 www.elsolucionario.org Physics Chapter 1: Introduction 36 (a) Strategy Replace each quantity in ma and kx with its dimensions Solution Show that the dimensions of ma and kx are equivalent [L] [M] [L] ma has dimensions [M] × × [L] = [M] × and kx has dimensions 2 [T] [T] [T]2 Since [M][L][T]−2 = [M][L][T]−2 , the dimensions are equivalent (b) Strategy Use the results of part (a) Solution Since F = ma and F = − kx, the dimensions of the force unit are [M][L][T]−2 37 Strategy Replace each quantity in T = 4π r (GM ) with its dimensions Solution Show that the equation is dimensionally correct 4π r [L]3 [L]3 [M][T]2 has dimensions = × = [T]2 T has dimensions [T]2 and 3 [L] [M] GM [L] × [M] [M][T]2 Since [T]2 = [T]2 , the equation is dimensionally correct 38 Strategy Determine the SI unit of momentum using a process of elimination Solution Find the SI unit of momentum p2 kg ⋅ m kg ⋅ m Since the SI unit for m is kg, the SI unit for p is Taking the square has units of K= 2m s s2 root, we find that the SI unit for momentum is kg ⋅ m ⋅ s −1 39 (a) Strategy Replace each quantity (except for V) in FB = ρ gV with its dimensions Solution Find the dimensions of V F [MLT −2 ] V = B has dimensions = [L3 ] −3 −2 ρg [ML ] × [LT ] (b) Strategy and Solution Since velocity has dimensions [LT −1 ] and volume has dimensions [L3 ], the correct interpretation of V is that is represents volume 40 Strategy Replace v, r, ω , and m with their dimensions Then use dimensional analysis to determine how v depends upon some or all of the other quantities [L] , [L], , and [M], respectively No combination of r, ω , and m [T] [T] [L] = and there is no dimensionless gives dimensions without [M], so v does not depend upon m Since [L] × [T] [T] Solution v, r , ω , and m have dimensions constant involved in the relation, v is equal to the product of ω and r , or v = ω r 173 www.elsolucionario.org Chapter 30: Particle Physics Physics 13 Strategy Use Newton’s second law The magnitude of the force on a charged particle moving at speed v perpendicular to a magnetic field of strength B is F = qvB Since the particle is extremely relativistic, use p ≈ E c Solution Find the magnetic field strength v2 mv p E E 7.0 × 1012 eV × 2π ΣFr = qvB = mar = m , so B = = ≈ = = = 5.4 T C r qr qr qrc q × 2π × c e × 27 × 103 m × 3.00 × 108 m s 14 Strategy The difference in the rest energy of the particles before and after the decay is related to the kinetic energy of the particles by Einstein’s mass-energy relation Use conservation of momentum Solution Compute the change in mass ∆m = 0.5110 MeV c − 106 MeV c = −105.49 MeV c The energy released in the decay is E = ∆m c = 105.49 MeV The electron will have maximum kinetic energy when the neutrino and antineutrino move in the same direction, and the electron moves in the opposite direction Conservation of momentum then requires that pe = pν + pν , so pe c = pν c + pν c , which means K e = Eν + Eν , since all three particles are extremely relativistic Find the total energy of the particles E 105.49 MeV = 52.7 MeV E = K e + Eν + Eν = K e + K e , so K e = = 2 15 Strategy The rest energy of the pion is 0.135 GeV Conservation of energy requires that the energy of the pion be equal to the energy of the two photons Conservation of momentum requires that the momenta of the photons are equal and opposite Solution Relate the energies Eπ = Eγ + Eγ Relate the wavelengths of the photons h h pγ = = pγ = , so λ1 = λ2 λ1 λ2 The photons are identical, so let Eγ = Eγ = Eγ Find the energy of each photon E 0.135 GeV Eπ = Eγ + Eγ = Eγ , so Eγ = π = = 67.5 MeV 2 16 Strategy The difference in the rest energy of the particles before and after the decay is related to the kinetic energy of the particles by Einstein’s mass-energy relation Conservation of momentum requires that the magnitude of the momenta of the two pions be the same Since the pions have the same mass, their velocities must be equal and opposite; thus, their kinetic energies are the same Solution Compute the change in mass ∆m = 2(0.14 GeV/c ) − 2(0.938 GeV/c ) = −1.6 GeV/c The energy released in the decay is E = ∆m c = 1.6 GeV Find the kinetic energies of the two pions Kπ − = Kπ + = E = (1.6 GeV) = 0.80 GeV 1132 www.elsolucionario.org Physics Chapter 30: Particle Physics 17 Strategy Use conservation of energy and momentum Solution Conservation of energy requires that E p1 + E p2 = E p3 + E p4 + E p5 + E p Assuming p1 and p2 have equal kinetic energies, and that p3, p4 , p5 , and p have zero kinetic energy, we have K + E0 + K + E0 = E0 where E0 is the rest energy of a proton (or antiproton) and K is the initial kinetic energy required Solve for K K = E0 , so K = E0 = 938 MeV 18 Strategy Mesons contain one quark and one antiquark Baryons contain three quarks Use Table 30.1 to determine the quark content of each particle Solution The charges of an up quark, a strange quark, and a strange antiquark are 23 e, − 13 e, and 13 e, respectively Therefore, the quark content of the meson with charge + e is us 19 Strategy Mesons contain one quark and one antiquark Baryons contain three quarks Use Table 30.1 to determine the quark content of each particle Solution The charge of an up quark is 23 e and a strange quark has charge − 13 e Therefore, the quark content of the baryon with charge is uss 20 Strategy Mesons contain one quark and one antiquark Baryons contain three quarks Use Table 30.1 to determine the quark content of each particle Solution The charge of the antibaryon + e Antibaryons are composed of three antiquarks The charge of an up antiquark is − 23 e The charge of a strange antiquark is 13 e It is not possible to combine any number of up antiquarks together or with strange antiquarks to give the antibaryon a charge of + e, so the antibaryon must be composed of three strange antiquarks Therefore, the quark content of the antibaryon with charge + e is s s s 21 Strategy Mesons contain one quark and one antiquark Baryons contain three quarks Use Table 30.1 to determine the quark content of each particle Solution The charge of an up antiquark is − 23 e, and that of a down quark is − 13 e Therefore, the quark content of the meson with charge − e is ud 22 Strategy Make an order of magnitude estimate Use Figure 30.2 Solution Between 10−37 s and 10−10 s after the big bang, the range of energies was 1024 eV and 1011 eV, respectively An energy of TeV = 1012 eV falls just inside the latter part of this time frame In this interval, the time increases by −10 − (−37) = 27 orders of magnitude and the energy decreases by 24 − 11 = 13 orders of magnitude 27 13 = 2.1, so roughly for a two-order increase in time there was a one-order decrease in energy Therefore, since TeV = 1012 eV is one order of magnitude greater than 1011 eV, TeV should correspond to two orders of time earlier than 10−10 s, or 10−12 s after the big bang 1133 www.elsolucionario.org Chapter 30: Particle Physics Physics 23 Strategy Replace each particle in the decay reaction with its corresponding antiparticle to write the two decay modes Solution Since π + is the antiparticle of π − , the decay products of π + must be antiparticles of the decay products of π − The decay modes of π + are then π + → µ + + ν µ and π + → e+ + ν e 24 Strategy Use conservation of charge and energy Solution (a) The net charge of a proton and antiproton is zero, so the three pions produced must have a net charge of zero There are two possible pion combinations: π + , π − , π and 3π (b) The possibilities for five pions are: 2π + , 2π − , π ; π + , π − , 3π ; 5π (c) The maximum number of pions produced occurs when their kinetic energies are zero, so the total rest energy of the pions must be less than or equal to the total rest energy of the proton and antiproton Since the neutral pion has the smallest mass, the greatest number of pions produced would all be neutral E p + E p × 938 MeV = = 13.9, so the maximum number is 13 pions Eπ 135 MeV 25 Strategy The pion was initially at rest Conservation of momentum requires that the momenta of the muon and antineutrino be equal in magnitude and opposite in direction, so pµ = pν For the muon, K = p /(2m), so pµ = 2mK For the antineutrino, E ≈ pc, so pν ≈ Eν c Solution Equate the momenta and solve for K E E E pµ = 2mK = pν = ν , so K = ν = ν c E0µ 2mc Use conservation of energy to find the energy of the antineutrino Ei = E0π = Ef = E0µ + K + Eν = E0µ + Eν Eν2 + Eν − ( E0π − E0µ ) + Eν , so = E0µ E0µ Solve using the quadratic formula Eν = ( −1 ± 12 + ⎛⎜ E1 ⎞⎟ E0π − E0µ ⎝ 0µ ⎠ E0µ −1 ) ⎡ ⎛ 0.140 ⎞ ⎤ = (0.106 GeV) ⎢ −1 ± + ⎜ − 1⎟ ⎥ ⎝ 0.106 ⎠ ⎦⎥ ⎣⎢ = 0.030 GeV or − 0.242 GeV, which is extraneous So, the kinetic energy of the muon is K = (0.030 GeV) = 0.0042 GeV = 4.2 MeV 2(0.106 GeV) 1134 www.elsolucionario.org Physics Chapter 30: Particle Physics 26 (a) Strategy The minimum total energy of the electron and the positron will just create the proton, kaon, and the sigma baryon without leaving any additional energy The products will be at rest Solution The minimum total kinetic energy is equal to the rest energies of the three product particles minus the rest energies of the electron and positron, so K total = 938 MeV + 498 MeV + 1197 MeV − × 0.511 MeV = 2.632 GeV Assuming the electron and positron have the same kinetic energy, each particle must have a minimum kinetic energy of (2.632 GeV) = 1.316 GeV (b) Strategy The rest energy of the sigma not used to create the neutron and the pion will become kinetic energy of the neutron and the pion Use conservation of momentum and energy Solution Find the energy not used to create the neutron and the pion K = 1197 MeV − (940 MeV + 140 MeV) = 117 MeV Let the momenta of the neutron and pion be pn and pp , respectively Then, according to conservation of momentum, pn = pp Let the kinetic energies of the neutron and pion be K n and K p , respectively Then, according to conservation of energy, K = K n + K p According to Equation (26-11) and conservation of momentum, ( pp c)2 = K p + K p Ep0 = ( pn c)2 = K n + K n En0 , where En0 and Ep0 are the rest energies of the neutron and pion, respectively Substituting for K p in ( pn c)2 = K p + K p Ep0 gives ( pn c)2 = ( K − K n )2 + 2( K − K n ) Ep0 Subtracting ( pn c)2 = K n + K n En0 from ( pn c)2 = ( K − K n )2 + 2( K − K n ) Ep0 gives = K − KK n + KEp0 − K n Ep0 − K n En0 Solving this equation for the kinetic energy of the neutron K n gives Kn = K + KEp0 2( K + Ep0 + En0 ) = (117 MeV)2 + 2(117 MeV)(140 MeV) = 19 MeV 2(117 MeV + 140 MeV + 940 MeV) Therefore, the kinetic energy of the pion is K p = K − K n = 117 MeV − 19 MeV = 98 MeV 27 Strategy Momentum must be conserved, so p = pγ = Eγ c , where p is the momentum of the lambda baryon and pγ and Eγ are the momentum and the total energy of the photon, respectively The difference in the rest energies of the sigma and lambda baryons is E = 1192 MeV − 1116 MeV = 76 MeV This energy is equal to the sum of the kinetic energy K of the lambda baryon and the total energy of the photon, so E = K + Eγ According to Equation (26-11), ( pc)2 = K + KE0 , where E0 is the rest energy of the lambda baryon Solution Substituting for p, we have Eγ = K + KE0 Substituting for K, we have Eγ = ( E − Eγ )2 + 2( E − Eγ ) E0 Solve for Eγ Eγ = E − EEγ + Eγ + EE0 − Eγ E0 = E − 2( E + E0 ) Eγ + EE0 Eγ = E + EE0 2( E + E0 ) Find the wavelength 2hc( E + E0 ) 2(1240 × 10−15 MeV ⋅ m)(76 MeV + 1116 MeV) hc E + EE0 = , so λ = = = 1.7 × 10−14 m Eγ = λ 2( E + E0 ) (76 MeV)2 + 2(76 MeV)(1116 MeV) E + EE0 1135 www.elsolucionario.org REVIEW AND SYNTHESIS: CHAPTERS 26–30 Review Exercises Strategy Use Eq (26-5) to find the velocity vpE of the escape pod relative to Earth Then, use Eq (26-4) to find how long the pod appears to the people on Earth Let the positive direction be toward the Earth Solution The velocity of the starship relative to Earth is vsE = 0.78c The velocity of the pod relative to the starship is vps = 0.63c Find the velocity of the pod relative to Earth vpE = vps + vsE + vps vsE c = 0.63c + 0.78c = 0.945c + (0.63)( 0.78) Find the length of the pod as it appears to the people on Earth L L = = L0 − v c = (12.0 m) − 0.9452 = 3.91 m γ Strategy When an electron is accelerated through a potential difference ∆V , it gains kinetic energy equal to e∆V Use the classical and relativistic expressions for kinetic energy to find the speeds Solution (a) Find the speed of the electron K= mv = e∆V , so v = 2e∆V = m 2(1.602 × 10−19 C)(25.00 × 106 V) 9.109 × 10−31 kg = 2.965 × 109 m s (b) Find the speed of the electron K = (γ − 1)mc = [(1 − v c ) −1 − 1]mc = e∆V e ∆V = (1 − v c ) −1 − mc −2 v ⎛ e∆V ⎞ 1− = ⎜1 + ⎟ c ⎝ mc ⎠ ⎛ e∆V ⎞ v = c − ⎜1 + ⎟ ⎝ mc ⎠ −2 ⎡ (1.602 × 10−19 C)(25.00 × 106 V) ⎤ = c − ⎢1 + ⎥ ⎢⎣ (9.109 × 10−31 kg)(2.998 × 108 m s )2 ⎥⎦ −2 = 0.999799c = 2.99740 × 108 m s (a) Strategy The period of the beacon is T = 6.00 s During one period, the light from the beacon sweeps out a circle of circumference 2π R, where R is the mean Earth-Moon distance Solution Compute the speed of the light from the beacon traveling across the surface of the Moon 2π R 2π (3.845 × 108 m) v= = = 4.03 × 108 m s T 6.00 s (b) Strategy and Solution You explain to Yoo Jin that this does not violate the results of the theory of special relativity because the disturbance that moves across the Moon’s surface is not an object that has mass, so there is no violation 1136 www.elsolucionario.org Physics Review and Synthesis: Chapters 26–30 (a) Strategy Use Eq (26-7) to find the amount of energy produced Solution Compute the rest energy of the bag of cookies 1.00 kg E0 = mc = (1.00 lb) (2.998 × 108 m s)2 = 4.09 × 1016 J 2.20 lb (b) Strategy The conversion of kilowatt-hours from joules is kW ⋅ h = 3.600 × 106 J Solution Compute the amount of electrical energy equivalent to the energy released by the total annihilation of the bag of cookies 1.00 kg kW ⋅ h = 1.13 × 1010 kW ⋅ h (1.00 lb) (2.998 × 108 m s)2 2.20 lb 3.600 × 10 J Strategy Use Eq (26-8) Solution Find the speed of the electron K = (γ − 1)mc K = γ −1 mc K = 1+ 2 mc − v2 c 1− K ⎞ ⎛ = ⎜1 + ⎟ c ⎝ mc ⎠ v −2 K ⎞ ⎛ v = c − ⎜1 + ⎟ ⎝ mc ⎠ −2 ⎡ ⎤ 1.02 × 10−13 J = c − ⎢1 + ⎥ − 31 kg)(2.998 × 10 m s ) ⎦⎥ ⎣⎢ (9.109 × 10 −2 = 0.895c = 2.68 × 108 m s Strategy The energy of a photon is given by E = hf = hc λ Use Eqs (27-2) and (27-7) Solution (a) Find the work function of the metal hc hc 1240.0 eV ⋅ nm K max = eVs = − φ , so φ = − eVs = − e(6.00 V) = 4.33 eV 120.0 nm λ λ Find the stopping potential required for photons with wavelength λ = 240.0 nm ⎛ hc ⎞ ⎛ 1240.0 eV ⋅ nm ⎞ − 4.333 eV ⎟ = 0.83 V Vs = ⎜ − φ ⎟ = ⎜ e⎝ λ e 240.0 nm ⎠ ⎝ ⎠ (b) Compute the maximum kinetic energy hc 1240 eV ⋅ nm K max = −φ = − 4.3 eV = − 0.9 eV < 360 nm λ The maximum kinetic energy is negative, which is impossible So, for 360 nm, no electrons will be emitted, since this energy is less than the work function 1137 www.elsolucionario.org Review and Synthesis: Chapters 26–30 Physics (a) Strategy The energy of a photon is given by E = hf = hc λ Use Eq (27-7) and K = 12 mv Solution Find the speed of the electrons hc − φ = mv , so K max = λ 2 ⎛ hc ⎞ ⎛ 1240 eV ⋅ nm ⎞ − 1.4 eV ⎟ (1.602 × 10−19 J eV) = 9.8 × 105 m s v= ⎜ −φ ⎟ = ⎜ − 31 m⎝ λ ⎠ ⎠ 9.109 × 10 kg ⎝ 300 nm (b) Strategy Use Eq (27-7) Solution Compute the maximum kinetic energy hc 1240 eV ⋅ nm K max = −φ = − 1.6 eV = − 0.1 eV < λ 800 nm The maximum kinetic energy is negative, which is impossible, so no electrons are ejected (c) Strategy and Solution Increasing the light intensity increases the number of photons incident on the metal, but does not increase the energy per photon, so doubling the intensity has no effect on the electron speed, nor does is cause electrons to be ejected if none were ejected prior to the doubling of the intensity (a) Strategy The energy of a photon is given by E = hc λ Solution Compute the energy of each photon in the laser beam hc 1240 eV ⋅ nm E= = = 1.8 eV λ 680 nm (b) Strategy The energy of a photon within this pulse is E photon = hc λ = E N , where E is the total energy of the pulse and N is the number of photons within the pulse The power of the laser pulse is related to the energy of the pulse by P = E ∆t Solution Find the number of photons in the pulse E P∆t Pλ ∆t (220 W)(680 nm)(0.250 × 10−3 s) N= = = = = 1.9 × 1017 Ephoton hc λ hc (1240 eV ⋅ nm)(1.602 × 10−19 J eV ) Strategy The width of the central maximum is the distance between the first minimum on either side of the central maximum Use conservation of energy to find the speed of the electrons Use the de Broglie wavelength for the wavelength of the electrons Solution Find the wavelength of the electrons h p2 h p = mv = and K = mv = = e∆V , so λ = 2m λ 2me∆V The first minimum on either side of the central maximum is given by a sin θ = λ Solve for θ θ = sin −1 λ = sin −1 h 6.626 × 10−34 J ⋅ s = sin −1 = 1.12555° (6.6 × 10−10 m) 2(9.109 × 10−31 kg)(1.602 × 10−19 C)(8950 V) If x is the distance to the first minimum on either side of the central maximum and D is the distance from the slit to the screen, then tan θ = x D Find 2x, the width of the central maximum a a 2me∆V x = D tan θ = 2(2.50 m) tan1.12555° = 9.8 cm 1138 www.elsolucionario.org Physics Review and Synthesis: Chapters 26–30 10 Strategy Use Eqs (28-1) and (28-2) Solution Find the minimum uncertainty in the speed of the particle h h h h v 6.50 × 106 m s ∆x ∆p = λ ∆ (mv) = m ∆v = m∆ v = ∆ v ≥ = , so (∆v)min = = = 5.17 × 105 m s p mv v 4π 4π 4π 11 (a) Strategy In beta-minus decay an electron is emitted, the atomic number Z increases by while the mass number A stays the same Use Eq (29-11) Solution In this case, the parent nuclide has Z = 38 and A = 90, so the daughter nuclide yttrium will have A = 90 and Z = 38 + = 39 Write down the decay scheme A ZP → 0 A Z +1 D + −1e + 0ν , so 90 38 Sr → 0 90 39 Y + −1 e + 0ν (b) Strategy The activity is related to the number of nuclei N and the time constant τ by R = N τ The number of nuclei is related to the mass by N = mN A M , where N A is Avogadro’s number and M is the molar mass The time constant is related to the half-life by T1 = τ ln Solution Compute the initial activity N mN A ln (2.0 × 103 g)(6.022 × 1023 mol−1 ) ln = = 1.0 × 1016 Bq R0 = = τ M T1 (89.907 737 g mol)(28.8 yr)(3.156 × 107 s yr) (c) Strategy Use Eqs (29-21) and (29-22) Solution Find the activity of the strontium-90 after 1000 years R = R0 e−t τ = R0 e −t ln T1 = (1.02 × 1016 Bq)e−1000 ln 28.8 = 3.6 × 105 Bq 12 Strategy Find the energy released in the fusion reaction H + H → He + n by finding the difference in the rest energies Solution Compute the energy released in the fusion reaction Efusion = ∆m c = [(2.014 101 u + 3.016 049 u) − (4.002 603 u + 1.008 664 u)](931.494 MeV u) = 17.5894 MeV 1139 www.elsolucionario.org Review and Synthesis: Chapters 26–30 Physics 13 Strategy The rest energy of the lambda particle not used to create the proton and the pion will become kinetic energy of the proton and the pion Use conservation of momentum and energy Solution Find the energy not used to create the proton and the pion K = 1116 MeV − (938 MeV + 139.6 MeV) = 38.4 MeV Let the magnitudes of the momenta of the proton and pion be pp and pπ , respectively Then, according to conservation of momentum, pp = pπ Let the kinetic energies of the proton and pion be K p and Kπ , respectively Then, according to conservation of energy, K = K p + Kπ According to Eq (26-11) and conservation of momentum, ( pπ c) = Kπ + Kπ Eπ = ( pp c) = K p + K p Ep0 , where Ep0 and Eπ are the rest energies of the proton and pion, respectively Substituting for Kπ in ( pp c) = Kπ + Kπ Eπ gives ( pp c)2 = ( K − K p ) + 2( K − K p ) Eπ Subtracting ( pp c)2 = K p + K p Ep0 from ( pp c)2 = ( K − K p ) + 2( K − K p ) Eπ gives = K − KK p + KEπ − K p Eπ − K p Ep0 Solving this equation for the kinetic energy of the proton K p gives Kp = K + KEπ 2( K + Eπ + Ep0 ) = (38.4 MeV)2 + 2(38.4 MeV)(139.6 MeV) = 5.5 MeV 2(38.4 MeV + 139.6 MeV + 938 MeV) Therefore, the kinetic energy of the pion is Kπ = K − K p = 38.4 MeV − 5.46 MeV = 33 MeV 14 Strategy Mesons are made up of quark-antiquark pairs and baryons are made up of three quarks Use Tables 30.1 and 30.3 Solution (a) The particle is a meson with charge − e + (−2e 3) = – e (b) The particle is a baryon with charge 2e + (− e 3) + 2e = + e (c) Two of the decay products are a positron and an electron neutrino, which are leptons Leptons are associated with the weak force 1140 www.elsolucionario.org Physics Review and Synthesis: Chapters 26–30 15 Strategy The magnitude of the magnetic force on an electron is given by F = evB Use Newton’s second law, Eq (27-7) for the work function, and K = 12 mv The energy of a photon is E = hf = hc λ Solution (a) Find the speed of the electrons mv eBr ΣFr = evB = mar = , so v = r m Find the work function of the metal 2 ⎛ eBr ⎞ e2 B r = mv = m ⎜ , so ⎟ λ 2 ⎝ m ⎠ 2m hc e2 B r 1240 eV ⋅ nm (1.602 × 10−19 C) (7.5 × 10−5 T)2 (0.067 m)2 φ= − = − = 4.7 eV λ 2m 180 nm 2(9.109 × 10−31 kg)(1.602 × 10−19 J eV) K max = hc −φ = (b) Since K ∝ v and v ∝ r , K ∝ r , so electrons with maximum kinetic energy follow a path with the maximum radius 16 Strategy The atomic number Z decreases by while the nucleon number A stays the same, so the isotope undergoes beta-plus decay Solution Since the isotope undergoes beta-plus decay, the charged particle emitted in this process is a positron with charge + e 17 (a) Strategy Use Eqs (29-21) and (29-22) Solution Find the activity of the sample of gold-198 after 8.10 days R = R0 e−t τ = R0 e −t ln T1 = (1.00 × 1010 Bq)e−8.10 ln 2.70 = 1.25 × 109 Bq (b) Strategy The atomic number Z increases by 1, while the mass number A stays the same, so the isotope undergoes beta-minus decay Solution Since the isotope undergoes beta-minus decay, the particles emitted in this process are an electron and an antineutrino 18 Strategy Use Eq (26-9) for the total energy Form a proportion Solution Find the new energy Ef when the particle has speed vf = 0.980c Ef γ f mc γ f − vi c − vi c − 0.6002 = = = , so Ef = Ei = (0.638 MeV) = 2.6 MeV Ei γ i mc γ i − vf c − vf c − 0.9802 1141 www.elsolucionario.org Review and Synthesis: Chapters 26–30 Physics 19 Strategy Use Eq (26-5) to find the speed vsE of the starship relative to Earth Solution Let the positive direction be away from Earth The velocity of the starship relative to us is vsu = 0.50c The velocity of Earth relative to the us is vEu = − 0.90c, so the velocity of us relative to the Earth is vuE = 0.90c Find the velocity of the starship relative to Earth vsu + vuE 0.50c + 0.90c vsE = = = 0.966c + (0.50)( 0.90) + vsu vuE c The speed measured by people on Earth is 0.966c 20 Strategy Refer to Table 28.1 for the quantum numbers for electron states in a hydrogen atom Solution Find the possible values of the orbital angular momentum quantum number The maximum value of is = n − = − = Since m = − , − + 1, , − 1, 0, 1, , − 1, , and m = for this electron, the minimum value of is = So, the possible values of are 4, 5, 6, and 21 Strategy Use Eq (28-2) Solution Find the order of magnitude of the minimum uncertainty in the momentum of the electron 10−34 J ⋅ s ∆x∆p x ≥ , so (∆p x ) = = 10− eV ⋅ s m ∼ 2∆x (10−11 m)(10−19 J eV ) 22 Strategy Use Eqs (26-6) and (28-1) Solution Find the de Broglie wavelength of the electron λ= h − v2 c2 h h (6.626 × 10−34 J ⋅ s) − 0.602 = = = = 0.0032 nm p γ mv mv (9.109 × 10−31 kg)0.60(2.998 × 108 m s) 23 Strategy Count the number of unique transitions between the four distinct energy levels Solution The transitions are: → 1, → 2, → 3, → 1, → 2, and → There are six transitions with distinct energies, so the number of spectral lines of different wavelengths emitted by this atom is six 24 Strategy The energy of an electron in a state n in the hydrogen atom is inversely proportional to the square of n Solution Form a proportion to find the kinetic energy of an electron in the first excited state K n12 n2 = , so K = K1 = K1 n22 n22 12 2 K= K 25 Strategy According to Eq (27-9), The cutoff frequency f max of x-rays produced by bremsstrahlung (braking radiation) is directly proportional to the kinetic energy of the incident electrons Since λmin ∝ f max , the minimum wavelength of the x-rays is inversely proportional to the kinetic energy of the electrons Now, by conservation of energy, an electron accelerated through a potential difference is given kinetic energy equal to e∆V Therefore, the minimum wavelength of the x-rays is inversely proportional to the potential difference through which the electrons are accelerated Solution Find the ratio of the minimum wavelength of x-rays in tube A to the minimum wavelength in tube B λA ∆VB 40 kV = = = ; the ratio is 4:1 λB ∆VA 10 kV 1142 www.elsolucionario.org Physics Review and Synthesis: Chapters 26–30 26 (a) Strategy and Solution The ionization energy is the energy needed to excite a ground-state electron (3s; E = −5.1 eV) to the n = ∞ state (E = 0) Therefore, the ionization energy is 5.1 eV (b) Strategy The energy of an emitted photon is equal to the energy difference between the levels Solution Compute the wavelength hc hc 1240 eV ⋅ nm E= , so λ = = = 890 nm λ E 3.0 eV − 1.6 eV (c) Strategy Find the energy of a photon Then compare this energy to the transitions in the diagram Solution The energy of a photon of wavelength λ = 589 nm is hc 1240 eV ⋅ nm E= = = 2.11 eV λ 589 nm The transition 3p → 3s has an energy difference of ∆ E = 5.1 eV − 3.0 eV = 2.1 eV So, the characteristic yellow light of sodium is generated by the p → 3s transition 27 Strategy Conservation of momentum requires that the momenta of the photon and the nucleus be equal in magnitude and opposite in direction The momentum of the photon is related to its energy by pp = E c The nucleus is assumed nonrelativistic, so its momentum is related to its kinetic energy by pTl = Km Solution Equate the momenta and solving for the kinetic energy of the nucleus E E2 (452 × 103 eV)2 pTl = Km = pp = , so K = = = 0.527 eV c 2mc × 208.0 u × 931.494 × 106 eV u 28 (a) Strategy The magnitude of the momentum is given by p = γ mv Solution Find v mv p = γ mv = = 2.5mv, so 2.5−2 = − v c and 2 1− v c v2 c2 = 1− v= 21 = 25 25 21 c ≈ 0.917c (b) Strategy Use the time dilation equation Solution Find the time between watch ticks, as measured by an Earth observer ∆t0 1s ∆t = γ ∆t0 = = = s − 21 25 − v2 c2 1143 www.elsolucionario.org Review and Synthesis: Chapters 26–30 Physics 29 Strategy The absorbed dose of ionizing radiation is the amount of radiation energy absorbed per unit mass of tissue Solution Find the equivalent dose of energy due to the x-rays 0.01 Gy J kg 65 kg × × = 0.0043 J E = (20 × 10−3 rad) × rad Gy Find the energy absorbed by the patient’s body biologically equivalent dose of energy 4.3 mJ = = 4.8 mJ energy absorbed = QF 0.90 30 (a) Strategy Use the relativistic energy equation Solution Compute the speed of the protons E = γ mc = v = c 1− mc − v2 c2 (mc )2 E2 , so − = c 1− v2 c2 = (mc )2 E2 and (938.272 × 106 eV) (7 × 1012 eV)2 = 0.999999991 c (b) Strategy Use the length contraction Solution Compute the time ∆t = L L0 (27 × 10 m) − 0.999999991 = = = 12 ns c cγ 2.998 × 108 m s (c) Strategy Distance equals rate times time Solution Compute the time L 27 × 103 m ∆t0 = = = 90 às c 2.998 ì 108 m s (d) Strategy Use λ = h p and the relativistic formula for momentum Solution Compute the de Broglie wavelength λ= (6.626 × 10−34 J ⋅ s) − 0.9999999912 h h = = = 1.8 × 10−19 m p γ mv (1.673 × 10−27 kg)0.999999991(2.998 × 108 m s) (e) Strategy Decide whether to use the relativistic formula or nonrelativistic formula for kinetic energy before finding the speed Solution In joules, TeV is (7 × 1012 eV)(1.602 × 10−19 J eV) = µJ The rest energy of the mosquito is E0 = mc = (1.0 × 10−6 kg)(2.998 × 108 m s) = 9.0 × 1010 J Since the rest energy is much greater than the kinetic energy, the mosquito’s speed is nonrelativistic Find its speed K= mv , so v = 2K = m 2(7 × 1012 eV)(1.602 × 10−19 J eV) 1.0 × 10−6 kg 1144 ≈ 1.5 m s www.elsolucionario.org Physics Review and Synthesis: Chapters 26–30 MCAT Review Strategy The rest energy of an alpha particle is approximately 3.7 GeV This is much greater than the kinetic energy of the alpha particle, so K = 12 mv is a reasonable approximation of the kinetic energy of the particle Solution Compute the approximate speed of the alpha particle K= mv , so v = 2K = m 2(4.8 × 106 eV)(1.602 × 10−19 J eV) (4 u)(1.66 × 10−27 kg u ) = 1.5 × 107 m s The correct answer is C Strategy For each alpha emitted, the nucleus loses two protons and two neutrons For each beta emitted, the nucleus gains a proton and loses a neutron Count the number of protons, neutrons, and betas emitted Solution A total of + + = alphas are emitted, so the nucleus loses 12 protons and 12 neutrons A total of + + = betas are emitted, so the nucleus gains protons and loses neutrons The new atomic number is Z = 90 − 12 + = 82 and the new nucleon number is A = 232 − 12 − 12 + − = 208 The correct answer is A Strategy There are three protons and − = neutrons in the nucleus of lithium-7 Use Eq (29-8) to find the binding energy Solution Compute the approximate binding energy EB = (∆m)c = [(3 × 1.0073 u + × 1.0087 u) − 7.014 u] × 931 MeV u = 40.0 MeV The correct answer is D Strategy The external magnetic field only exerts forces on moving charged particles Solution Alphas are positively charged, betas are negatively charged, and gammas have no charge Therefore, gamma rays travel straight and alpha and beta rays are bent in opposite directions The correct answer is D Strategy The nucleus decreases by ∆m = E c , where E is the energy of the gamma ray Solution Compute the decrease in mass of the nucleus E (2.5 × 106 eV)(1.602 × 10−19 J eV) ∆m = = = 4.5 × 10−30 kg c2 (2.998 × 108 m s)2 The correct answer is C Strategy Use Eqs (29-21) and (29-22) Solution Find the original activity R0 of the sodium-24 sample R = R0 e−t τ = R0 e −t ln T1 , so R0 = Re t ln T1 = (100 mCi)e24 ln 15 = 300 mCi The correct answer is B 1145 www.elsolucionario.org Review and Synthesis: Chapters 26–30 Physics Strategy Subtract the mass equivalent of the energy required to break the nucleus of the neon-20 atom into its constituent parts from the masses of Z = 10 protons and N = A − Z = 20 − 10 = 10 neutrons to find the atomic mass of the atom Solution Find the atomic mass of the atom 10 × 1.0073 u + 10 × 1.0087 u − 0.173 u = 19.987 u The correct answer is A Strategy The intermediate product uranium-234 has 238 − 234 = fewer neutrons than does uranium-238 Count the number of neutrons lost in each sequence Solution Beta emission does not effect the nucleon number A, but does increase the atomic number by one, so choice A is not the correct answer Alpha emission reduces the nucleon number by four and the atomic number by two Evaluate the remaining three sequences B: A = 238 − = 234 and Z = 92 − + = 93 C: A = 238 − − = 230 and Z = 92 − − + + = 90 D: A = 238 − = 234 and Z = 92 − + + = 92 The correct answer is D Strategy Every months, the sample has been reduced by half Solution Find the fraction of the sample still remaining after years yr = 24 mo and 24 = 3, so the sample has been reduced to ⎛1⎞ ⎜ ⎟ = of its original amount ⎝2⎠ The correct answer is C 10 Strategy and Solution In gamma decay, the atomic number Z and the mass number A stay the same The correct answer is B 11 Strategy The nucleon number A is the sum of the total number of protons Z and neutrons N Solution The nucleon number of thallium-201 is A = 201, which is equal to the total number of protons and neutrons in the nucleus The atomic number is Z = 81, which is equal to the total number of protons in the nucleus In a neutral atom, the number of electrons is equal to the number of protons; in this case, there are 81 electrons There are N = A − Z = 201 − 81 = 120 neutrons The correct answer is D 12 Strategy The activity R is given by R = R0 e−t τ , where R0 is the initial activity, t is the time elapsed, and τ is the time constant Solution The activity decreases exponentially with time The correct answer is C 13 Strategy The energy of a photon is related to its wavelength by E = hc λ Solution Compute the wavelength of the gamma ray photon hc (4.15 × 10−15 eV ⋅ s)(3.0 × 108 m s) (4.15 × 10−15 )(3.0 × 108 ) = = m λ= E 1.35 × 105 eV 1.35 × 105 The correct answer is A 1146 ... direction be to the right Draw a diagram Solution Find the chipmunk’s total displacement 80 cm right 30 cm left 90 cm right 310 cm left 170 cm left 80 cm − 30 cm + 90 cm − 310 cm = −170 cm The total... +x-direction Solution Draw a vector diagram; then compute the sum of the three displacements The vector diagram: 32 cm 48 cm 64 cm N 16 cm The sum of the three displacements is (32 cm + 48 cm − 64 cm)... positive direction Solution Find the constant acceleration required to stop the airplane The acceleration must be opposite to the direction of motion of the airplane, so the direction of the acceleration