Chapter 46 Solutions

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1 Chapter 46 Solutions 46.1 Assuming that the proton and antiproton are left nearly at rest after they are produced, the energy of the photon E, must be E = E 0 = (938.3 MeV) = MeV = J Thus, E = hf = J f = J J s = Hz λ = c f = m/s Hz = m 46. The minimum energy is released, and hence the minimum frequency photons are produced, when the proton and antiproton are at rest when they annihilate. That is, E = E 0 and K = 0. To conserve momentum, each photon must carry away one-half the energy. Thus, E min = hf min = (E 0) = E 0 = MeV Thus, f min = (938.3 MeV)( J/MeV) J s = Hz λ = c f min = m/s Hz = m *46.3 In γ p + + p, we start with energy.09 GeV we end with energy MeV MeV MeV + K where K is the kinetic energy of the second proton. Conservation of energy gives K = 118 MeV 000 by Harcourt, Inc. All rights reserved.

2 Chapter 46 Solutions Goal Solution A photon with an energy E γ =.09 GeV creates a proton-antiproton pair in which the proton has a kinetic energy of 95.0 MeV. What is the kinetic energy of the antiproton? ( m p c = MeV). G : O : A : An antiproton has the same mass as a proton, so it seems reasonable to expect that both particles will have similar kinetic energies. The total energy of each particle is the sum of its rest energy and its kinetic energy. Conservation of energy requires that the total energy before this pair production event equal the total energy after. E γ = ( E Rp + K p )+ ( E Rp + K p ) The energy of the photon is given as E γ =.09 GeV = MeV. From Table 46., we see that the rest energy of both the proton and the antiproton is E Rp = E Rp = m p c = MeV If the kinetic energy of the proton is observed to be 95.0 MeV, the kinetic energy of the antiproton is K p = E γ E Rp E Rp K p = MeV ( MeV) 95.0 MeV = 118 MeV L : The kinetic energy of the antiproton is slightly (~0%) greater than the proton. The two particles most likely have different shares in momentum of the gamma ray, and therefore will not have equal energies, either. *46.4 The reaction is µ + + e ν + ν muon-lepton number before reaction: ( 1) + (0) = 1 electron-lepon number before reaction: (0) + (1) = 1 Therefore, after the reaction, the muon-lepton number must be 1. Thus, one of the neutrinos must be the anti-neutrino associated with muons, and one of the neutrinos must be the neutrino associated with electrons: ν µ and ν e Then µ + + e ν µ + ν e 46.5 The creation of a virtual Z 0 boson is an energy fluctuation E = ev. It can last no longer than t = h E and move no farther than c( t)= hc 4π E ( )( ms) 4π ( ev) J s = 1 ev = m = ~10 18 m J

3 Chapter 46 Solutions (a) E = (m n m p m e )c From Table A-3, E = ( )931.5 = 0.78 MeV (b) Assuming the neutron at rest, momentum is conserved, p p = p e relativistic energy is conserved, ( m p c ) + p p c + ( m e c ) + p e c = m n c Since p p = p e, ( 938.3) + ( pc) + ( 0.511) + ( pc) = MeV Solving the algebra If p e c = γ m e v e c = 1.19 MeV, then pc = 1.19 MeV γ ve = c 1.19 MeV MeV = x 1 x =.33 where x = v e c Solving, x = (1 x )5.43 and x = v e /c = v e = 0.919c Then m p v p = γ e m e v e : v p = γ e m e v e c m p c = (1.19 MeV)( J/MeV) ( )( ) = m/s = 380 km/s (c) The electron is relativistic, the proton is not. *46.7 The time for a particle traveling with the speed of light to travel a distance of m is t = d v = m m/s = ~ 10 3 s *46.8 With energy MeV, the time that a virtual proton could last is at most t in E t ~ h. The distance it could move is at most c t ~ hc E = ( J s)( m/s) (938.3)( J) = ~ m 000 by Harcourt, Inc. All rights reserved.

4 4 Chapter 46 Solutions 46.9 By Table 46., M π 0 = 135 MeV/c Therefore, Eγ = 67.5 MeV for each photon p = E γ c = 67.5 MeV c and f = E γ h = Hz *46.10 In? + p + n + µ +, charge conservation requires the unknown particle to be neutral. Baryon number conservation requires baryon number = 0. The muon-lepton number of? must be 1. So the unknown particle must be ν µ Ω + Λ 0 + K + K S 0 π + + π (or π 0 + π 0 ) Λ 0 p + π + n p + e + + ν e 46.1 (a) p + p µ + + e L e and L µ (b) π + p p + π + charge (c) p + p p + π + baryon number (d) p + p p + p + n baryon number (e) γ + p n + π 0 charge *46.13 (a) Baryon number and charge are conserved, with values of = and = in both reactions. (b) Strangeness is not conserved in the second reaction.

5 000 by Harcourt, Inc. All rights reserved. Chapter 46 Solutions 5

6 6 Chapter 46 Solutions Baryon number conservation allows the first and forbids the second (a) π µ + ν µ L µ : (b) K+ + µ + ν µ L µ : (c) ν e + p n+ e + + (d) ν e + n p + e L e : L e : (e) ν µ + + n p + µ L µ : (f) µ e + ν e + ν µ L µ : and L e : *46.16 Momentum conservation requires the pions to have equal speeds. The total energy of each is MeV so E = p c + (mc ) gives (48.8 MeV) = (pc) + (139.6 MeV) Solving, pc = 06 MeV = γ mvc = mc v ( ) c 1 vc pc 06 MeV = mc MeV = 1 1 vc [ ] ( vc)= ( vc) and ( vc) =.18 1 ( vc) 3.18( vc) =.18 so v ( ) c = = ( vc ) v c =.18 = 0.88 and v = 0.88c (a) p + π + + π 0 Baryon number is violated: (b) p + + p + p + + p + + π 0 This reaction can occur. (c) p + + p + p + + π + Baryon number is violated: (d) π + µ + + ν µ This reaction can occur. (e) n 0 p + + e + ν e This reaction can occur. (f) π + µ + + n Violates baryon number :

7 Chapter 46 Solutions 7 Violates muon-lepton number : by Harcourt, Inc. All rights reserved.

8 8 Chapter 46 Solutions (a) p e + + γ Baryon number: B 0, so baryon number is violated. (b) From conservation of momentum: p e = p γ Then, for the positron, E e = ( p e c) + E 0, e becomes E e = ( p γ c) + E 0, e = E γ + E 0, e From conservation of energy: E 0, p = E e + E γ or E e = E 0, p E γ so Equating this to the result from above gives E e = E 0, p E γ + E 0, e E 0, p E γ + E γ. = E 0, p E 0,p E γ + E γ, or E γ = E 0, p E 0, e E 0, p MeV = ( ) MeV MeV ( ) ( ) = 469 MeV Thus, E e = E 0, p E γ = MeV 469 MeV = 469 MeV Also, p γ = E γ = 469 MeV/c c and p e = p γ (c) The total energy of the positron is E e = 469 MeV. = 469 MeV/c But, E e = γ E 0, e = E 0, e so 1 ( vc) = E 0, e MeV = = (v / c) 103 E e 469 MeV which yields: v = c *46.19 The relevant conservation laws are: L e = 0, L µ = 0, and L τ = (a) π π + e +? L : L L = 1 e e e and we have a ν e + (b)? + p µ + p+ π Lµ : Lµ Lµ = 1 and we have a ν µ (c) Λ 0 p + µ +? Lµ : Lµ Lµ = 1 and we have a ν µ + + (d) τ µ +? +? Lµ :0 1+ Lµ Lµ = 1 and we have a ν µ Lτ : Lτ Lτ = 1 and we have a ν τ Conclusion for (d): L µ = 1 for one particle, and L τ = 1 for the other particle. We have ν µ and ν τ The ρ 0 + π + π decay must occur via the strong interaction. + The K 0 S π + π decay must occur via the weak interaction.

9 Chapter 46 Solutions (a) Λ 0 p + π Strangeness: (strangeness is not conserved ) (b) π + p Λ 0 + K 0 Strangeness: (0 = 0 and strangeness is conserved ) (c) p + p Λ 0 + Λ 0 Strangeness: (0 = 0 and strangeness is conserved ) (d) π + p π + Σ + Strangeness: (0 1: strangeness is not conserved ) (e) Ξ Λ 0 + π Strangeness: ( 1 so strangeness is not conserved ) (f) Ξ 0 p + π Strangeness: ( 0 so strangeness is not conserved ) 46. (a) µ e + γ L e : , and L µ : 1 0 (b) n p + e + ν e L e : (c) Λ 0 p + π 0 Strangeness: , and charge: (d) p e + + π 0 Baryon number: (e) Ξ 0 n + π 0 Strangeness: *46.3 (a) π + p η violates conservation of baryon number as not allowed (b) K + n Λ 0 + π Baryon number = Charge = Strangeness, Lepton number, 0 0 The interaction may occur via the strong interaction since all are conserved. (c) K π + π 0 Strangeness, Baryon number, 0 0 Lepton number, 0 0 Charge, 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 0 Baryon number, Lepton number, 0 0 Charge, Strangeness, May occur by weak interaction, but not by strong or electromagnetic. (e) η γ Baryon number, 0 0 Lepton number, 0 0 Charge, 0 0 Strangeness, by Harcourt, Inc. All rights reserved.

10 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. 0 *46.4 (a) Ξ Λ + µ + ν µ Baryon number: Charge: L e : L µ : L τ : Strangeness: Conserved quantities are: B, charge, Le, and Lτ (b) K 0 S π 0 Baryon number: 0 0 Charge: 0 0 L e : 0 0 L µ : 0 0 L τ : 0 0 Strangeness: +1 0 Conserved quantities are: B, charge, L, L, and L e µ τ (c) K + p Σ 0 + n Baryon number: Charge: L e : L µ : L τ : Strangeness: Conserved quantities are: S, charge, L, L, and L e µ τ (d) Σ0 Λ 0 + γ Baryon number: Charge: 0 0 L e : L µ : L τ : Strangeness: Conserved quantities are: B, S, charge, L, L, and L e µ τ + + (e) e + e µ + µ Baryon number: Charge: L e : L µ : L τ : Strangeness: Conserved quantities are: B, S, charge, L, L, and L e µ τ 0 (f) p+ n Λ + Σ Baryon number: Charge: L e : L µ : L τ : Strangeness: Conserved quantities are: B, S, charge, L, L, and L e µ τ

11 Chapter 46 Solutions 11 *46.5 (a) K + p? + p The strong interaction conserves everything. Baryon number, B + 1 so B = 0 Charge, Q + 1 so Q =+1 Lepton numbers, L + 0 so Le L = L = Strangeness, S + 0 so S = 1 µ τ 0 The conclusion is that the particle must be positively charged, a non-baryon, with strangeness of +1. Of particles in Table 46., 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, +1 B + 0 so B = 1 Charge, 1 Q 1 so Q = 0 Lepton numbers, 0 L + 0 so L e = L µ = L τ = 0 Strangeness, 3 S + 0 so S = 1: S = The particle must be a neutral baryon with strangeness of. Thus, it is the Ξ 0. (c) + + K? + µ + ν µ : Baryon number, 0 B so B = 0 Charge, +1 Q so Q = 0 Lepton Numbers Le, 0 Le so L e = 0 Lµ, 0 Lµ 1+ 1 so L µ = 0 Lτ, 0 Lτ so L τ = 0 Strangeness: 1 S so S =±1 (for weak interaction): S = 0 The particle must be a neutral meson with strangeness = 0 π 0. *46.6 (a) proton u u d total strangeness baryon number 1 1/3 1/3 1/3 1 charge e e/3 e/3 e/3 e (b) neutron u d d total strangeness baryon number 1 1/3 1/3 1/3 1 charge 0 e/3 e/3 e/3 0 *46.7 (a) The number of protons N p = 1000 g molecules 18.0 g 10 protons molecule = protons 000 by Harcourt, Inc. All rights reserved.

12 1 Chapter 46 Solutions and there are N n = 1000 g ( ) molecules 18.0 g 8 neutrons molecule = neutrons So there are for electric neutrality The up quarks have number and there are electrons = up quarks = down quarks (b) Model yourself as 65 kg of water. Then you contain ~ 10 8 electrons ~ 10 9 up quarks ~ 10 9 down quarks Only these fundamental particles form your body. You have no strangeness, charm, topness or bottomness (a) K 0 d s total strangeness baryon number 0 1/3 1/3 0 charge 0 e/3 e/3 0 (b) Λ 0 u d s total strangeness baryon number 1 1/3 1/3 1/3 1 charge 0 e/3 e/3 e/ Quark composition of proton = uud and of neutron = udd. Thus, if we neglect binding energies, we may write m p = m u + m d (1) and m n = m u + m d () Solving simultaneously, we find m u = 1 3 (m p m n ) = 1 3 [ ( MeV/ c ) MeV/ c ] = 31 MeV/c and from either (1) or (), m d = 314 MeV/c *46.30 In the first reaction, π + + p K 0 Λ 0, the quarks in the particles are: ud + uud ds + uds. There is a net of 1 up quark both before and after the reaction, a net of down quarks both

13 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. 0 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 1 up and down quarks before the reaction but a net of 1 up, 3 down, and 1 anti-strange quark after the reaction. Thus, the reaction does not conserve the net number of each type of quark (a) π p Κ + Λ In terms of constituent quarks: ud + uud ds + uds. up quarks: , or 1 1 down quarks: , or strange quarks: , or 0 0 (b) π + + p Κ + + Σ + ud + uud us + uus up quarks: , or 3 3 down quarks: , or 0 0 strange quarks: , or (c) Κ + p Κ + Κ + Ω us + uud us + ds + sss up quarks: , or 1 1 down quarks: , or 1 1 strange quarks: , or 1 1 (d) p + p K 0 + 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: + = ? (has 1 u quark) down quarks: = ? (has 1 d quark) strange quarks: = ? (has 1 s quark) quark composite = uds = Λ 0 or Σ Σ p Σ + γ + X dds + uud uds + 0 +? The left side has a net 3d, u and 1s. The right-hand side has 1d, 1u, and 1s leaving d and 1u missing. The unknown particle is a neutron, udd. Baryon and strangeness numbers are conserved. 000 by Harcourt, Inc. All rights reserved.

14 14 Chapter 46 Solutions *46.33 Compare the given quark states to the entries in Tables 46.4 and (a) suu = Σ + (b) ud = π (c) sd = Κ 0 (d) ssd = Ξ *46.34 (a) u ud : charge = ( 3 e )+( 3 e )+ ( 3 1 e )= e. This is the antiproton. (b) u d d : charge = ( 3 e )+ ( 3 1 e )+ ( 3 1 e )= 0. This is the antineutron. *46.35 Section 39.4 says f observer = f source 1 + v a c 1 v a c The velocity of approach, v a, is the negative of the velocity of mutual recession: v a =v. Then, c c vc = 1 λ λ 1 + vc + and λ = λ 1 vc 1 vc v = HR (Equation 46.7) H = ( m/s) ly (a) v ( ly) = m/s λ' = λ 1 + v/ c 1 v/ c = 590( ) = nm (b) v ( ly) = m/s λ' = (c) v ( ly) = m/s λ' = = 597 nm = 661 nm (a) λ' λ 650 nm = 434 nm = 1.50 = 1 + v/ c 1 v/ c 1 + v/c 1 v/c =.4 v = 0.383c, 38.3% the speed of light (b) Equation 46.7, v = HR R = v H = (0.383)( ) ( ) = light years

15 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 9 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) λ 650 nm 1 + vc = = = λ 434 nm 1 vc or squared, 1 + vc 1 vc =.43 Therefore, v = 0.383c or 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, R = v H = ( ) ( m/s ly) ( ) m/s = ly 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 1 v / c = (Z + 1)λ n 1 + v/c 1 v/c = ( Z + 1) 1 + v c = ( Z + 1) v c ( Z + 1) v ( Z c + Z + ) = Z + Z v = c Z + Z Z + Z + (b) R = v H = c H Z + Z Z + Z by Harcourt, Inc. All rights reserved.

16 16 Chapter 46 Solutions *46.39 The density of the Universe is ρ = 1. 0ρ c = 1.0( 3H 8π G). Consider a remote galaxy at distance r. The mass interior to the sphere below it is 4π r3 3H 4π r 3 M = ρ 3 = π G 3 = 0.600H r 3 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: 1 mv GmM = 0 GmM r R so 1 mh r Gm r 0.600H r 3 G = 0 Gm R 0.600H r 3 G r = R so R = 6.00r The Universe will expand by a factor of 6.00 from its current dimensions. ( ) *46.40 (a) k B T m p c mc p MeV so T = k JK B ( ) MeV J ~10 13 K (b) kt B mc e ( ) mc e MeV T = k JK B ( ) MeV J ~10 10 K *46.41 (a) Wien s law: λ max T = m K m K m K 3 Thus, λ max = = = m = T 73. K 1.06 mm (b) This is a microwave. *46.4 (a) L = hg c 3 = J s ( )( N m kg ) ( ms) 3 = m (b) This time is given as T = L c = m ms 44 = s, which is approximately equal to the ultra-hot epoch.

17 Chapter 46 Solutions (a) E t h, and t r c = m ms = s E h t = J s s m = E c MeV c ~10 MeV c = MeV J = MeV J (b) From Table 46., mc π = MeV, a pi-meson *46.44 (a) π + p Σ + + π 0 is forbidden by charge conservation (b) µ π + ν e 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)( ) = W Over 10 9 years, the Sun emits J in neutrinos. This represents an annihilated mass mc = J m = kg About 1 part in 50,000,000 of the Sun's mass, over 10 9 years, has been lost to neutrinos. 000 by Harcourt, Inc. All rights reserved.

18 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 10 9 years due to the radiation of neutrinos. (The mass of the Sun is kg. The Earth-Sun distance is m.) G : O : A : 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%). 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. Since the neutrino flux from the Sun reaching the Earth is 0.4 W/m, 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) ( m) = W In a period of 10 9 yr, the Sun emits a total energy of ( J/s) ( 10 9 yr) ( s / yr)= J in the form of neutrinos. This energy corresponds to an annihilated mass of E = m ν c = J so m ν = kg Since the Sun has a mass of about 10 3 kg, this corresponds to a loss of only about 1 part in of the Sun's mass over 10 9 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! 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 M p c + M π c + M X c = ( M p c + K p )+ ( M p c + K p ) M X c = M p c + K p M π c = MeV + ( 70.4 MeV) MeV = MeV X must be a neutral baryon of rest energy MeV. Thus X is a neutron.

19 Chapter 46 Solutions 19 *46.47 We find the number N of neutrinos: J = N(6 MeV) = N( J) The intensity at our location is N = neutrinos N A = N 4πr = ly 4π ( ly) ( m/s)( s) = /m The number passing through a body presenting 5000 cm = 0.50 m is then m ( )= or ~ m *46.48 By relativistic energy conservation, E γ + m e c = 3m e c 1 v / c (1) By relativistic momentum conservation, E γ c = 3m e v () 1 v / c Dividing () by (1), X = E γ E γ + m e c = v c Subtracting () from (1), m e c = 3m e c 3m e c X 1 X 1 X Solving, 1 = 3 3X 1 X and X = 4 5 so E γ = 4m e c =.04 MeV m Λ c = MeV Λ 0 p + π m p c = MeV mc π = MeV The difference between starting mass-energy and final mass-energy is the kinetic energy of the products. K p + K π = 37.7 MeV and p p = p π = p Applying conservation of relativistic energy, ( ) + pc ( ) pc = 7 MeV Solving the algebra yields p π c = p p c = MeV Then, K p = (m p c ) + (100.4) m p c = 5.35 MeV K π = (139.6) + (100.4) = 3.3 MeV 000 by Harcourt, Inc. All rights reserved.

20 0 Chapter 46 Solutions Momentum of proton is qbr = ( C) ( 0.50 kg / C s) ( 1.33 m) 0 kg m p p = s Therefore, p p = 99.8 MeV c 11 kg m cp p = s = J = 99.8 MeV The total energy of the proton is E p = E 0 + (cp) = (938.3) + (99.8) = 944 MeV For pion, the momentum qbr is the same (as it must be from conservation of momentum in a -particle decay). p π = 99.8 MeV c E 0π = MeV E π = E 0 + (cp) = (139.6) + (99.8) = 17 MeV Thus, E Total after = E Total before = Rest Energy Rest Energy of unknown particle = 944 MeV + 17 MeV = 1116 MeV (This is a Λ 0 particle!) Mass = 1116 MeV/c Σ 0 Λ 0 +γ From Table 46., m Σ = MeV/c and mλ = MeV c Conservation of energy requires E0, Σ = ( E0, Λ + KΛ)+ E γ, p Λ or MeV = MeV + m + E Λ γ Momentum conservation gives p p Λ = γ, so the last result may be written as p γ MeV = MeV + m + E Λ pc γ or MeV = MeV + m c + E Λ γ γ Recognizing that mλc = MeV and pc γ = Eγ, we now have MeV MeV γ = + ( ) + E MeV E γ Solving this quadratic equation, E γ 74.4 MeV

21 Chapter 46 Solutions 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 ( mpc + Kp)= mpc + mnc + m π c so the kinetic energy of each of the incident protons is K p ( ) mc n + mc π mc p MeV = = = 70.4 MeV Time-dilated lifetime: T = γ T 0 = s s = = s 1 v /c 1 (0.960) distance = (0.960)( m/s)( s) = 9.6 cm π µ + ν µ π µ ν From the conservation laws, mc = MeV = E + E [1] and pµ = pν, Eν = pνc µ µ ν Thus, E p c MeV p c MeV = ( ) + ( ) = ( ) + ( ) or E µ E ν = ( MeV) [] Since Eµ + Eν = MeV [1] ( µ ν )( µ ν)= ( ) and E + E E E MeV [] then E E µ ν MeV MeV = ( ) = [3] Subtracting [3] from [1], E ν = 594. MeV and E ν = 9. 7 MeV 000 by Harcourt, Inc. All rights reserved.

22 Chapter 46 Solutions *46.55 The expression e Ek B T de gives the fraction of the photons that have energy between E and E + de. The fraction that have energy between E and infinity is E 0 e EkBT de = e EkBT de Ek e BT E Ek e BT 0 ( de k B T) = de k B T ( ) e Ek BT e Ek BT E = ee k B T 0 We require T when this fraction has a value of (i.e., 1.00%) and E = ev = J Thus, ( ) ( ) = J J K T e or ln ( 19 J )= ( JK)T 10 4 K =1.16 T giving T = K (a) 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 +. (a) (b) (a) The mediator of this weak interaction is a Z 0 boson. (b) The Feynman diagram shows a down quark and its antiparticle annihilating each other. They can produce a particle carrying energy, momentum, and angular momentum, but zero charge, zero baryon number, and, it may be, no color charge. In this case the product particle is a photon. For conservation of both energy and momentum, we would expect two photons; but (a) (b) 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 P Depending on the color charges of the d and d quarks,

23 Chapter 46 Solutions 3 the ephemeral particle could also be a gluon, as suggested in the discussion of Figure 46.13(b). 000 by Harcourt, Inc. All rights reserved.

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