Wolfgang bauer ,gary westfall instructors solution manuals to university physics with modern physics mcgrawhill (2013)

Instructor Solutions Manual to accompany University Physics Second Edition Wolfgang Bauer Michigan State University Gary D Westfall Michigan State University www.elsolucionario.org Instructor Solutions Manual to accompany UNIVERSITY PHYSICS, Second Edition Table of Contents PART MECHANICS OF POINT PARTICLES Overview Motion in a Straight Line Motion in Two and Three Dimensions Force Kinetic Energy, Work, and Power Potential Energy and Energy Conservation Momentum and Collisions PART EXTENDED OBJECTS, MATTER, AND CIRCULAR MOTION Systems of Particles and Extended Objects Circular Motion 10 Rotation 11 Static Equilibrium 12 Gravitation 13 Solids and Fluids PART OSCILLATIONS AND WAVES 14 Oscillations 15 Waves 16 Sound PART THERMAL PHYSICS 17 Temperature 18 Heat and the First Law of Thermodynamics 19 Ideal Gases 20 The Second Law of Thermodynamics PART ELECTRICITY 21 Electrostatics 22 Electric Fields and Gauss’s Law 23 Electric Potential 24 Capacitors 25 Current and Resistance 26 Direct Current Circuits PART MAGNETISM 27 Magnetism 28 Magnetic Fields of Moving Charges 29 Electromagnetic Induction 30 Alternating Current Circuits 31 Electromagnetic Waves PART OPTICS 32 Geometric Optics 33 Lenses and Optical Instruments 34 Wave Optics PART RELATIVITY AND QUANTUM PHYSICS 35 Relativity 36 Quantum Physics 37 Quantum Mechanics 38 Atomic Physics 39 Elementary Particle Physics 40 Nuclear Physics 45 108 163 223 255 308 380 430 474 521 574 628 673 713 747 783 806 835 870 898 934 973 1007 1046 1075 1113 1141 1171 1197 1224 1248 1270 1304 1324 1354 1382 1419 1444 1464 Chapter 1: Overview Chapter 1: Overview Concept Checks 1.1 a 1.2 a) b) c) d) e) 1.3 a, c and e 1.4 b 1.5 e 1.6 a) 4th b) 2nd c) 3rd d) 1st Multiple-Choice Questions 1.1 c 1.2 c 1.3 d 1.4 b 1.5 a 1.6 b 1.7 b 1.8 c 1.9 c 1.10 b 1.11 d 1.12 b 1.13 c 1.14 a 1.15 e 1.16 a Conceptual Questions 1.17 (a) In Europe, gas consumption is in L/100 km In the US, fuel efficiency is in miles/gallon Let’s relate these two: mile = 1.609 km, gal = 3.785 L mile 1.609 km 1.609   km 1 100 ) = = ( 0.00425 )  =   (= gal 3.785 L 3.785  100  L  L/100 km  235.24 L/100 km Therefore, mile/gal is the reciprocal of 235.2 L/100 km 1L 12.2 L (b) Gas consumption is Using from part (a), = 100 km 235.24 miles/gal 100 km 1   12.2 L  1L  = 12.2 =   12.2  235.24 miles/gal  = 19.282 miles/gal 100 km  100 km    Therefore, a car that consumes 12.2 L/100 km of gasoline has a fuel efficiency of 19.3 miles/gal (c) If the fuel efficiency of the car is 27.4 miles per gallon, then 27.4 miles 27.4 = = gal 235.24 L/100 km 8.59 L/100 km Therefore, 27.4 miles/gal is equivalent to 8.59 L/100 km (d) 1.18 A vector is described by a set of components in a given coordinate system, where the components are the projections of the vector onto each coordinate axis Therefore, on a two-dimensional sheet of paper there are two coordinates and thus, the vector is described by two components In the real three-dimensional world, there are three coordinates and a vector is described by three components A four-dimensional world would be described by four coordinates, and a vector would be described by four components 1.19 A vector contains information about the distance between two points (the magnitude of the vector) In contrast to a scalar, it also contains information direction In many cases knowing a direction can be as important as knowing a magnitude Bauer/Westfall: University Physics, 2E 1.20 In order to add vectors in magnitude-direction form, each vector is expressed in terms of component vectors which lie along the coordinate axes The corresponding components of each vector are added to obtain the components of the resultant vector The resultant vector can then be expressed in magnitudedirection form by computing its magnitude and direction 1.21 The advantage to using scientific notation is two-fold: Scientific notation is more compact (thus saving space and writing), and it also gives a more intuitive way of dealing with significant figures since you can only write the necessary significant figures and extraneous zeroes are kept in the exponent of the base 1.22 The SI system of units is the preferred system of measurement due to its ease of use and clarity The SI system is a metric system generally based on multiples of 10, and consisting of a set of standard measurement units to describe the physical world In science, it is paramount to communicate results in the clearest and most widely understood manner Since the SI system is internationally recognized, and its definitions are unambiguous, it is used by scientists around the world, including those in the United States 1.23 It is possible to add three equal-length vectors and obtain a vector sum of zero The vector components of the three vectors must all add to zero Consider the following arrangement with T= T= T3 : The horizontal components of T1 and T2 cancel out, so the sum T1 + T2 is a vertical vector whose magnitude is T cosθ + T cosθ = 2T cosθ The vector sum T1 + T2 + T3 is zero if 2T cosθ − T = θ= 60° Therefore it is possible for three equal-length vectors to sum to zero cosθ = 1.24 Mass is not a vector quantity It is a scalar quantity since it does not make sense to associate a direction with mass 1.25 The volume of = 2V 2= ( / )π r a sphere is given by ( / 3= )π (23/3 )r ( / )π (21/3 r )3 V = ( / )π r Doubling the volume gives Now, since the distance between the flies is the diameter of the sphere, d = 2r , and doubling the volume increases the radius by a factor of 21/3 , the 1/3 1/3 distance between the flies is then increased to 2(2 = r ) 2= (2r ) 21/3 d Therefore, the distance is increased by a factor of 21/3 1.26 The volume of a cube of side r is Vc = r , and the volume of a sphere of radius r is Vsp = ( / )π r The ratio of the volumes is: Vc r3 = = Vsp π r 4π The ratio of the volumes is independent of the value of r www.elsolucionario.org Chapter 1: Overview 1.27 The surface area of a sphere is given by 4π r A cube of side length s has a surface area of 6s To determine s set the two surface areas equal: 4π r 2π = r s 4π r ⇒= s 6= 1.28 The mass of Sun is ⋅ 1030 kg, the number of stars in the Milky Way is about 100 ⋅ 109 = 1011 , the number of galaxies in the Universe is about 100 ⋅ 109 = 1011 , and the mass of an H-atom is ⋅ 10 −27 kg (a) The total mass of the Universe is roughly equal to the number of galaxies in the Universe multiplied by the number of stars in a galaxy and the mass of the average star: M universe = (1011 )(1011 )(2 ⋅ 1030 ) = ⋅ 10(11+11+ 30) kg = ⋅ 1052 kg (b) nhydrogen ≈ 1.29 M universe ⋅ 1052 kg = = 1079 atoms M hydrogen ⋅ 10 −27 kg The volume of teaspoon is about 4.93 ⋅ 10 −3 L , and the volume of water in the oceans is about 1.35 ⋅ 1021 L 1.35 ⋅ 1021 L = 2.74 ⋅ 1023 tsp 4.93 ⋅ 10 −3 L/tsp There are about 2.74 ⋅ 1023 teaspoons of water in the Earth’s oceans 1.30 The average arm-span of an adult human is d = m Therefore, with arms fully extended, a person takes up a circular area = of π r π (= d / ) π= (1 m)2 π m Since there are approximately 6.5 ⋅ 109 humans, the amount of land area required for all humans to stand without being able to touch each other is 6.5 ⋅ 109 m (π ) = 6.5 ⋅ 109 m (3.14) = 2.0 ⋅ 1010 m The area of the United States is about 3.5 ⋅ 106 square miles or 9.1 ⋅ 1012 m In the United States there is almost five hundred times the amount of land necessary for all of the population of Earth to stand without touching each other 1.31 The diameter of a gold atom is about 2.6 ⋅ 10 −10 m The circumference of the neck of an adult is roughly 0.40 m The number of gold atoms necessary to link to make a necklace is given by: circumference of neck 4.0 ⋅ 10 −1 m = = 1.5 ⋅ 109 atoms diameter of atom 2.6 ⋅ 10 −10 m/atom The Earth has a circumference at the equator of about 4.008 ⋅ 107 m The number of gold atoms necessary to link to make a chain that encircles the Earth is given by: = n = N circumference of Earth 4.008 ⋅ 107 m = = 1.5 ⋅ 1017 atoms diameter of a gold atom 2.6 ⋅ 10 −10 m Since one mole of substance is equivalent to about 6.022 ⋅ 1023 atoms , the necklace of gold atoms has (1.5 ⋅10 (1.5 ⋅10 17 1.32 )( atoms ) / ( 6.022 ⋅ 10 ) atoms/mol ) = 2.5 ⋅10 moles of gold atoms / 6.022 ⋅ 1023 atoms/mol = 2.5 ⋅10 −15 moles of gold 23 The gold chain has −7 The average dairy cow has a mass of about 1.0 ⋅ 103 kg Estimate the cow’s average density to be that of water, ρ = 1000 kg/m volume = mass = ρ 1.0 ⋅ 103 kg = 1.0 m 1000 kg/m 3 Bauer/Westfall: University Physics, 2E Relate this to the volume of a sphere to obtain the radius ( ) 1/3  1.0 m    ≈ 0.62 m 4π     A cow can be roughly approximated by a sphere with a radius of 0.62 m 1/3  3V  r  = volume = π r ⇒=  4π  1.33 The mass of a head can be estimated first approximating its volume A rough approximation to the shape of a head is a cylinder To obtain the volume from the circumference, recall that the circumference is C = 2π r , which gives a radius of r = C / 2π The volume is then: C 2h  C  V = πr2 h = π  h =  4π  2π  The circumference of a head is about 55 cm = 0.55 m, and its height is about 20 cm = 0.20 m These values can be used in the volume equation: ( ) ( 0.55 m ) ( 0.20 m=) 4.8 ⋅10−3 m3 4π Assuming that the density of the head is about the same as the density of water, the mass of a head can then be estimated as follows: V = ( )( ) 4.8 kg mass = density ⋅ volume = 1.0 ⋅ 103 kg/m 4.8 ⋅ 10 −3 m = 1.34 The average adult human head is roughly a cylinder 15 cm in diameter and 20 cm in height Assume about 1/3 of the surface area of the head is covered by hair 1 2π 2π  2π r + 2π rh= Ahair= Acylinder= r + rh= ( 7.5 cm ) + ( 7.5 cm )( 20 cm ) 3 3  ≈ 4.32 ⋅ 102 cm On average, the density of hair on the scalp is ρhair = 2.3 ⋅ 102 hairs/cm Therefore, you have Ahair × ρhair hairs on your head ( ) ( ) ( ( ) )( ) Ahair ρhair = 4.32 ⋅ 102 cm 2.3 ⋅ 102 hairs/cm = 9.9 ⋅ 10 hairs Exercises 1.35 (a) Three (b) Four (c) One (d) Six (e) One (f) Two (g) Three 1.36 THINK: The known quantities are: F1 = 2.0031 N and F2 = 3.12 N Both F1 and F2 are in the same direction, and act on the same object The total force acting on the object is Ftotal SKETCH: RESEARCH: Forces that act in the same direction are summed, Ftotal = ∑ Fi SIMPLIFY: Ftotal= ∑ F= i F1 + F2 CALCULATE: Ftotal= 2.0031 N + 3.12 N= 5.1231 N ROUND: When adding (or subtracting), the precision of the result is limited by the least precise value used in the calculation F1 is precise to four places after the decimal and F2 is precise to only two places after the decimal, so the result should be precise to two places after the decimal: Ftotal = 5.12 N DOUBLE-CHECK: This result is reasonable as it is greater than each of the individual forces acting on the object Chapter 1: Overview 1.37 The result should have the same number of decimal places as the number with the fewest of them Therefore, the result is 2.0600 + 3.163 + 1.12 = 6.34 1.38 In a product of values, the result should have as many significant figures as the value with the smallest number of significant figures The value for x only has two significant figures, so w = (1.1 ⋅ 103 )(2.48 ⋅ 10 −2 )(6.000) = 1.6 ⋅ 102 1.39 Write “one ten-millionth of a centimeter” in scientific notation One millionth is 1/106 = ⋅ 10−6 Therefore, one ten-millionth is 1/ 10 ⋅ 106  = 1/107 = ⋅ 10 −7 cm 1.40 153,000,000 = 1.53 ⋅ 108 1.41 There are 12 inches in a foot and 5280 feet in a mile Therefore there are 63,360 inch/mile 30.7484 miles · 63,360 inch / mile = 1948218.624 inches Rounding to six significant figures and expressing the answer in scientific notation gives 1.94822·106 inches 1.42 (a) kilo (b) centi (c) milli 1.43  1000 m  1000 mm  km = km    = 1,000,000 mm = ⋅ 10 mm  km  m  1.44 hectare = 100 ares, and are = 100 m , so:  (1000 )2 m km2 = km   km   are  hectare      = 100 hectares   100 m  100 ares   1.45 milliPascal 1.46 THINK: The known quantities are the masses of the four sugar cubes Crushing the sugar cubes doesn’t change the mass Their masses, written in standard SI units, using scientific notation are m1 2.53 ⋅ 10 −2 kg , = m2 2.47 ⋅ 10 −2 kg , = m4 2.58 ⋅ 10 −2 kg = m3 2.60 ⋅ 10 −2 kg and = SKETCH: A sketch is not needed to solve this problem RESEARCH: (a) The total mass equals the sum of the individual masses: M total = ∑ m j j =1 (b) The average mass is the sum of the individual masses, divided by the total number of masses: m + m2 + m3 + m4 M average = SIMPLIFY: (a) M total = m1 + m2 + m3 + m4 M total CALCULATE: (a) M total = 2.53 ⋅ 10 −2 kg + 2.47 ⋅ 10 −2 kg + 2.60 ⋅ 10 −2 kg + 2.58 ⋅ 10 −2 kg (b) M average = = 10.18 ⋅ 10 −2 kg =1.018 ⋅ 10 −1 kg (b) M average = 10.18 ⋅ 10 −2 kg = 2.545 ⋅ 10 −2 kg ROUND: (a) Rounding to three significant figures, M total = 1.02 ⋅ 10 −1 kg (b) Rounding to three significant figures, M average = 2.55 ⋅ 10 −2 kg www.elsolucionario.org Bauer/Westfall: University Physics, 2E DOUBLE-CHECK: There are four sugar cubes weighing between 2.53 ⋅ 10 −2 kg and 2.60 ⋅ 10 −2 kg, so it is reasonable that their total mass is M total = 1.02 ⋅ 10 −1 kg and their average mass is 2.55 ⋅ 10 −2 kg 1.47 THINK: The cylinder has height h = 20.5 cm and radius r = 11.9 cm SKETCH: RESEARCH: The surface area of a cylinder is= A 2π rh + 2π r SIMPLIFY: = A 2π r (h + r ) CALCULATE: = A 2π (11.9 cm)(20.5 cm = + 11.9 cm) 2422.545 cm ROUND: Three significant figures:= A 2.42 ⋅ 103 cm DOUBLE-CHECK: The units of area are a measure of distance squared so the answer is reasonable 1.48 THINK: When you step on the bathroom scale, your mass and gravity exert a force on the scale and the scale displays your weight The given quantity is your mass m1 = 125.4 lbs Pounds can be converted to SI units using the conversion lb = 0.4536 kg Let your mass in kilograms be m2 SKETCH: A sketch is not needed to solve this problem  0.4536 kg  RESEARCH: m2 = m1    lb  SIMPLIFY: It is not necessary to simplify  0.4536 kg  CALCULATE: m2 = 125.4 lbs   = 56.88144 kg lb   ROUND: The given quantity and conversion factor contain four significant figures, so the result must be rounded to 56.88 kg DOUBLE-CHECK: The SI units of mass are kg, so the units are correct 1.49 THINK: The orbital distance from the center of the Moon to the center of the Earth ranges from 356,000 km to 407,000 km Recall the conversion factor mile = 1.609344 kilometer SKETCH: RESEARCH: Let d1 be a distance in kilometers, and d2 the equivalent distance in miles The formula to convert from kilometers to miles is d2 = d1 /1.609344 SIMPLIFY: It is not necessary to simplify Chapter 1: Overview mile   CALCULATE: 356,000 km   = 221208.144 miles  1.609344 km  mile   407,000 km   = 252898.0752 miles  1.609344 km  ROUND: The given quantities have three significant figures, so the calculated values must be rounded to 221,000 miles and 253,000 miles respectively DOUBLE-CHECK: A kilometer is roughly 2/3 of a mile, and the answers are roughly 2/3 of the given values, so the conversions appear correct 1.50 THINK: It is a distance d = 60 feet, inches from the pitcher’s mound to home plate Recall the conversion factors: foot = 12 inches, inch = 2.54 cm, 100 cm = m SKETCH: RESEARCH: If the distance is x in meters and y in feet, then using the conversion factor c, x = cy  12 inches  2.54 cm  m  c = foot     / foot  foot  inch  100 cm  SIMPLIFY: c = 0.3048 meters/foot CALCULATE: 60 feet plus inches = 60.5 feet Then, converting 60.5 feet to meters:  0.3048 m  d 60.5 ft  = =  18.440 m  ft  ROUND: Rounding to three significant figures, the distance is 18.4 m DOUBLE-CHECK: The answer is a reasonable distance for a pitcher to throw the ball 1.51 THINK: The given quantities, written in scientific notation and in units of meters, are: the starting position, xo = ⋅ 10−3 m and the lengths of the flea’s successive hops, x= 6.5 ⋅ 10 −2 m , 3.2 ⋅ 10−2 m , x= x= 8.3 ⋅ 10 −2 m , = x6 15.5 ⋅ 10 −2 m The flea makes six jumps in x 10.0 ⋅ 10 −2 m , = x5 11.5 ⋅ 10 −2 m and = total SKETCH: RESEARCH: The total distance jumped is x total = ∑ xn The average distance covered in a single hop is: n =1 xavg = ∑ xn 6n =1 Bauer/Westfall: University Physics, 2E 40.42 THINK: Plot Sn and S2n for element Sn (Z = 50) versus neutron number, N Since N is an integer, simply examining the plot will be sufficient to determine when each plot crosses into the negative The following constants are given on page = 13.47: aV 15.85 = MeV, as 18.34 = MeV, ac 0.71 MeV, aa = 92.86 MeV, aP = 11.46 MeV Use the fact that A= N + Z to write the formulas that will be obtained in the RESEARCH step in terms of N and Z SKETCH: The plots will be in the CALCULATE step RESEARCH: The Bethe-Weizsäcker formula is given in Equation 40.37: ( −1) + ( −1) Z2  Z 1 =av − as A − ac 4/3 − aa  −  + ap A A A3/2  A 2 Page 1333 discusses the separation energy, S, required to separate some part of an isotope away from the remainder of the nucleus It gives the formula S = B ( N1 + N , Z1 + Z2 ) − B ( N1 , Z1 ) − B ( N , Z2 ) To B(N,Z ) Z N −1/3 compute Sn , = let N 1,= let N 2,= and Z2 To compute S2n ,= and Z2 As mentioned in Example 40.2, the binding energy of two neutrons is zero Also, the binding energy for a single neutron is zero, as there is nothing to bind This means B= (1,0 ) B= ( 2,0 ) SIMPLIFY: Sn = B ( N , Z ) − B ( N − 1, Z ) − B (1,0 ) = B ( N , Z ) − B ( N − 1, Z ) , and S2n = B ( N , Z ) − B ( N − 2, Z ) − B ( 2,0 ) = B ( N , Z ) − B ( N − 2, Z ) B ( N − 1, Z ) , and B ( N − 2, Z ) B ( N , Z ) =av A − as A ( −1) + ( −1) Z2 A  − ac 1/3 − aa  Z −  + ap 2 A A1/2  Z 2/3 = av ( Z + N ) − as ( Z + N ) 2/3 − ac (Z + N ) B ( N − 1, Z= ) av ( Z + N − 1) − as ( Z + N − 1) 1/3 2/3 B ( N − 2, Z= ) av ( Z + N − ) − as ( Z + N − ) − ac 2/3 N  ( Z + N )  + a ( −1) + ( −1) − aa  Z −  p 1/2   (Z + N )   Z2 Sn= ( N , Z ) B ( N , Z ) − B ( N − 1, Z ) Z 1/3 1/3 N −1 Z  ( Z + N − )  + a ( −1) + ( −1) − aa  Z −  p 1/2   ( Z + N − 2)   Z2 ( Z + N − 2) N  ( Z + N − 1)  + a ( −1) + ( −1) − aa  Z −  p 1/2   ( Z + N − 1)   Z2 ( Z + N − 1) − ac B ( N , Z ), Now find expressions for Z N Z N   −1) + ( −1)  Z +N) ( ( 2/3 Z2   = av ( Z + N ) − as ( Z + N ) − ac − aa  Z −  + ap 1/3 1/2    ( Z + N )  (Z + N )    Z N −1   Z + N − 1)  ( ( −1) + ( −1)  2/3 Z a Z a − av ( Z + N − 1) − as ( Z + N − 1) − ac − − +   a p 1/3 1/2    ( Z + N − 1) ( Z + N − 1)     S2n= ( N , Z ) B ( N , Z ) − B ( N − 2, Z ) Z N   −1) + ( −1)  Z +N) ( ( 2/3 Z2   = av ( Z + N ) − as ( Z + N ) − ac − aa  Z −  + ap 1/3 1/2     Z +N) Z + N ( ) (     Z N   Z + N − 2)  −1) + ( −1)  ( ( 2/3 Z2   − av ( Z + N − ) − as ( Z + N − ) − ac − aa  Z −  + ap 1/ 1/     Z + N − 2) Z N + − ( ( )     At this point, further algebraic simplification will only make the functions messier Move on to the following step where the substitution of some values will give the explicit function 1388 Chapter 40: Nuclear Physics CALCULATE: These formulas are unpleasant, and clearly impossible to graph without the aid of a computer graphing utility Sn ( N ,50 ) = 15.85 − 18.34 ( 50 + N ) +18.34 ( 49 + N ) 2/3 + 2/3 − 1775 ( 50 + N ) 1/3 1775 ( 49 + N ) 1/3 N   25 − − 92.86   50 + N  N   25.5 − + 92.86   49 + N   N  + ( −1)  + 11.46 50 + N    N −1  + ( −1)  − 11.46 49 + N   Here is the same function plotted again zoomed in on where it crosses the x-axis: S2n ( N ,50 ) = 31.70 − 18.34 ( 50 + N ) 2/3 + 18.34 ( 48 + N ) 2/3 − + 1775 ( 50 + N ) 1/3 N   25 − − 92.86   50 + N   N  + ( −1)  + 11.46 50 + N   N   26 − + 92.86   48 + N   N  + ( −1)  − 11.46 48 + N   1775 ( 48 + N ) 1/3 1389 Bauer/Westfall: University Physics, 2E Again, the function is replotted, zoomed in on where it crosses the x-axis: ROUND: Sn becomes negative for the first time at N = 99 S2n becomes negative at N = 113 DOUBLE-CHECK: As the number of neutrons in the atom increases, they help keep the atom together so that the binding energy decreases, which is evident in the plots Also, when the separation energy is negative, energy is necessary to separate the atom 40.43 Given that the power plant with P = 1.50 GW has an efficiency of ε = 0.350, the total power that is created is P0 = P / ε The energy that the plant produces in ∆t =1 day is given by E= P0 ∆t Since each reaction is ∆E = 200 MeV, the number of 235 U consumed per day is given by N = E / ∆E Since a molar mass of mM = 235.0439299 g/mol, the mass of Em M P ∆tm M N mM = = = NA ∆EN A ε∆EN A m = 235 235 235 U U has U consumed in one day is: (1.50 ⋅10 W ) ( 86400 s/day )(235.0439299 g/mol ) ( 0.350 ) ( 200 ⋅10 eV )(1.602 ⋅10 J/eV )( 6.022 ⋅10 atoms/mol ) −19 23 = 4.51 kg 40.44 The reaction is 21 H+ 21 H → 42 He + Q The atomic masses of 21 H and 42 He are m (1, 1) = 2.014101778 u and m ( 2, ) = 4.002603254 u (a) The energy released is ∆E = Qc , where Q is the change in mass; therefore, = ∆E 2))c (2m(1, 1) − m(2,= ( ) = c 23.8465 MeV 2 ( 2.014101778 ) − ( 4.002603254 )  c 931.4940 MeV/ (b) The total mass of water in the ocean is approximately = M 1.50 ⋅ 10 kg Since every reaction needs 16 two 21 H atoms, and 0.0300% of the mass is 21 H atoms, then the number of reactions that could occur is: = N ( ) 1.50 ⋅ 1016 kg ( 0.000300 ) M ( 0.000300 ) = = 6.7256 ⋅ 1038 reactions 2m (1, 1) ( 2.014101778 u ) 1.661 ⋅ 10 −27 kg/u Therefore, the total energy is: ( ( ) )( ) E = N ∆E = ( 23.8465 MeV ) 1.602 ⋅ 10 −19 J/eV 6.7256 ⋅ 1038 = 2.57 ⋅1027 J 1390 www.elsolucionario.org Chapter 40: Nuclear Physics (c) If the world uses= P 1.00 ⋅ 1013 W, then the energy would last for: ∆t = 40.45 E = P ( 2.569 ⋅10 J ) = (1.00 ⋅10 J/s ) 27  hr   day   yr  2.57 ⋅ 1014 s   = 8.15 ⋅ 10 yr    3600 s  24 hr   365 day  13 (a) The Sun radiates energy at a rate of= P 3.85 ⋅ 1026 W, giving off energy ∆E in time ∆t : P = ∆E / ∆t The change in energy is related to the change in mass by ∆E =∆mc Therefore, the mass loss rate is: ( 3.85 ⋅ 1026 W ∆E ∆mc ∆m P ⇒ =2 = P= = ∆t ∆t ∆t c 3.00 ⋅ 108 m/s ( ) ) =4.2778 ⋅ 109 kg/s ≈ 4.28 ⋅ 109 kg/s (b) This result is the rate at which mass itself is converted into energy In Example 40.6, the result was the rate of proton mass required as protons fuse to form helium-4 (only a fraction of the proton’s mass gets converted into energy) (c) If a star loses mass at a constant rate, ∆m / ∆t , for = T 4.50 ⋅ 109 yr, and its current mass is = M 1.99 ⋅ 1030 kg, then the percent change in mass over the star’s lifetime, T, is: ( ∆m / ∆t )T 100% %∆m = ( ) M + ( ∆m / ∆t )T ( 4.2778 ⋅10 kg/s )( 4.50 ⋅10 yr ) (365 days/yr )(24 hr/day )( 3600 s/hr ) = (100% ) (1.99 ⋅10 kg ) + ( 4.2778 ⋅10 kg/s )( 4.50 ⋅10 yr ) (365 days/yr )( 24 hr/day )( 3600 s/hr ) 9 30 9 = 0.0305% 40.46 The reaction is: He+ 42 He → 74 Be + γ , where m ( 2, 1) = 3.016029 u, m ( 2, ) = 4.002603 u, m ( 4, ) = 7.0169298 u, and the energy of the photon is given by E = ∆mc Therefore, the minimum possible energy of the photon that is emitted is: ( ) E = m ( 2, 1) + m ( 2, ) − m ( 4, ) c ( ) = ( 3.016029 + 4.002603 − 7.0169298 ) c 931.4940 MeV/c = 1.58559 MeV The maximum possible wavelength of this photon is: λ= max 40.47 hc = Emin ( 6.626 ⋅10 −34 )( )( ) J s 3.00 ⋅ 108 m/s = 783 fm 1.58558 ⋅ 106 eV 1.602 ⋅ 10 −19 J/eV ( ) THINK: In order for the two 23 He atoms to bind, they must come close enough for the strong force to overcome the Coulomb repulsion The closest the two can get is when their centers are separated by the sum of the radii of the atoms, i.e the diameter of one atom Assuming that one is at rest, the kinetic energy of the other atom must be greater than the potential barrier due to the repulsion force The kinetic energy is directly proportional to the temperature of the surroundings SKETCH: 1391 Bauer/Westfall: University Physics, 2E RESEARCH: Thermal energy of a particle is given by K = kBT The Coulomb potential is given by U C = kq / d , where q = 2e for 23 He The diameter of 23 He is d = R(A) k ( 2e ) 4ke ⇒ kBT = ⇒ T= 2R ( A ) 3kB R0 A1/3 SIMPLIFY: The temperature is given by: K = U C ( ( )( ( ) ) 8.99 ⋅ 109 N m / C 1.602 ⋅ 10 −19 C CALCULATE: = T = 13.790 ⋅ 109 K 1/3 −15 −23 1.12 ⋅ 10 m ( ) 1.381 ⋅ 10 J/K ) ROUND: To three significant figures, the temperature required to make the fusion occur is T = 13.8 GK DOUBLE-CHECK: This result is about 1000 times hotter than the core of the Sun However, the temperature really only needs to be a fraction of this because there will be nuclei in the high energy “tail” of the energy distribution of a lower temperature 40.48 THINK: From the principle of conservation of energy, the net Coulomb repulsion can be determined SKETCH: RESEARCH: The conservation of energy is given by Ei = Ef ⇒ K i + m ( 68, 52 ) c = 2m ( 34, 26 ) c + U + K f The Coulomb potential energy is given by: kZ e R SIMPLIFY: Letting K= K= 0, the potential energy U of the two Fe nuclei is given by i f UC = = U (m(68, 52 ) − 2m(34, 26 ) ) c This energy must be equal to the Coulomb potential energy, U = U C , so the separation between the two iron nuclei is given by R = kZ e (m( 68, 52 ) − 2m(34, 26 ) ) c CALCULATE: Substituting the values for the masses and Z = 26 gives: = R ( 8.99 ⋅10 ) ( N m / C ( 26 ) 1.602 ⋅ 10 −19 C ( )( ) (119.904040 ) − ( 59.934078 )  c 931.494 ⋅10 eV/c 1.602 ⋅10 ROUND: Rounding the result to three significant figures gives R = 29.1 fm −19 = 2.9126 ⋅ 10 −14 m J/eV ) DOUBLE-CHECK: A typical distance between nucleons is of the order of 10 fm, so the answer is reasonable 40.49 THINK: From the definition of the mass excess, the atomic mass can be determined For this problem, use the energy conversion, u = 931.49 MeV/c SKETCH: A sketch is not necessary RESEARCH: (a) The definition of the mass excess is= ∆m m ( N , Z ) − A u 1392 Chapter 40: Nuclear Physics (b) Using the atomic masses, the mass-energy difference between the initial and final states for a fission reaction is ∆E =( mi − ∑ mf ) c =( ∆mi − ∑ ∆mf ) c Since mass number is conserved, the mass excess can be used SIMPLIFY: (a) The atomic mass is m ( N , Z= ) A u + ∆m (b) The mass-energy difference between the initial and the final states of two fission reactions are: ∆ECf = ∆EFm = ( ∆m(154, 98 ) − ∆m(84, 56 ) − ∆m(67, 42 ) − 3∆m ) c ( ∆m(156, 100 ) − ∆m(86, 54 ) − ∆m( 46, 66 ) − 4∆m ) c n n 2 (c) Since ∆ECf and ∆EFm are larger than zero, the reactions can occur spontaneously Therefore, energy is released in both reactions CALCULATE: (a) Using the energy conservation and u = 931.49 MeV/c , the atomic masses are determined and given in the following table (rounding to five significant figures) No Nuclide 1 252 98 256 100 140 56 Mass number, A Mass excess, ( ∆m keV/c ) Atomic mass (u) n 8071.3 1.0087 Cf 252 76034 252.08 Fm 256 85496 256.09 Ba 140 -83271 139.91 140 54 Xe 140 -72990 139.92 112 46 Pd 112 -86336 111.91 109 42 Mo 109 -67250 108.93 (b) ∆E= = 202.3411 MeV Cf 76034 + 83271 + 67250 − ( 8071.3 )  keV ∆EFm = 85496 + 72990 + 86336 − ( 8071.3 )  keV = 212.5368 MeV (c) Not necessary ROUND: (a) Not necessary (b) ∆E Cf = 202.34 MeV and ∆E Fm = 212.54 MeV (c) Not necessary DOUBLE-CHECK: As a comparison, the mass-energy difference is ∆EU = 173.3 MeV for the reaction: n + 235 U → 141 Ba + 92 Kr + 3n, so the values obtained here are of a reasonable magnitude 40.50 It is known that the nuclear radius is proportional to A1/3 , as given by R = R0 A1/3 So, A = ( R / R0 ) , where R0 = 1.12 fm and A is the number of nucleons in a nucleus Therefore, the number of nucleons in a 10.0 km diameter star is:  5.00 ⋅ 103 m  55 = = A   8.90 ⋅ 10 nucleons −15 ⋅ 1.12 10 m   1393 www.elsolucionario.org Bauer/Westfall: University Physics, 2E 40.51 The average kinetic energy is related to the temperature by K ave = 3kT / Substituting the numerical values gives: 2.07 ⋅ 10 −16 J K ave = 1.38 ⋅ 10 −23 J/K 1.00 ⋅ 107 K = This corresponds to an average velocity of: ( )( ) ( ) ) 2.07 ⋅ 10 −16 J 2K ave K ave = mv ave ⇒ v ave = = = 4.98 ⋅ 105 m/s −27 m 1.67 ⋅ 10 kg ( 40.52 THINK: It is possible to make a rough estimate of the age of the solar system by comparing the current ratio of 235 U to 238 U isotopes to the ratio which is believed to have been present at the time of the solar system’s formation SKETCH: Not applicable RESEARCH: It is known that the half-life of 235 U and 238 U are 2.22 ⋅ 1016 s and 1.41 ⋅ 1017 s, respectively Using the exponential decay law, N = N e −t ln 2/t1/2 , the ratio of the numbers of the two isotopes yields: N 238 ( t ) N 235 ( t ) = N 0,238 e − t ln 2/ t1/2, 238 N 0,235 e − t ln 2/ t1/2, 235 SIMPLIFY: Since initially, N 0,238 = N 0,235 , the ratio becomes: N 238 ( t )   1 exp  −t ln2  = −  t1/2, 238 t1/2, 235 N 235 ( t )      ⇒ t = −  ln2 1/ t1/2, 238 − 1/ t1/2, 235  ((  0.9928  ln    0.0072  CALCULATE: t = − ln2 1/1.41 ⋅ 1017 s − 1/ 2.22 ⋅ 1016 s ((  N (t )  ln  238  N 235 ( t )    ) ( )) ) ( )) =⋅ 1.873 1017 s = 5.938 ⋅ 109 yr ROUND: The answer should be rounded to two significant figures The explosion was about 5.9 billion years ago DOUBLE CHECK: The formation of our solar system is believed to have occurred approximately 4.5 billion years ago The calculated answer is of the correct order of magnitude 40.53 The activity at time, t, is related to the initial activity by A ( t ) = A0 (1/ ) t / t1/2 ( 2.50 h ) /t1/2 = A ( t 2.50 = h ) A0 (1/ μCi ) = 1.50 = A0 40.54 The activity after 2.50 h is Thus, the initial activity should be: 2.50 h / 6.05 h = μCi )2 ( ) ( ) (1.50 2.00 μCi The frequency of a photon is given to be = f 40.58 ⋅ 106 Hz The wavelength of the photon is: ( 2.998 ⋅10 m/s=) 7.388 m ( 40.58 ⋅10 s ) E= hf = 2.689 ⋅10 ( 6.626 ⋅10 J s )( 40.58 ⋅10 s ) = c λ= = f The energy of the photon is −34 −1 −1 −26 J This photon lies in the radio wave spectrum Since the energy is much less than the energy to ionize electrons from an atom, it is not harmful to the human body Radio waves are constantly around us 1394 Chapter 40: Nuclear Physics 40.55 The exponential decay law can be expressed as N = N (1/ ) t / t1/2 It is given that N / N = 0.1000 The time required is determined using: ln ( N / N )  N  1 t ln= ⇒ t t1/2 =   ln   = ln (1/ )   t1/2  N0  ( 3.825 days ) ln ( 0.1000 ) = 12.71 days ln (1/ ) 40.56 The absorbed dose is given by: amount of radiation energy 0.180 J = dose = = 3.60 = J/kg 3.60 = Gy 360 rd mass of absorbing material 0.0500 kg 40.57 Using the mass-energy relation, E = mc , the energy required to take apart the nucleus into its constituent pieces is: E = Ef − Ei = N p mp c + N n mn c − mi c ( ) = 11(1.007276 ) + 11(1.008665 ) − ( 21.994435 )  c 931.494 MeV/c = 168.522 MeV 40.58 The initial measurement is N = 7210 count/s After 45 min, the measurement is N = 4585 counts/s Using the exponential decay law, N = N 0e − t /τ 1 or N = N   2 t / t1/2 , it is found that the half-life of the material is:  ln (1/ )   N  1 t ⇒= ln = t1/2 t    ln     t1/2  N0   ln ( N / N )    ln (1/ ) = 45 )  t1/2 ( =  69  ln ( 4585 / 7210 )  40.59 Initially, all of the energy is in the form of kinetic energy (assume the potential is zero at a large distance) When the particles are at their closest approach, all of the energy is potential energy Therefore, Ki = Uf ⇒ Ki = 40.60 kZ1 Z2 qp2 Ki −19 −19 53.0 fm The activity of a radioactive material is given by A = λ N The number of nuclei in the sample of mass, m, is N = mN A / M , where M is the molar mass of the material Therefore, the activity is: mN A ln2 m A λ N = = = A M t1/2 M 40.61 r ⇒ r= 8.99 ⋅ 10 N m / C ) ( )( 92 ) (1.602 ⋅ 10 C ) (= ( 5.00 ⋅10 eV )(1.602 ⋅10 J ) = r kZ1 Z2 qp2 (1.00 kg ) ( 6.02 ⋅1023 mol −1 ) ln2 = ( 24100 yr ) ( 3.1536 ⋅107 s/yr ) ( 0.239052 kg/mol ) 2.30 ⋅ 1012 Bq The activity of a sample is determined using the exponential decay law, A = A0 e − λt or A = A0 (1/ ) t / t1/2 After one year, the activity is = A (1.000 μCi )   ( 365 days ) / ( 5.01 days )   1395 = 1.17 ⋅10 -22 μCi Bauer/Westfall: University Physics, 2E 40.62 Denote the fraction of mass of carbon atoms in a human body by f C = 0.14 For a living object, it is known that the abundance of 14 C to 12 C is= r N C) ( C) / N ( = 14 1.20 ⋅ 10 −12 The activity of 12 14 C is given ( C ) The decay constant is given by λ = ln2 / t and the number of C nucleons is m( C ) N N ( C) = , M ( C) M ( C ) is the molar mass of C For this person of mass m = 75 kg, the mass of C present is by A = λ N 14 14 1/2 14 A 14 14 where 14 14 given by: m N C = N  ( ) 12 14 ( C ) M ( C )  m C ( ) ( C ) M ( C )  12 12 14 14 14 m ( C) =  14 and m f m ( C ) + m( C ) = 12 fC m ( C ) M ( C )  +   N ( C) M ( C)    N 12 12 14 14 14 C Combining the above equations and simplifying gives the activity: N A f C rm ln2 A= t1/2 M 12 C + rM 14 C ( ( ) ( )) ( 6.02 ⋅10 mol ) ( 0.14 ) (1.20 ⋅10 )(75 ⋅10 g ) ln2 = (1.81 ⋅10 s ) ( (12.000000 g/mol ) + (1.20 ⋅10 ) (14.003242 g/mol ) ) −1 23 −12 −12 11 = 2.4 ⋅ 103 Bq 40.63 The binding energy of a nucleus is given by B ( N ,= Z )  Zm ( 0, 1) + Nmn − m ( N , Z )  c N = 5, m ( 0,1) = 1.007825032 u, mn = 1.008664916 u, m ( 5,3 ) = 8.022485 u and Substituting Z = 3, u = 931.494 MeV/c gives: ( ) B ( 5,3 ) = 3 (1.007825032 ) + (1.008664916 ) − ( 8.022485 )  c 931.494 MeV/c = 41.279 MeV 40.64 The energy released in decay n → p + e − + ν e , is equal to the difference between the initial and final energies, E= Ei − Ef Since the mass of a neutrino is negligible, the total energy released is: ( ) (m E = m nc − m pc + m ec = = 40.65 n ) − mp − me c = c ) ((1.008664916 ) − (1.007276467 ) − (5.4858 ⋅10 ))c (931.494 MeV/ −4 2 The rest mass energy is defined by the mass-energy relation, E = mc Therefore, the rest mass energy is ( ) ( )( E= ρVc = 737 kg/m ( 3.785 L ) 10 −3 m / L 3.00 ⋅ 108 m/s 40.66 0.782 MeV The exponential decay law is given by N ( t ) = N e − λt or N ( t ) = N (1/ ) ) t / t1/2 = 2.51 ⋅1011 MJ If t = 10t1/2 , the number of remaining atoms is = N ( t 10 = t1/2 ) N (1/ ) 10 If t = 20t1/2 , the number of remaining atoms is 20 10 10 10 N= N (1/ 2= (t 20t= ) N (1/ )  (1/ 2= ) N= (t 10t1/2 )(1/ ) 1/2 ) 1396 www.elsolucionario.org Chapter 40: Nuclear Physics Thus, = N ( t 20 = t1/2 ) 40.67 2) (10 ) (1/= 30 10 1027 atoms The binding energy per nucleon is: B(N, Z )  Zm ( 0, 1) + Nmn − m ( N , Z )  c =  A A (a) For 24 He, B ( 2, ) 2 (1.007825032 ) + (1.008664916 ) − ( 4.002603 )  c 931.494 MeV/c = 4 = 7.074 MeV (b) For He, ( B (1, ) 2 (1.007825032 ) + (1.008664916 ) − ( 3.016030 )  c 931.494 MeV/c = 3 = 2.572 MeV ( (c) For 13 H, B ( 2, 1) (d) For 12 H, = (1.007825032 ) + (1.008664916 ) − ( 3.016050 )  c 931.494 MeV/c 3 = 2.827 MeV ( B (1, 1) (1.007825032 ) + (1.008664916 ) − ( 2.014102 )  c 931.494 MeV/c = 2 = 1.112 MeV ( 40.68 ) ) ) The mean lifetime is related to the half-life by t1/2 = τ ln2 Therefore, the half-life is: t1/2 = 40.69 ) = ( 4300 s ) ln2 3.0 ⋅ 103 s The exponential decay law is given by N = N e − λt or N = N (1/ ) t / t1/2 Since 90% of the sample has decayed, there is 10.0% of the sample remaining Therefore, N / N = 0.100 The time required to reach N / N = 0.100 is: N 1 =   N0   t / t1/2 ⇒ = t 40.70 THINK: The concentrations of 238 t t1/2 = = t t1/2 ln (1/ ) ln ( N / N ) ⇒ ln ( 0.100 ) 26.8 ) (= ln (1/ ) ln ( N / N ) ln (1/ ) 89.0 U and its first five daughters are in equilibrium, that is each daughter is produced as fast as it decays This means the rates of decay (or activities) of the same SKETCH: 238 U and its daughters are RESEARCH: The rate of decay or activity is defined as A = −dN / dt = λN = λnN A , where n is the number of moles For simplicity, denote 238 U and its daughters by letters A, B, C, D, E, F, and G Since the mixture is in equilibrium, the rates of decay are the same for all nuclei, except for Rn that is: 1397 Bauer/Westfall: University Physics, 2E λ= λ= λD= nD λ= λFnF Therefore, the amount of A= A= A= A= A= AF ≡ A or λ= A B C D E A nA B nB C nC E nE each species is ni = λA nA / λi , where i = B, C, D, E, and F The masses of all the species must add up to the total mass, mtot This means: mtot = ∑ ni Mi , where Mi is the molar mass of ith species i SIMPLIFY: M F i A A A A i A =i A= i (a) = mtot F ∑λ n = λ Mi ∑λ λ n ⇒ λ= A nA i mtot F ∑M i=A i / λi Using λ = ln2 / t1/2 , the above equation becomes: λA nA = mtot ln2 F ∑ Mi t1/2,i or A = mtot N A ln2 Then, the rate, in mass per unit time, that ∑ Mi t1/2,i i=A i=A 222 F Rn is produced is: AM G r= NA (b) The rate of activity of radon is: dARn  dN Rn  A ln2 = λRn  =  λ= Rn A dt dt t1/2,G   CALCULATE: (a) The following values are found in Appendix B: 238 U: M A = 238.0508 g/mol and t1/2,A = 1.41 ⋅ 1017 s, 234 Th: M B = 234.0436 g/mol and t1/2,B = 2.08 ⋅ 106 s, 234 Pa: M C = 234.0433 g/mol and t1/2,C = 2.41 ⋅ 10 s, 234 U: M D = 234.0410 g/mol and t1/2,D = 7.74 ⋅ 1012 s, 230 Th: M E = 230.0331 g/mol and t1/2,E = 2.38 ⋅ 1012 s, 226 Ra: M F = 226.0254 g/mol and t1/2,F = 5.05 ⋅ 1010 s, and 222 Rn: M G = 222.0176 g/mol and t1/2,G = 3.30 ⋅ 105 s Therefore, the sum is: F ∑M t i=A = A i 1/2,i = 3.35675 ⋅ 1019 g s/mol = 3.35675 ⋅ 1016 kg s/mol The activity is given by: (1.00 kg ) ln ( ) ( 6.02 ⋅1023 mol −1 ) ( 3.35675 ⋅10 16 The rate, in mass per unit time, that ( 222 kg s/mol = 1.2431 ⋅ 107 decay/s ) Rn is produced is: ) 1.2431 ⋅ 10 decay/s ( 222.0176 g/mol ) r= =4.5845 ⋅ 10 −15 g/s = 0.1446 μg/yr 6.02 ⋅ 1023 mol −1 (b) ( ( ) ) 1.2431 ⋅ 10 decay/s ln2 dARn 26.11 decays/s = 26.11 Bq/s = 8.234 ⋅ 108 Bq/yr = 22.25 mCi/yr = = dt 3.30 ⋅ 105 s ( ) ROUND: Round the results to three significant figures (a) r = 0.145 μg/yr (b) dARn = 22.3 mCi/yr dt 1398 Chapter 40: Nuclear Physics DOUBLE-CHECK: The activity of radon can be approximated by assuming that the sample is composed of only 238 U That is, (1.00 kg ) ( 6.02 ⋅1023 mol −1 ) ln2 = (1.41 ⋅1017 s ) ( 0.2380508 kg/mol ) mN A ln2 = t1/2,A M A A λ= N = 1.24 ⋅ 107 decays/s This is in agreement with the activity calculated above 40.71 14 THINK: The radioactive decay of C follows an exponential decay law, while the number of isotopes stays constant in time because this isotope is stable It can be assumed that 12 ( ) 12 C C comprises all of the mass of the ash; that= is, m m= C 50.0 g The activity, A = 20.0 decays/hr, can be used along with 12 the half-life, t1/2 = 5730 yr, to determine the current number of will provide an approximate age for the tree SKETCH: 14 C atoms Using all of this information RESEARCH: The exponential decay law for the number of atoms remaining as a function of time is given by N (t ) = N e − λt , where λ = ln2 / t1/2 The activity of ratio of 14 C to atoms is N 12 14 is r N C atoms = C is given by A = λ N C) ( C) / N ( = 14 14 N where M 12 12 ( C ) is the molar mass of 12 SIMPLIFY: The decay law for 14 12 14 C = 12 m 12 ( ) ( ) 14 = N 14 C N= C e − λt rN A rm = λ ( C)N e M ( C) 12 A 12 − λt C A , C C is Simplifying and solving for t gives: 14 C atoms is given by ( C) ( ) M CN ( ) 12 As stated in the question, the initial 1.300 ⋅ 10 −12 Therefore, the number of initial 12 ( C ) = rN ( C ) The number of ( C ) ( C )e 12 − λt ( C)N e e ⇒= M ( C)  AM ( C ) t  t  t= − ln  ln2  rm ( C ) N ln2    At1/2 rm ⇒ = ln2 12 A − ln 2t / t1/2 − ln 2t / t1/2 12 12 12 12 1/2 A 1/2 1/2 12 ( C )t rm ( C ) N ln2 AM A ( 5730 yr ) ln  ( 20.0 decays/hr )( 8760 hr/yr ) (12.000000 g/mol )( 5730 yr )  = CALCULATE: t = 63813 yr −   ln2 1.300 ⋅ 10 −12 ( 50.0 g ) 6.02 ⋅ 1023 mol −1 ln2   ROUND: Rounding to three significant figure yields t = 63800 yr ( ) 1399 www.elsolucionario.org ( ) Bauer/Westfall: University Physics, 2E DOUBLE-CHECK: This is a reasonable age for a campfire, considering that fossil evidence indicates that modern humans originated in Africa 200,000 years ago 40.72 THINK: It is known that protons decay according to an exponential decay law To determine the number of protons that would decay during 70.0 years, the initial number of protons inside a human body is needed It is assumed that the mass of the human body is 70.0 kg and the human body is made entirely of water SKETCH: RESEARCH: The exponential decay law is given by N = N e − λt or N = N (1/ ) t / t1/2 The initial number of protons, N , is determined by assuming the human body is composed entirely of water The number of water molecules in the human body is N w = mN A / M w , where m is the mass and M w is the molar mass Each water molecule contains ten protons, so the initial number of protons is= N 10 = N w 10mN A / M w SIMPLIFY: The number of protons that decay during an interval of time, t, is:   t /t1/2 N decay = N − N = N  −    2    10mN A =    Mw  CALCULATE: Substituting the numerical values gives: ( )(     1 −       t / t1/2    ) 30 10 70.0 ⋅ 103 g 6.022 ⋅ 1023 mol −1   (70.0 yr ) /(1.00⋅10 yr )  1 −    N decay =  2  (18.015 g/mol )     7.00⋅10 yr   = 2.33993994 ⋅ 10  −    2    Evaluating this expression is tricky on a handheld calculator, because the first factor is very large, while the second factor is indistinguishable from zero when evaluated The product is not easily determined To deal with this complication, we rewrite the exponential term using the approximation ex ≈ + x, which is good for small x (as is certainly the case here): ( = N decay 28 -29 ) ( 2.33993894 ⋅10 ) (1 − exp −7.00 ln(2) ⋅10  ) ≈ ( 2.33993894 ⋅ 10 ) (1 − 1 − 7.00 ln(2) ⋅ 10  ) = ( 2.33993894 ⋅10 )( 7.00 ln(2) ⋅10 ) = ( 2.33993894 ) ( 7.00 ln(2) ) (10 ) 28 -29 28 -29 28 -29 −1 = 1.1353 ROUND: Rounding to three significant figure gives N decay = 1.14 decays DOUBLE-CHECK: The activity of the protons is: A= λN ≈ ( ( ) 2.34 ⋅ 1028 ln2 N ln2 = = 0.01622 decays/yr t1/2 1.00 ⋅ 1030 yr ) Therefore, during 70.0 years, the number of decays is approximately (since N ≈ N ) : = N decay ≈ A∆t = decays/yr )( 70.0 yr ) ( 0.01622 1.14 decays This agrees with the above result This is expected since the half-life of the proton is very large 1400 Chapter 40: Nuclear Physics 40.73 THINK: Since the theory predicts that protons never get any older it can be assumed that the activity is constant The half-life of the proton, t= 1.80 ⋅ 1029 yr, can be used to determine how many protons, in a 1/2 tank with 1.00 ⋅ 10 tons of water, will decay over two years SKETCH: Not necessary RESEARCH: Assuming that the activity is constant, the number of decays over a given time period is N decay ≈ λ N ∆t , where λ = ln2 / t1/2 Since there are ten protons in a water molecule, the number of protons, N, is given by N = 10mN A / M w , where M w = 18.015 g/mol is the molar mass of water SIMPLIFY: The number of proton decays is N decay = ( )( 10mN A ∆t ln2 M w t1/2 ) 10 1.0010 g 6.02 ⋅ 1023 mol −1 ( yr ) ln2 CALCULATE: N = = 2.5736 ⋅ 10 decays decay (18.015 g/mol ) 1.80 ⋅1029 yr ( ) ROUND: Rounding to three significant figures, N decay = 2.57 ⋅ 10 decays DOUBLE-CHECK: Although the half-life of the proton is very large, a large number decay over two years ( ) because there are so many N = ⋅ 1033 in the tank of water 40.74 THINK: When a proton is placed in a magnetic field, the magnetic dipole moment of the proton can have only two directions: parallel or anti-parallel to the external field By introducing a time-varying electric field at a proper frequency, it can induce a proton to flip its magnetic dipole moment SKETCH: RESEARCH: The energy required to flip the dipole moment of the proton is equal to ∆E = µ B This corresponds to a frequency given by f = ∆E / h SIMPLIFY: The magnetic field is given by: B = hf 2µ 6.626075 ⋅ 10 J s )(15.35850 ⋅ 10 s ) (= (1.410608 ⋅ 10 J/T ) −34 CALCULATE: B = −26 −1 0.360718828 T ROUND: To seven significant figures, B = 0.3607188 T DOUBLE-CHECK: This result is a typical value used in NMR spectroscopy of a proton 40.75 THINK: Radioactive decay follows an exponential decay law In this problem, two species of radioactive nuclei, A and B, are compared After a time of 100 s, it is observed that N A = 100N B with τ A = 2τ B The initial number of nuclei for both species is N SKETCH: A sketch is not necessary RESEARCH: The exponential decay law is given by N = N e −t /τ After an interval of time, t, the number of nuclei A and B are N A = N e −t /τ A and N B = N e −t /τ B 1401 Bauer/Westfall: University Physics, 2E SIMPLIFY: Taking the ratio of N A to N B and using τ A = 2τ B gives N  NA t t − ( t /τ ) (1/2 ) −1 − t 1/τ −1/τ = e ( A B) = e B  ⇒ ln  A  = ⇒ τB = NB 2ln ( N A / N B )  N B  2τ B (100 s ) CALCULATE: = τ B = 10.86 s 2ln (100 ) ROUND: Rounding to three significant figure, τ B = 10.9 s DOUBLE-CHECK: If τ B = 10.86 s, then τ A = 21.72 s Inserting these results into the equation N ( t ) = N e −t /τ gives = = N B (t 100 = s ) 0.0001N N 100 s ) 0.01N and = A (t N A (t = 100 s ) is larger than N B (t = 100 s ) by a factor of 100, as required Multi-Version Exercises 40.76 = λ ln2 ln2 = = 3.833 ⋅ 10 −12 s −1 t1/2 5730 y ( ) ( )  A ⋅ M 12 C  /λ t = − ln   r ⋅ m 12 C N A λ      (105 decays/min )(1 min/60 s ) (12 g/mol )  / (3.833 ⋅ 10 −12 s −1 ) = − ln   1.20 ⋅ 10 −12 (12.43 g ) 6.022 ⋅ 1023 mol −1 (3.833 ⋅10 −12 s −1 )    = 4100 yr ( 40.77 = λ ) ( ) ln2 ln2 = = 3.833 ⋅ 10 −12 s −1 t1/2 5730 y ( ) ( ) ( ) ( ) ( )  A ⋅ M 12 C  A ⋅ M 12 C  / λ ⇒ e − λt = t= − ln  ⇒  r ⋅ m 12 C N A λ  r ⋅ m 12 C N A λ   A ⋅ M 12 C e λt m 12 C = r ⋅ NAλ ( ) (107 decays/min )(1 min/60 s ) (12 g/mol ) (1.20 ⋅10 )( 6.022 ⋅10 −12 23 mol −1 )(3.833 ⋅10 −12 −1 s ) exp (3.833 ⋅ 10 −12 s −1 )(4384 y) = 13.1 g 40.78 = λ ln2 ln2 = = 3.833 ⋅ 10 −12 s −1 t1/2 5730 y ( ) ( ) ( ) ( ) (1.20 ⋅10 )( 6.022 ⋅10 = ( ) ( )  A ⋅ M 12 C  A ⋅ M 12 C  / λ ⇒ e − λt = t= − ln  ⇒  r ⋅ m 12 C N A λ  r ⋅ m 12 C N A λ   r ⋅ N A m 12 C λe − λt A= M 12 C −12 23 ) mol −1 (13.83 g)(3.833 ⋅10 −12 s −1 )exp  −(3.833 ⋅10 −12 s −1 )(4814 y)  (12 g/mol ) = 107 /min 1402 www.elsolucionario.org ...Instructor Solutions Manual to accompany University Physics Second Edition Wolfgang Bauer Michigan State University Gary D Westfall Michigan State University www.elsolucionario.org Instructor Solutions... vector B onto vector A times the length of vector A , or the area of a rectangle with one side    the length of vector A and the other side the length of the projection of vector B onto vector... and = total SKETCH: RESEARCH: The total distance jumped is x total = ∑ xn The average distance covered in a single hop is: n =1 xavg = ∑ xn 6n =1 Bauer/ Westfall: University Physics, 2E x total