www.elsolucionario.org Chapter 23 Solutions 23.1 (a) 10.0 grams electrons 24 23 atoms 47.0 = 2.62 × 10 N= 6.02 × 10 atom mol 107.87 grams mol (b) # electrons added = or 23.2 Q 1.00 × 10 −3 C = = 6.25 × 1015 e 1.60 × 10 -19 C electron 2.38 electrons for every 10 already present ( )( ) (a) 8.99 × 10 N ⋅ m 2/ C 1.60 × 10 −19 C k qq Fe = e 21 = r (3.80 × 10 − 10 m)2 (b) 6.67 × 10 −11 N ⋅ m kg (1.67 × 10 − 27 kg)2 G m1m2 Fg = = = 1.29 × 10 − 45 N r2 (3.80 × 10 −10 m)2 ( = 1.59 × 10 − N (repulsion) ) The electric force is larger by 1.24 × 10 36 times (c) If ke q = m 23.3 q1q2 mm = G 12 2 r r G = ke with q1 = q2 = q and m1 = m2 = m, then 6.67 × 10 −11 N ⋅ m / kg = 8.61 × 10 −11 C / kg 8.99 × 10 N ⋅ m / C If each person has a mass of ≈ 70 kg and is (almost) composed of water, then each person contains 70, 000 grams protons molecules 10 N≈ ≈ 2.3 × 10 28 protons 6.02 × 10 23 mol molecule 18 grams mol With an excess of 1% electrons over protons, each person has a charge q = (0.01)(1.6 × 10 −19 C)(2.3 × 10 28 ) = 3.7 × 107 C So F = ke q1q2 (3.7 × 10 ) = (9 × 10 ) N = × 10 25 N ~ 1026 N r2 0.6 This force is almost enough to lift a "weight" equal to that of the Earth: Mg = (6 × 10 24 kg)(9.8 m s ) = × 10 25 N ~ 1026 N © 2000 by Harcourt, Inc All rights reserved Chapter 23 Solutions We find the equal-magnitude charges on both spheres: 23.4 F = ke q1q2 q2 = ke 2 r r q=r so F 1.00 × 10 N = (1.00 m ) = 1.05 × 10 −3 C ke 8.99 × 10 N ⋅ m 2/ C The number of electron transferred is then ( N xfer = 1.05 × 10 −3 C ) (1.60 × 10 −19 ) C / e − = 6.59 × 1015 electrons The whole number of electrons in each sphere is 10.0 g 23 − 24 − Ntot = 6.02 × 10 atoms / mol 47 e / atom = 2.62 × 10 e 107.87 g / mol ( )( ) The fraction transferred is then f= Ntot 6.59 × 1015 = = 2.51 × 10–9 2.62 × 1024 ( = 2.51 charges in every billion )( 8.99 × 10 N ⋅ m C 1.60 × 10 −19 C qq F = ke 2 = r 2(6.37 × 106 m) ) (6.02 × 10 ) 23 www.elsolucionario.org [ ] 23.5 *23.6 N xfer (a) The force is one of attraction The distance r in Coulomb's law is the distance between centers The magnitude of the force is F= (b) = 514 kN ( )( ) 12.0 × 10 −9 C 18.0 × 10 −9 C ke q1q2 N⋅m = 8.99 × 10 = 2.16 × 10 − N r2 C2 (0.300 m)2 The net charge of − 6.00 × 10 −9 C will be equally split between the two spheres, or − 3.00 × 10 −9 C on each The force is one of repulsion, and its magnitude is ( )( ) 3.00 × 10 −9 C 3.00 × 10 −9 C ke q1q2 N⋅m F= = 8.99 × 10 = r2 C2 (0.300 m)2 8.99 × 10 −7 N Chapter 23 Solutions 23.7 F1 = ke q1q2 (8.99 × 10 N ⋅ m 2/ C )(7.00 × 10 −6 C)(2.00 × 10 −6 C) = = 0.503 N r2 (0.500 m)2 F2 = k e q1q2 (8.99 × 10 N ⋅ m / C )(7.00 × 10 −6 C)(4.00 × 10 −6 C) = = 1.01 N (0.500 m)2 r2 Fx = (0.503 + 1.01) cos 60.0° = 0.755 N Fy = (0.503 − 1.01) sin 60.0° = − 0.436 N F = (0.755 N)i − (0.436 N)j = 0.872 N at an angle of 330° Goal Solution Three point charges are located at the corners of an equilateral triangle as shown in Figure P23.7 Calculate the net electric force on the 7.00− µ C charge G: Gather Information: The 7.00− µ C charge experiences a repulsive force F1 due to the 2.00− µ C charge, and an attractive force F due to the −4.00− µ C charge, where F2 = 2F1 If we sketch these force vectors, we find that the resultant appears to be about the same magnitude as F2 and is directed to the right about 30.0° below the horizontal O: Organize : We can find the net electric force by adding the two separate forces acting on the 7.00− µ C charge These individual forces can be found by applying Coulomb’s law to each pair of charges A: Analyze: F1 The force on the 7.00− µ C charge by the 2.00 C charge is (8.99 ì 10 = )( )( N ⋅ m 2/ C 7.00 × 10 −6 C 2.00 × 10 −6 C (0.500 m) ) (cos60°i + sin 60°j) = F = (0.252i + 0.436j) N Similarly, the force on the 7.00− µ C by the −4.00− µ C charge is ( )( ) −6 −6 N ⋅ m 7.00 × 10 C − 4.00 × 10 C F = − 8.99 × 10 (cos60°i − sin 60° j) = (0.503i − 0.872j) N C2 (0.500 m)2 Thus, the total force on the 7.00− µ C , expressed as a set of components, is F = F1 + F = (0.755 i − 0.436 j) N = 0.872 N at 30.0° below the +x axis L: Learn: Our calculated answer agrees with our initial estimate An equivalent approach to this problem would be to find the net electric field due to the two lower charges and apply F=qE to find the force on the upper charge in this electric field © 2000 by Harcourt, Inc All rights reserved www.elsolucionario.org Chapter 23 Solutions Let the third bead have charge Q and be located distance x from the left end of the rod This bead will experience a net force given by *23.8 F= ke ( 3q)Q x2 i+ ke ( q)Q ( d − x )2 ( −i) The net force will be zero if x = , or d − x = x ( d − x )2 This gives an equilibrium position of the third bead of stable if the third bead has positive charge The equilibrium is *23.9 k ee (1.60 × 10–19 C)2 = (8.99 × 109 N ⋅ m2/C 2) = 8.22 × 10–8 N r (0.529 × 10–10 m)2 (a) F= (b) We have F = mv r from which v = Fr = m (8.22 × 10 The top charge exerts a force on the negative charge 23.10 to the left, at an angle of tan −1 ( d / 2x ) to the x-axis force k qQ e d + x2 ( (a) (− x)i d2 + x2 ) ( = ma 1/2 ) −8 )( N 0.529 × 10 −10 m 9.11 × 10 −31 ke qQ ( d )2 + x or for x 1, the components perpendicular to the x-axis add to zero The total field is (b) nke (Q/n)i keQxi cos θ = 2 R +x (R + x 2)3/2 A circle of charge corresponds to letting n grow beyond all bounds, but the result does not depend on n Smearing the charge around the circle does not change its amount or its distance from the field point, so it does not change the field Chapter 23 Solutions 23.23 E=∑ 23.24 E= − ke qi π ke q ke q kq kq kq 1 ~ = e2 (− i)+ e (− i) + e (− i) + = 1+ + + = − i 2 r a (2a) (3a) a2 a (8.99 × 109)(22.0 × 10–6) k (Q / l)l ke λ l keQ = e = = (0.290)(0.140 + 0.290) d(l+ d) d(l+ d) d(l+ d) E = 1.59 × 106 N/C , E=∫ 23.25 ke dq x2 directed toward the rod where dq = λ0 dx ∞ dx 1 = k e – x x0 x ⌠ E = ke λ0 ⌡ E = ∫ dE = ∫ 23.26 ∞ x0 E= 23.27 ∞ = x0 k eλ x0 The direction is –i or left for λ0 > ∞ −3 ke λ x0 dx( −i ) = − k λ x i x dx = − k λ x i − e 0 e 0 ∫x0 x3 2x (8.99 × 109)(75.0 × 10–6)x 6.74 × 105 x k exQ = 2 3/2 = 3/2 (x + a ) (x + 0.100 ) (x + 0.0100)3/2 (a) At x = 0.0100 m, E = 6.64 × 106 i N/C = 6.64 i MN/C (b) At x = 0.0500 m, E = 2.41 × 107 i N/C = 24.1 i MN/C (c) At x = 0.300 m, E = 6.40 × 106 i N/C = 6.40 i MN/C (d) At x = 1.00 m, E = 6.64 × 105 i N/C = 0.664 i MN/C © 2000 by Harcourt, Inc All rights reserved ∞ x0 = ke λ (− i) 2x0 Chapter 46 Solutions 46.21 (a) Λ0 → p + π – Strangeness: –1 → + (b) π – + p → Λ0 + K Strangeness: + → –1 + (0 = and strangeness is conserved ) (c) – – p + p → Λ + Λ0 Strangeness: + → +1 – (0 = and strangeness is conserved ) (strangeness is not conserved ) (d) π – + p → π – + Σ + Strangeness: + → – (0 ≠ –1: strangeness is not conserved ) Ξ – → Λ0 + π – Strangeness: –2 → –1 + (–2 ≠ –1 so strangeness is not conserved Strangeness: –2 → + (–2 ≠ so strangeness is not conserved ) (e) ) 46.22 (f) Ξ0 → p + π – (a) µ – → e– + γ Le : → + 0, (b) n → p + e– + ν e Le : → + + (c) Λ0 → p + π Strangeness: –1 → + 0, (d) p → e+ + π (e) *46.23 (a) Ξ0 → n + π0 and Lµ : → and charge: → +1 + Baryon number: +1 → + Strangeness: –2 → + π − + p → η violates conservation of baryon number as + → not allowed (b) K − + n → Λ0 + π − Baryon number = + → + Charge = –1 + → – Strangeness, – + → –1 + Lepton number, → The interaction may occur via the strong interaction since all are conserved (c) K− → π − + π Strangeness, –1 → + Baryon number, → Lepton number, → Charge, –1 → –1 + Strangeness is violated by one unit, but everything else is conserved Thus, the reaction can occur via the weak interaction , but not the strong or electromagnetic interaction (d) Ω − → Ξ− + π Baryon number, → + Lepton number, → Charge, –1 → –1 + Strangeness, –3 → –2 + May occur by weak interaction , but not by strong or electromagnetic (e) η → 2γ Baryon number, → Lepton number, → Charge, → Strangeness, → © 2000 by Harcourt, Inc All rights reserved www.elsolucionario.org 10 Chapter 46 Solutions No conservation laws are violated, but photons are the mediators of the electromagnetic interaction Also, the lifetime of the η is consistent with the electromagnetic interaction *46.24 (a) Ξ− → Λ0 + µ − + ν µ Baryon number: + → + + + Le : → + + Charge: –1 → – + Lµ : → + + Lτ : → + + Strangeness: –2 → –1 + + Conserved quantities are: (b) K 0S → 2π Baryon number: → Le : → Charge: → Lµ : → Lτ : → Strangeness: +1 → Conserved quantities are: (c) Charge: –1 + → + Lµ : + → + Lτ : + → + Strangeness: –1 + → –1 + Charge: → Lµ : → + Lτ : → + Strangeness: –1 → –1 + B, S, charge , Le , Lµ , and Lτ e+ + e− → µ + + µ − Baryon number: + → + Le : –1 + → + Charge: +1 – → +1 – Lµ : + → + – Lτ : + → + Strangeness: + → + Conserved quantities are: (f) S, charge , Le , Lµ , and Lτ Σ → Λ0 + γ Baryon number: + → + Le : → + Conserved quantities are: (e) B, charge , Le , Lµ , and Lτ K− + p → Σ0 + n Baryon number: + → + Le : + → + Conserved quantities are: (d) B, charge , Le , and Lτ B, S, charge , Le , Lµ , and Lτ p + n → Λ + Σ− Baryon number: –1 + → –1 + Le : + → + Charge: –1 + → – Lµ : + → + Lτ : + → + Strangeness: + → +1 – Conserved quantities are: B, S, charge , Le , Lµ , and Lτ Chapter 46 Solutions *46.25 (a) K+ + p → ? + p The strong interaction conserves everything +1→ B +1 so Baryon number, +1 + → Q + so Charge, so Lepton numbers, + → L + +1 + → S + so Strangeness, 11 B=0 Q = +1 Le = Lµ = Lτ = S=1 The conclusion is that the particle must be positively charged, a non-baryon, with strangeness of +1 Of particles in Table 46.2, it can only be the K + Thus, this is an elastic scattering process The weak interaction conserves all but strangeness, and ∆S = ±1 (b) Ω− → ? + π − Baryon number, Charge, Lepton numbers, Strangeness, +1 → B + −1 → Q − 0→L+0 −3 → S + so so so so ∆S = 1: B=1 Q=0 Le = Lµ = L τ = S = −2 The particle must be a neutral baryon with strangeness of –2 Thus, it is the Ξ (c) K+ → ? + µ+ + νµ : Baryon number, 0→B+0+0 +1 → Q + + Charge, Lepton Numbers Le , → Le + + Lµ , → Lµ − + Lτ , → Lτ + + 1→ S + + Strangeness: B=0 Q=0 Le = Lµ = Lτ = ∆S = ±1 (for weak interaction): S = so so so so so so www.elsolucionario.org The particle must be a neutral meson with strangeness = ⇒ π *46.26 (a) strangeness baryon number charge proton e u 1/3 2e/3 u 1/3 2e/3 d 1/3 –e/3 total e strangeness baryon number charge neutron u 1/3 2e/3 d 1/3 –e/3 d 1/3 –e/3 total (b) *46.27 (a) The number of protons 6.02 × 10 23 molecules 10 protons 26 N p = 1000 g molecule = 3.34 ì 10 18.0 g â 2000 by Harcourt, Inc All rights reserved protons 12 Chapter 46 Solutions 6.02 × 10 23 molecules neutrons 26 N n = (1000 g ) molecule = 2.68 × 10 18.0 g and there are So there are for electric neutrality (b) neutrons 3.34 × 1026 electrons The up quarks have number × 3.34 × 1026 + 2.68 × 1026 = 9.36 × 1026 up quarks and there are × 2.68 × 1026 + 3.34 × 1026 = 8.70 × 1026 down quarks Model yourself as 65 kg of water Then you contain 65 × 3.34 × 1026 ~ 1028 electrons 65 × 9.36 × 1026 ~ 1029 up quarks 65 × 8.70 × 1026 ~ 1029 down quarks Only these fundamental particles form your body You have no strangeness, charm, topness or bottomness 46.28 strangeness baryon number charge K0 0 d 1/3 –e/3 s –1/3 e/3 total 0 strangeness baryon number charge Λ0 –1 u 1/3 2e/3 d 1/3 –e/3 s –1 1/3 –e/3 (a) (b) 46.29 total –1 Quark composition of proton = uud and of neutron = udd Thus, if we neglect binding energies, we may write m p = 2m u + m d (1) and m n = m u + 2m d (2) Solving simultaneously, we find 1 [ ] m u = (2m p – m n) = 2(938.3 MeV / c ) − 939.6 MeV / c = 312 MeV/c2 and from either (1) or (2), m d = 314 MeV/c *46.30 In the first reaction, π − + p → K + Λ0 , the quarks in the particles are: ud + uud → ds + uds There is a net of up quark both before and after the reaction, a net of down quarks both www.elsolucionario.org Chapter 46 Solutions 13 before and after, and a net of zero strange quarks both before and after Thus, the reaction conserves the net number of each type of quark In the second reaction, π − + p → K + n , the quarks in the particles are: ud + uud → ds + udd In this case, there is a net of up and down quarks before the reaction but a net of up, down, and anti-strange quark after the reaction Thus, the reaction does not conserve the net number of each type of quark 46.31 (a) π − + p → Κ + Λ0 In terms of constituent quarks: ud + uud → ds + uds – + → + 1, or → 1 + → + 1, or → + → –1 + 1, or → up quarks: down quarks: strange quarks: (b) π + + p → Κ+ + Σ+ ud + uud → us + uus ⇒ + → + 2, or → –1 + → + 0, or → 0 + → –1 + 1, or → up quarks: down quarks: strange quarks: (c) Κ − + p → Κ + + Κ + Ω− us + uud → us + ds + sss ⇒ –1 + → + + 0, or → + → + + 0, or → 1 + → –1 – + 3, or → up quarks: down quarks: strange quarks: (d) p + p → K0 + p + π + + ? uud + uud → ds + uud + ud + ? ⇒ The quark combination of ? must be such as to balance the last equation for up, down, and strange quarks up quarks: down quarks: strange quarks: 2+2=0+2+1+? 1+1=1+1–1+? + = –1 + + + ? (has u quark) (has d quark) (has s quark) quark composite = uds = Λ0 or Σ 46.32 Σ0 + p → Σ+ + γ + X dds + uud → uds + + ? The left side has a net 3d, 2u and 1s The right-hand side has 1d, 1u, and 1s leaving 2d and 1u missing The unknown particle is a neutron, udd Baryon and strangeness numbers are conserved © 2000 by Harcourt, Inc All rights reserved 14 Chapter 46 Solutions Compare the given quark states to the entries in Tables 46.4 and 46.5 *46.33 (a) suu = Σ + (b) ud = π − (c) sd = Κ (d) ssd = Ξ − *46.34 (a) (b) ) ( ) ( ) ( ) ( ) ( ) u d d : charge = − 23 e + 31 e + 31 e = This is the antineutron Section 39.4 says *46.35 ( u ud : charge = − 23 e + − 23 e + 31 e = −e This is the antiproton f observer = f source + va c − va c The velocity of approach, va , is the negative of the velocity of mutual recession: va = −v Then, and λ′ = λ 1+ v c 1− v c (1.7 × 10–2 m/s) ly v = HR (Equation 46.7) H= (a) v (2.00 × 106 ly) = 3.4 × 104 m/s λ' = λ (b) v (2.00 × 108 ly) = 3.4 × 106 m/s λ' = 590 + 0.01133 = 597 nm − 0.01133 (c) v (2.00 × 109 ly) = 3.4 × 107 m/s λ' = 590 + 0.1133 = 661 nm − 0.1133 (a) λ' 650 nm = 434 nm = 1.50 = λ + v/c – v/c = 2.24 46.36 46.37 c c 1− v c = λ′ λ + v c v = 0.383c, (b) Equation 46.7, v = HR + v/c − v/c + v/c = 590(1.0001133) = 590.07 nm − v/c 38.3% the speed of light v (0.383)(3.00 × 108) R= H = = 6.76 × 109 light years (1.7 × 10–2) Chapter 46 Solutions 15 Goal Solution A distant quasar is moving away from Earth at such high speed that the blue 434-nm hydrogen line is observed at 650 nm, in the red portion of the spectrum (a) How fast is the quasar receding? You may use the result of Problem 35 (b) Using Hubble's law, determine the distance from Earth to this quasar G: The problem states that the quasar is moving very fast, and since there is a significant red shift of the light, the quasar must be moving away from Earth at a relativistic speed (v > 0.1c) Quasars are very distant astronomical objects, and since our universe is estimated to be about 15 billion years old, we should expect this quasar to be ~10 light-years away O: As suggested, we can use the equation in Problem 35 to find the speed of the quasar from the Doppler red shift, and this speed can then be used to find the distance using Hubble’s law A: (a) 1+ v c λ ′ 650 nm = = 1.498 = λ 434 nm 1− v c Therefore, v = 0.383c or or squared, 1+ v c = 2.243 1− v c 38.3% the speed of light (b) Hubble’s law asserts that the universe is expanding at a constant rate so that the speeds of galaxies are proportional to their distance R from Earth, v = HR so, ( ) v (0.383) 3.00 × 10 m / s R= = = 6.76 × 10 ly H 1.70 × 10 -2 m / s ⋅ ly ( ) www.elsolucionario.org L: The speed and distance of this quasar are consistent with our predictions It appears that this quasar is quite far from Earth but not the most distant object in the visible universe *46.38 (a) λ ′n = λ n 1+ v/c = (Z + 1)λ n 1− v/c + v/c – v/c = ( Z + 1) v v + c = ( Z + 1)2 – c ( Z + 1) v ( Z + Z + 2) = Z + Z c Z2 + Z v = c Z + Z + 2 (b) v c Z2 + Z R=H = H Z + Z + 2 © 2000 by Harcourt, Inc All rights reserved www.elsolucionario.org 16 Chapter 46 Solutions ( ) The density of the Universe is ρ = 1.20ρc = 1.20 3H π G *46.39 Consider a remote galaxy at distance r The mass interior to the sphere below it is 4π r 3H π r 0.600H r M = ρ = 1.20 = G 8π G both now and in the future when it has slowed to rest from its current speed v = Hr The energy of this galaxy is constant as it moves to apogee distance R: mv 2 − GmM GmM =0− r R so mH r 2 r R so R = 6.00 r − 0.100 = − 0.600 The Universe will expand by a factor of 6.00 *46.40 (a) kBT ≈ 2mp c (b) kBT ≈ 2me c *46.41 (a) Wien’s law: Thus, (b) *46.42 (a) (b) T≈ so T≈ mp c kB = − Gm 0.600H r Gm 0.600H r = − R r G G from its current dimensions (938.3 MeV ) 1.60 × 10 −13 1.38 × 10 −23 J K MeV ( ) (0.511 MeV ) 1.60 × 10 −13 2me c = kB 1.38 × 10 −23 J K MeV ( ) J 13 ~ 10 K J λ maxT = 2.898 × 10 − m ⋅ K λ max = 2.898 × 10 − m ⋅ K 2.898 × 10 − m ⋅ K = = 1.06 × 10 − m = 1.06 mm T 2.73 K This is a microwave hG L= = c3 ~ 1010 K (1.055 × 10 −34 )( J ⋅ s 6.67 × 10 −11 N ⋅ m kg (3.00 × 10 This time is given as T = ms ) )= 1.61 × 10 − 35 m L 1.61 × 10 −35 m = = 5.38 × 10 − 44 s , c 3.00 × 108 m s which is approximately equal to the ultra-hot epoch Chapter 46 Solutions 46.43 (a) ∆E ∆t ≈ h, and ∆E ≈ m= (b) *46.44 (a) 46.45 ∆t ≈ 17 r 1.4 × 10 −15 m = = 4.7 × 10 −24 s c 3.0 × 108 m s h 1.055 × 10 −34 J ⋅ s = = 2.3 × 10 −11 ∆t 4.7 × 10 −24 s MeV J = 1.4 × 10 MeV 1.60 × 10 −13 J ∆E ≈ 1.4 × 10 MeV c ~ 10 MeV c c2 From Table 46.2, mπ c = 139.6 MeV , a pi-meson π – + p → Σ + + π is forbidden by charge conservation (b) µ – → π – + νe is forbidden by energy conservation (c) p → π + + π + + π – is forbidden by baryon number conservation The total energy in neutrinos emitted per second by the Sun is: (0.4)(4π)(1.5 × 1011)2 = 1.1 × 1023 W Over 109 years, the Sun emits 3.6 × 1039 J in neutrinos This represents an annihilated mass mc = 3.6 × 1039 J m = 4.0 × 1022 kg About part in 50,000,000 of the Sun's mass, over 109 years, has been lost to neutrinos © 2000 by Harcourt, Inc All rights reserved 18 Chapter 46 Solutions Goal Solution The energy flux carried by neutrinos from the Sun is estimated to be on the order of 0.4 W/m at Earth's surface Estimate the fractional mass loss of the Sun over 109 years due to the radiation of neutrinos (The mass of the Sun is × 1030 kg The Earth-Sun distance is 1.5 × 1011 m.) G: Our Sun is estimated to have a life span of about 10 billion years, so in this problem, we are examining the radiation of neutrinos over a considerable fraction of the Sun’s life However, the mass carried away by the neutrinos is a very small fraction of the total mass involved in the Sun’s nuclear fusion process, so even over this long time, the mass of the Sun may not change significantly (probably less than 1%) O: The change in mass of the Sun can be found from the energy flux received by the Earth and Einstein’s famous equation, E = mc A: Since the neutrino flux from the Sun reaching the Earth is 0.4 W/m 2, the total energy emitted per second by the Sun in neutrinos in all directions is (0.4 W / m )(4π r ) = (0.4 W / m )(4π )(1.5 × 10 2 11 m ) = 1.13 × 10 23 W In a period of 10 yr, the Sun emits a total energy of (1.13 × 10 23 )( )( ) J / s 10 yr 3.156 × 107 s / yr = 3.57 × 10 39 J in the form of neutrinos This energy corresponds to an annihilated mass of E = mν c = 3.57 × 10 39 J mν = 3.97 × 10 22 kg so Since the Sun has a mass of about × 10 kg, this corresponds to a loss of only about part i n 50 000 000 of the Sun's mass over 10 yr in the form of neutrinos L: It appears that the neutrino flux changes the mass of the Sun by so little that it would be difficult to measure the difference in mass, even over its lifetime! 46.46 p + p → p +π+ + X We suppose the protons each have 70.4 MeV of kinetic energy From conservation of momentum, particle X has zero momentum and thus zero kinetic energy Conservation of energy then requires ( ) ( Mp c + M π c + M X c = Mp c + K p + Mp c + K p ) MX c = Mp c + 2Kp − Mπ c = 938.3 MeV + (70.4 MeV ) − 139.6 MeV = 939.5 MeV X must be a neutral baryon of rest energy 939.5 MeV Thus X is a neutron www.elsolucionario.org Chapter 46 Solutions *46.47 19 We find the number N of neutrinos: 1046 J = N(6 MeV) = N(6 × 1.6 × 10–13 J) N = 1.0 × 1058 neutrinos The intensity at our location is ly N 1.0 × 10 58 N 14 = = = 3.1 × 10 / m 2 A 4π r π (1.7 × 10 ly) (3.0 × 10 m / s)(3.16 × 10 s) The number passing through a body presenting 5000 cm2 = 0.50 m2 is then ( ) 3.1 × 1014 0.50 m = 1.5 × 1014 or m *46.48 ~ 1014 E γ + me c = By relativistic energy conservation, By relativistic momentum conservation, c = − v2 / c2 3me v X= Subtracting (2) from (1), me c = E γ + me c = 3me c − X2 (1) (2) − v2 / c2 Eγ Dividing (2) by (1), Solving, 46.49 Eγ 3me c v c − 3me c X − X2 www.elsolucionario.org 1= − 3X 1− X X= and so E γ = 4me c = 2.04 MeV m Λc2 = 1115.6 MeV Λ0 → p + π – mπ c = 139.6 MeV m p c = 938.3 MeV The difference between starting mass-energy and final mass-energy is the kinetic energy of the products and p p = pπ = p K p + K π = 37.7 MeV Applying conservation of relativistic energy, (938.3)2 + p c − 938.3 + (139.6)2 + p c − 139.6 = 37.7 MeV Solving the algebra yields p π c = ppc = 100.4 MeV Then, Kp = (m p c 2)2 + (100.4)2 – m pc = 5.35 MeV Kπ = (139.6)2 + (100.4)2 – 139.6 = 32.3 MeV © 2000 by Harcourt, Inc All rights reserved 20 46.50 Chapter 46 Solutions ( pp = 5.32 × 10–20 ) qBr = 1.60 × 10 −19 C (0.250 kg / C ⋅ s)(1.33 m ) Momentum of proton is kg · m s cpp = 1.60 × 10–11 kg m2 = 1.60 × 10–11 J = 99.8 MeV s2 MeV c Therefore, pp = 99.8 The total energy of the proton is Ep = E02 + (cp)2 = (938.3)2 + (99.8)2 = 944 MeV For pion, the momentum qBr is the same (as it must be from conservation of momentum i n a 2-particle decay) pπ = 99.8 MeV c E0π = 139.6 MeV Eπ = E02 + (cp)2 = (139.6)2 + (99.8)2 = 172 MeV Thus, ETotal after = ETotal before = Rest Energy Rest Energy of unknown particle = 944 MeV + 172 MeV = 1116 MeV (This is a Λ0 particle!) Mass = 1116 MeV/c 46.51 Σ → Λ0 + γ From Table 46.2, mΣ = 1192.5 MeV/c and ( mΛ = 1115.6 MeV c ) Conservation of energy requires E0 , Σ = E0 , Λ + K Λ + Eγ , or p2 1192.5 MeV = 1115.6 MeV + Λ + Eγ 2mΛ Momentum conservation gives pΛ = pγ , so the last result may be written as pγ2 1192.5 MeV = 1115.6 MeV + + Eγ 2mΛ or pγ2 c 1192.5 MeV = 1115.6 MeV + + Eγ 2mΛ c Recognizing that mΛ c = 1115.6 MeV we now have 1192.5 MeV = 1115.6 MeV + Solving this quadratic equation, Eγ ≈ 74.4 MeV and pγ c = Eγ , Eγ2 (1115.6 MeV ) + Eγ Chapter 46 Solutions 46.52 21 p + p → p + n +π+ The total momentum is zero before the reaction Thus, all three particles present after the reaction may be at rest and still conserve momentum This will be the case when the incident protons have minimum kinetic energy Under these conditions, conservation of energy gives ( ) mp c + K p = mp c + mnc + mπ c so the kinetic energy of each of the incident protons is Kp = 46.53 mnc + mπ c − mp c 2 Time-dilated lifetime: = (939.6 + 139.6 − 938.3) MeV = T = γ T0 = 0.900 × 10–10 s – v2/c2 = 70.4 MeV 0.900 × 10–10 s – (0.960)2 = 3.214 × 10–10 s distance = (0.960)(3.00 × 108 m/s)(3.214 × 10–10 s) = 9.26 cm 46.54 π − → µ− + νµ From the conservation laws, mπ c = 139.5 MeV = Eµ + Eν and pµ = pν , Eν = pν c Thus, Eµ2 = pµ c or Eµ2 − Eν2 = (105.7 MeV ) Since Eµ + Eν = 139.5 MeV [1] and (Eµ + Eν )(Eµ − Eν ) = (105.7 MeV)2 [2] then Eµ − Eν = Subtracting [3] from [1], 2Eν = 59.4 MeV ( ) [1] + (105.7 MeV ) = ( pν c) + (105.7 MeV ) 2 (105.7 MeV)2 139.5 MeV [2] = 80.1 and © 2000 by Harcourt, Inc All rights reserved [3] Eν = 29.7 MeV www.elsolucionario.org 22 Chapter 46 Solutions The expression e −E kBT dE gives the fraction of the photons that have energy between E and E + dE The fraction that have energy between E and infinity is *46.55 ∞ ∞ B −E k B T ∞ B B −E k T dE ∫ e −E k T ( − dE kBT ) e ∫E e E = E∞ = = e −E k T ∞ −E k T −E k T −E k T ∞ dE ∫ e (− dE kBT ) e ∫0 e 0 B B B We require T when this fraction has a value of 0.0100 (i.e., 1.00%) and E = 1.00 eV = 1.60 × 10 −19 J Thus, 0.0100 = e or 46.56 (a) ( − 1.60 × 10 −19 J ) (1.38 × 10 −23 J K )T ln (0.0100) = − ( 1.60 × 10 −19 J ) 1.38 × 10 −23 J K T = − 1.16 × 10 K T giving T = 2.52 × 10 K This diagram represents the annihilation of an electron and an antielectron From charge and leptonnumber conservation at either vertex, the exchanged particle must be an electron, e – (b) This is the tough one A neutrino collides with a neutron, changing it into a proton with release of a muon This is a weak interaction The exchanged particle has charge +1e and is a W + 46.57 (a) (b) (a) The mediator of this weak interaction is a Z0 boson (b) The Feynman diagram shows a down quark and its antiparticle annihilating each other They can produce a particle carrying energy, momentum, and angular momentum, but zero charge, zero baryon number, and, it may be, no color charge In this case the product (a) (b) particle is a photon For conservation of both energy and momentum, we would expect two photons; but momentum need not be strictly conserved, according to the uncertainty principle, if the photon travels a sufficiently short distance before producing another matter-antimatter pair of particles, as shown in Figure P46.57 Depending on the color charges of the d and d quarks, Chapter 46 Solutions 23 the ephemeral particle could also be a gluon , as suggested in the discussion of Figure 46.13(b) www.elsolucionario.org © 2000 by Harcourt, Inc All rights reserved ... F= k eQ 2a F = b ∫ x = b – 2a x(x + 2a) dx = k eQ 2a + x b – ln 2a 2a x b – 2a +k eQ 2a + b b k eQ b2 k e Q 2? ?? b2 – ln + ln = ln = ln 2 2 (b – 2a)(b + 2a) b b – 2a 4a... =0 3 /2 2 5 /2 dx (x + a ) (x + a ) a x= or Substituting into the expression for E gives E= 2ke Q ke Qa k Q = e = 3 a2 2( 23 a2 )3 /2 3 a2 Q π e0 a2 x E = π ke σ − 2 x +R 23 .29 ... spheres, with charge Q /2 on each Then they repel like point charges at their centers: 24 .43 F= k e Q2 ke(Q /2) (Q /2) 8.99 × 109 N · m2(60.0 × 10-6 C )2 = = = 2. 00 N 4(L + 2R )2 C 2( 2.01 m )2 (L + R + R)2