Unformatted text preview: 1190 CHAPTER 45 QUARKS, LEPTONS, AND THE BIG BANG CHAPTER 45 Answer to Checkpoint Questions 1. 2. 3. (a) the muon family; (b) a particle; (c) L (b) and (e) (c) Answer to Questions 1. 2. 3. 4. 5. 6. 7. 8. 9. 10. 11. 12. 13. (d) into the + pion, whose track curves downward at the left (b), (c), and (d) (a), (b), (c), and (d) (f) (c) and (f) (c) 1d, 2e, 3a, 4b, 5c (a) lepton; (b) antiparticle; (c) fermion; (d) yes 1b, 2c, 3d, 4e, 5a (b), (f), (c), (d), (a), (g), (e) (a) 0; (b) +1; (c) 1; (d) +1; (e) 1 CHAPTER 45 QUARKS, LEPTONS, AND THE BIG BANG 1191 Solutions to Exercises & Problems 1E The dierence in mass is E 60 m = 2 = (33:9 MeV)(1: 10810 2J/MeV) = 6:03 10 c (3:00 m/s)
13
2E 29 kg : Conservation of momentum requires that the gammaray particles move in opposite directions with momenta of the same magnitude. Since the magnitude p of the momentum of a gammaray particle is related to its energy by p = E=c, the particles have the same energy E . Conservation of energy yields m c2 = 2E , where m is the mass of a neutral pion. According to Table 454 the rest energy of a neutral pion is m c2 = 135:0 MeV. Hence E = (135:0 MeV)=2 = 67:5 MeV. Use the result of 1E of Chapter 39 to obtain the wavelength of the gammaray waves: 1240 nm = 67:5 eV 6 eV = 1:84 10 5 nm = 18:4 fm : 10
3E Fgrav = Gm2 =r2 = 40 Gm2 e e 2 =40 r2 2 Fele e e 11 N m2 = C2 )(9:11 10 31 kg)2 67 = (6:(9:0 10 9 N m2 =C2 )(1:60 10 19 C)2 10 = 2:4 10 43 :
Since Fgrav Fele we can neglect the gravitational force acting between particles in a bubble chamber.
4E Since the antiparticle of + is and that of is , the decay scheme of is ! +. The energy released would be twice the rest energy of the Earth, or E = 2Me c2 = 2(5:98 1024 kg)(3:00 108 m/s)2 = 1:08 1042 J. 5E 1192 CHAPTER 45 QUARKS, LEPTONS, AND THE BIG BANG 6P Thus v2 1=2 = 1 + K = 1 + 80 MeV = 1:593 ; = 1 c2 mc2 135 MeV p p we get v = c 1 2 = 1 (1:593) 2 (3:00 108 m/s) = 2:335 108 m/s. Since Lmax = v = v
0 = (2:335 108 m/s)(1:593)(8:3 10 = 3:1 10 8 m = 31 nm :
p 17 s) 7P Use v = c 1 2 and = 1 + K=mc2 to obtain
2 s v = c v = c 41 1
2 K 1 + mc2
1 2 3 5 s = (3:00 108 m/s) 41 : MeV 1 + 1250 eV : 2 3 5 = 0:27 m/s : Note that, since the quantity x = (1 + 1:5 MeV=20 eV) 2 1:78 10 10 1, your calculator may not give you the correct numerical answer to v due to errors in rounding p o digits. Try use the approximation 1 x 1 1 x. 2 The mass of the water in the pool is m = V = (1:00 103 kg/m3 )(114; 000 gal)(231 in.3 =gal)(1:639 10 5 m3 =in.3 ) = 4:316 105 kg. The number of protons in this much water is 105 N = (4:316 16:7 kg)(10=18) = 1:44 1032 : kg Thus the number of proton decay in one year is 32 n = Rt = Nt = Ntln 2 = (1:44 10 32ln 2)(1:0 y) 1 : 10 y
9P 8P (a) From K = mc2 mc2 we get = 1 + K=mc2 . Also, v = 1 2 c. Thus p K 2 K p =
mv = mc 1 + mc2 1 1 + mc2 s 13 J/MeV) 60 2200 = (1784 MeV)(1: 10810 1 + 1784 1 3:00 m/s 18 kg m/s : = 1:90 10 s 1 + 2200 1784 2 CHAPTER 45 QUARKS, LEPTONS, AND THE BIG BANG 1193 (b) p :90 10 18 kg R = mv = qB = (1:160 10 19 C)(1m/sT) = 9:90 m : qB :20 Table 454 gives the rest energy of each pion as 139:6 MeV. The magnitude of the momentum of each pion is p = (358:3 MeV)=c. Use the relativistic relationship between energy and momentum to nd the total energy of each pion: 10P E = (p c)2 + (m c2 )2 = (358:3 MeV)2 + (139:6 MeV)2 = 384:5 MeV :
Conservation of energy then yields m c2 = 2E = 2(384:5 MeV) = 769 MeV. Use the relativistic energymomentum relationship for both particles 1 and 2: (E1 + K1 )2 = 2 2 p2 c2 + E1 , (E2 + K2 )2 = p2 c2 + E2 . Note that p2 = p2 from momentum conservation. 1 2 1 2 Eliminate p1 and p2 to obtain
2 2 (E1 + K1 )2 (E2 + K2 )2 = E1 E2 :
11P p p Now write down the conservation of energy equation: E0 E1 E2 = K1 + K2 . Solve for K2 : K2 = E0 E1 E2 K1 . Substitute this into the equation above:
2 2 (E1 + K1 )2 (E0 E1 K1 )2 = E1 E2 : Some simple algebra then leads to 1 2 K1 = 2E (E0 E1 )2 E2 : 0 Consider the decay scheme p ! e+ + . (a) Qi = qp = e = qe+ = Qf . (b) The conservation of energy equation Ep = Ee+ + E can be satised with the proper choice of E , a continuous variable. (c) Again, the conservation of linear momentum equation pp = pe + p may be satised with the proper choice of p . (d) The initial spin angular momentum is h=2. If the angular momentum of e+ and are antiparallel to each other then the magnitude of the nal angular momentum is jh=2 hj = h=2, the same as the initial value. 12E 1194 CHAPTER 45 QUARKS, LEPTONS, AND THE BIG BANG 13E (a) The conservation laws considered so far are associated with energy, momentum, angular momentum, charge, and baryon number. The rest energy of the muon is 105:7 MeV, the rest energy of the electron is 0:511 MeV, and rest energy of the neutrino is zero. Thus the total rest energy before the decay is greater than the total rest energy after. The excess energy can be carried away as the kinetic energies of the decay products and energy can be conserved. Momentum is conserved if the electron and neutrino move away from the decay in opposite directions with equal magnitudes of momenta. Since the orbital angular momentum is zero, we consider only spin angular momentum. All the particles have spin h=2. The total angular momentum after the decay must be either h (if the spins are aligned) or zero (if the spins are antialigned). Since the spin before the decay is h=2, angular momentum cannot be conserved. The muon has charge e, the electron has charge e, and the neutrino has charge zero, so the total charge before the decay is e and the total charge after is e. Charge is conserved. All particles have baryon number zero, so baryon number is conserved. The only conservation law not obeyed is the conservation of angular momentum. (b) Analyze the decay in the same way. You should nd that only charge is not conserved. (c) Here you should nd that energy and angular momentum cannot be conserved. (a) Follow the decay schemes shown in the problem statement. You will nd that the decay schemes are equivalent to the following:
14P A+ ! 2e+ + e + 4 + 4 : 2
Thus the decay products consist of four neutrinos and four antineutrinos, an electronpositron pair, plus an extra positron. Since the electronpositron pair is unstable and will decay into gammaray photons, the only stable, massive decay product left is a positron (e+ ). (b) A+ is a boson, since both 0 and + are bosons. It is also a meson, since BA+ = 2 2 B0 + B+ = 2B+ + B = 0.
15E For purposes of deducing the properties of the antineutron, cancel a proton from each side of the reaction and write the equivalent reaction as n + ! p + : Particle properties can be found in Tables 453 and 4. The pion and proton each have charge +e, so the antineutron must be neutral. The pion has baryon number zero (it is a meson) and the proton has baryon number +1, so the baryon number of the antineutron must be 1. The pion and the proton each have strangeness zero, so the strangeness of the antineutron must also be zero. In summary, Q = 0, B = 1, and S = 0 for the antineutron. CHAPTER 45 QUARKS, LEPTONS, AND THE BIG BANG 1195 16E Refer to Tables 453 and 454 for the strangeness of the particles involved. (a) The strangeness of K0 is +1, while it is zero for both + and . So the strangeness is not conserved here and therefore the decay K0 ! + + is not via strong interaction. (b) The strangeness of either sides is 1 so the decay is via strong interaction. (c) The strangeness or 0 is 1 while that of p + is zero so the decay is not via strong interaction. (d) The strangeness of either side is 1 so the decay is via strong interaction. 17E (a) See the solution to 13E for the quantities to be considered, but add strangeness to the list. The lambda has a rest energy of 1115:6 MeV, the proton has a rest energy of 938:3 MeV, and the kaon has a rest energy of 493:7 MeV. The rest energy before the decay is less than the total rest energy after, so energy cannot be conserved. Momentum can be conserved. The lambda and proton each have spin h=2 and the kaon has spin zero, so angular momentum can be conserved. The lambda has charge zero, the proton has charge +e, and the kaon has charge e, so charge is conserved. The lambda and proton each have baryon number +1 and the kaon has baryon number zero, so baryon number is conserved. The lambda and kaon each have strangeness 1 and the proton has strangeness zero, so strangeness is conserved. Only energy cannot be conserved. (b) The omega has a rest energy of 1680 MeV, the sigma has a rest energy of 1197:3 MeV, and the pion has a rest energy of 135 MeV. The rest energy before the decay is greater than the total rest energy after, so energy can be conserved. Momentum can be conserved. The omega and sigma each have spin h=2 and the pion has spin zero, so angular momentum can be conserved. The omega has charge e, the sigma has charge e, and the pion has charge zero, so charge is conserved. The omega and sigma have baryon number +1 and the pion has baryon number , so baryon number is conserved. The omega has strangeness 3, the sigma has strangeness 1, and the pion has strangeness zero, so strangeness is not conserved. (c) The kaon and proton can bring kinetic energy to the reaction, so energy can be conserved even though the total rest energy after the collision is greater than the total rest energy before. Momentum can be conserved. The proton and lambda each have spin h=2 and the kaon and pion each have spin zero, so angular momentum can be conserved. The kaon has charge e, the proton has charge +e, the lambda has charge zero, and the pion has charge +e, so charge is not conserved. The proton and lambda each have baryon number +1 and the kaon and pion each have baryon number zero, so baryon number is conserved. The kaon has strangeness 1, the proton and pion each have strangeness zero, and the lambda has strangeness 1, so strangeness is conserved. Only charge is not conserved. 1196 CHAPTER 45 QUARKS, LEPTONS, AND THE BIG BANG 18E (a) E = m c2 = (m+ + mK+ m+ mp )c2 = 1189:4 MeV + 493:7 MeV 139:6 MeV 938:3 MeV = 605:2 MeV : (b) E = m c2 = (m0 + m0 mK mp )c2 = 1115:6 MeV + 135:0 MeV 493:7 MeV 938:3 MeV = 181:4 MeV : 19P Kf = m c2 + Ki = (m m mn )c2 + Ki = 1197:3 MeV 139:6 MeV 939:6 MeV + 220 MeV = 338 MeV : (a) As far as the conservation laws are concerned, we may cancel a proton from each side of the reaction equation and write the reaction as p ! 0 + x. Since the proton and the lambda each have a spin angular momentum of h=2, the spin angular momentum of x must be either zero or h. Since the proton has charge +e and the lambda is neutral, x must have charge +e. Since the proton and the lambda each have a baryon number of +1, the baryon number of x is zero. Since the strangeness of the proton is zero and the strangeness of the lambda is 1, the strangeness of x is +1. Take the unknown particle to be a spin zero meson with a charge of +e and a strangeness of +1. Look at Table 454 to identify it as a K+ particle. (b) Similar analysis tells us that x is a spin 1 antibaryon (B = 1) with charge and 2 strangeness both zero. Inspection of Table 453 reveals it is an antineutron. (c) Here x is a spin0 (or spin1) meson with charge zero and strangeness 1. According to Table 454 it could be a K 0 particle. 20P CHAPTER 45 QUARKS, LEPTONS, AND THE BIG BANG 1197 21P n p Y = +1  0  + Y =0 Tz = 1 0 Tz = 0 Tz = +1 Y = 1 22P (a) E = (m0 mp m )c2 = 1115:6 MeV 938:3 MeV 139:6 MeV = 37:7 MeV : (b) Use the formula obtained in 11P: 1 2 Kp = K1 = 2E [(E0 E1 )2 E2 ] 0 2 2 938:3 = (1115:6 MeV 2(1115MeV) (139:6 MeV) = 5:35 MeV : :6 MeV) (c) K = K2 = E0 E1 E2 K1 = 1115:6 MeV 938:3 MeV 139:6 MeV 5:35 MeV = 32:4 MeV : 23E (a) Since p=uud we have p = uud. (b) Since n=uud we have n = udd. 1198 CHAPTER 45 QUARKS, LEPTONS, AND THE BIG BANG (a) The combination ddu has a total charge of ( 1 1 + 2 ) = 0, and a total strangeness 3 3 3 of zero. From Table 453 we nd it to be a neutrons (n). (b) For the combination uus we have Q = + 2 + 2 1 = 1 and S = 0 + 0 1 = 1. This is 3 3 3 a + . (c) For the combination ssd we have Q = 1 1 1 = 1 and S = 1 1 + 0 = 2. 3 3 3 This is a . (a) and (b) From Tables 453 and 455 you can easily check that the combinations sud and uss give the correct Q and S quantum numbers of 0 and 0 , respectively.
26E 25E 24E (a) Look at the rst three lines of Table 455. Since the particle is a baryon it must consist of three quarks. To obtain a strangeness of 2, two of them must be s quarks. Each of these has a charge of e=3, so the sum of their charges is 2e=3. To obtain a total charge of e, the charge on the third quark must be 5e=3. There is no quark with this charge, so the particle cannot be constructed. In fact, such a particle has never been observed. (b) Again the particle consists of three quarks (and no antiquarks). To obtain a strangeness of zero, none of them may be s quarks. We must nd a combination of three u and d quarks with a total charge of 2e. The only such combination consists of three u quarks. 27E  0 + ++ S=0  0 + S = 1  0 S = 2 Q = 1 Q= 0 Q = +1 Q = +2 S = 3 CHAPTER 45 QUARKS, LEPTONS, AND THE BIG BANG 1199 We see from the diagram above that an inverted equilateral triangle is formed. Since only the strange quark (s) has nonzero strangeness, in order to obtain S = 1 we need to combine s with another quark listed in Table 455. But none of the quarks in the table has Q = + 3 , which is what would be needed to get the total charge of 1 (note that 4 Qs = 1 ). So a meson with S = 1 and Q = +1 cannot be formed with the quarks in 3 Table 455. Similarly you can show that since no quark has Q = 4 there cannot be a 3 meson with S = +1 and Q = 1.
29P 28P From = 1 + K=mc2 and v = c = c 1 2 , we get
s p v=c 1
So for 0
s K 1 + mc2 2 :
2 v = (3:00 108 m/s)
and for 0
s 1 1000 MeV 1 + 1385 MeV = 2:442 108 m/s ;
2 v0 = (3:00 108 m/s) 1
Thus 0 moves faster than 0 by 1000 1 + 1192:5MeV MeV = 2:517 108 m/s : v = v0 v = (2:517 2:442)(108 m/s) = 7:54 106 m/s :
30E Let v = Hr = c to obtain c 108 r = H = 3:00 mm=m/s = 1:2 1010 ly : 24:54 s ly 31E Apply the formula for the Doppler shift in wavelength: = v ; c 1200 CHAPTER 45 QUARKS, LEPTONS, AND THE BIG BANG where v is the recessional speed of the galaxy. Use Hubble's law to nd the recessional speed: v = Hr, where H is the Hubble constant and r is the distance to the galaxy. Thus v = (24:54 mm/s ly)(2:40 108 ly) = 5:89 106 m/s and 5 89 106 = v = 3::00 108 m/s (656:3 nm) = 12:9 nm : c m/s Since the galaxy is receding the observed wavelength is longer than the wavelength in the rest frame of the galaxy. Its value is 656:3 nm + 12:9 nm = 669 nm. First, nd the speed of the receding galaxy from Eq. 3825:
2 1 2 1 = 1 + (f=f0 )2 = 1 + (0 =)2 (f=f0 ) (0 =) 2 1 = 1 + (590::0 nm=602::0 nm)2 = 0:02013 ; (590 0 nm=602 0 nm)
32E where we used f = c= and f0 = c=0 . Then from Eq. 4515 v :00 8 r = H = c = (0:02013)(3mm/s10 m/s) = 2:46 108 ly : H 24:54 ly
(Note: if you used the classical Doppler shift formula instead of the relativistic version in Eq. 3825, you would get r = 2:49 108 ly, which is reasonably clsoe to the relativistic value we obtained above. This is expected since 0:02 1:) (a) From f = c= and Eq. 3825 we get
33P 0 = 1 + = (0 + ) 1 + : 1 1 Divide both sides by 0 to obtain 1 = (1 + z ) 1 + : 1 Solve for :
2 z (1 + z )2 = (1 + z)2 + 1 = z2z+ +z2+ 2 : 1 2
s s s CHAPTER 45 QUARKS, LEPTONS, AND THE BIG BANG 1201 (b) Now z = 4:43 so (c) 2 = (4(4:43)+ + 2(4:43) 2 = 0:934 : :43)2 2(4:43) + 8 v r = H = c = (0:943)(3:00 10lym/s) = 1:15 1010 ly : H 24:54 mm/s (a) Let v(r) = Hr ve = 2GM=r, we get M=r3 H 2 =2G. Thus 3H 2 M = 4r3 =3 = 43 M 8G : r3 (b) The density of unit expressed in Hatoms/m3 is equivalent to 0 = mH =m3 = 1:67 10 27 kg/m3 . Thus in terms of 0 we have
3 mm/s 2 00 ly=9:460 1015 2 3H 2 = 8G = 3(24:854(6:67 ly) (1:N m2 =kg2 )(1:67 m) (Hatoms/m ) 10 11 10 27 kg/m3 ) 34P p = 7:21 Hatoms/m3 : 35P (a) From Eq. 4121 we know that n2 =n1 = e E=kT . Solve for E : 25% E = kT ln n1 = (8:62 10 5 eV/K)(2:7 K) ln 1 25% n2 = 2:56 10 4 eV = 256 eV : (b) hc = E = 2:1240 eV nm = 4:84 mm : 56 10 4 eV 36P From Fgrav = GMm=r2 = mv2 =r we nd M / v2 . Thus the mass of the Sun would be 2 2 0 = vMercury Ms = 47:9 km/s Ms = 102Ms : Ms vPluto 4:74 km/s 1202 CHAPTER 45 QUARKS, LEPTONS, AND THE BIG BANG (a) The mass M within the Earth's orbit is used to calculate the gravitational force on the Earth. If r is the radius of the orbit, R is the radius of the new Sun, and Ms is the mass of the Sun, then 37P r 3 M = 1:50 1011 m 3 (1:99 1030 kg) = 3:27 1025 kg : M= R s 5:90 1012 m The gravitational force on the Earth is given by GMMe =r2 , where Me is the mass of the Earth and G is the universal gravitational constant. Since the centripetal acceleration is given by v2 =r, where v is the speed of the Earth, GMMe =r2 = Me v2 =r and GM = (6:67 10 11 m3 /s2 kg)(3:27 1025 kg) = 1:22 102 m/s : v= r 1:50 1011 m
(b) The period of revolution is m) T = 2r = 2(1:50 10m/s = 7:73 109 s : 2 v 1:22 10
11 r s This is 246 y.
38P (a) The mass of the portion of the galaxy within the radius r from its center is given by M 0 = (r=R)3 M . Thus from GM 0 m=r2 = mv2 =r (m is the mass of the star) we get v=
(b) In the case M 0 = M so r GM 0 r = GM R r r
r s 3 = r GM : R3 r 3=2 r p T = 2r = 2r GM = 2r : v GM 39E (a) (b) T = 2898 m K = 2898 m K = 2:6 K : 1:1 mm
max max = 2898T m K = 2898 m K = 29 nm : 105 K CHAPTER 45 QUARKS, LEPTONS, AND THE BIG BANG 1203 (a) Substitute = (2898 m K)=T into the result of Exercise 1 of Chapter 43: E = (1240 nm eV)=. First convert units: 2898 m K = 2:898 106 nm K and 1240 nm eV = 1:240 10 3 nm MeV. Hence 40E E = (1:24089810 10nm MeV)T = (4:28 10 2: 6 nm K
3 10 MeV/K) T : (b) The minimum energy required to create an electronpositron pair is twice the rest energy of an electron, or 2(0:511 MeV) = 1:022 MeV. Hence 1 022 MeV T = 4:28 10 E10 MeV/K = 4:28 :10 10 MeV/K = 2:39 109 K : 41 (a) and (b) Use the relativistic relationship between speed and momentum: p = mv = p mv 2 ; 1 (v=c)
which we solve for the speed v: v = u1 u c t v pc 2 + 1 : mc2 1 Here m is the rest mass of a particle. For an antiproton mc2 = 938:3 MeV and pc = 1:19 GeV, so s 1 v=c 1 = 0:785c : (1:19 GeV=938:3 MeV)2 + 1 For a negative pion mc2 = 193:6 MeV and pc = 1:19 GeV, so v=c 1 s 1 = 0:993c : (1:19 GeV=193:6 MeV)2 + 1 (c) Since the speed of the antiprotons is about 0:78c but not over 0:79c, an antiproton will trigger C1. (d) Since the speed of the negative pions exceeds 0:79c, a negative pion will trigger C2. (e) and (f) Use t = d=v, where d = 12 m. For an antiproton 12 m t = 0:785(3:00 108 m/s) = 5:1 10 8 s = 51 ns ; 1204 CHAPTER 45 QUARKS, LEPTONS, AND THE BIG BANG and for a negative pion 12 m t = 0:993(3:00 108 m/s) = 4:0 10 8 s = 40 ns :
42 t = c r r : (b) The detected wavelength 0 is longer then by t due to the expansion of the universe: 0 = + t. Thus = 0 = t = r : (a) During the time interval t the light emitted from galaxy A has traveled a distance ct. Meanwhile the distance between the Earth and the galaxy has expanded from r to r0 = r + rt. Let ct = r0 = r + rt to obtain c r (c) Use the binomial expansion formula (1 x)n = 1 nx + n(n 2! 1)x + 1! to obtain
2 (x2 < 1) = r = r 1 r 1 c r c c r 1 + 1 r + ( 1)( 2) r 2 + = c 1! c 2! c 2 3 r + r + r : c c c (d) When only the rst term in the expansion for = is retained we have r : c = v = Hr c c and compare with the result of (d) to obtain = H . (f) Use the formula = = r=(c r) to solve for r: (e) Set c (3:00 108 m/s)(0 r = (1(=) ) = (24:54 mm/s ly)(1 +:050) = 5:82 108 ly : + = 0:050) CHAPTER 45 QUARKS, LEPTONS, AND THE BIG BANG 1205 (i) The distance is given by r = ct = c(6:10 108 y) = 6:10 108 ly : (j) From the result of part (f) c (3:00 108 m/s)(0 rB = (1(=) ) = (24:54 mm/s ly)(1 +:080) = 9:06 108 ly : + = 0:080) (k) From the formula obtained in part (a) 8 1015 (9 tB = c rB = 3:00 108:06 10(9ly)(9:46 ly)(24m/ly) ly) = 3:09 1016 s ; rB m/s :06 108 :54 mm/s which is equivalent to 9:76 108 y. (l) At the present time the separation between the two galaxies A and B is given by rnow = ctB ctA = c(9:76 108 y 6:098 108 y) = 3:66 109 ly. Since rnow = rthen + rthen t, we get r rthen = 1 +now t 8 ly 3 66 = 1 + (24:54 mm/s ly)(1::927 10 16 s)(1 ly=9:46 1015 m) 10 8 ly : = 3:49 10 (a) Since the rate of expansion of the universe is assumed to be uniform, we have / t. Thus now = tnow = 14 109 y = 4:67 104 : then tthen 3 105 y (b) The wavelength of a photon corresponding to the Lyman series limit, when just emitted 300,000 years ago, was given by hc = 1240 eV nm = (13:6 eV) 1 1 ; 12 12
then then
43 (g) From the result of part (a) 8 ly)(9 46 15 (5 t = c r r = 3:00 108:82 10(24:54 :mm/s10 m/ly) 108 ly) = 1:93 1016 s ; m/s ly)(5:82 which is equivalent to 6:10 108 y. (h) Let r = ct and solve for t: 8 t = r = 5:82 10 ly = 5:82 108 y : c c which gives then = 91:4 nm. Thus from part (a) above the wavelength detected today would be now = (4:67 104 )(91:4 nm) = 4:27 mm : (c) In this case now = 1:1 mm, so the corresponding value of then is given by then = 4:67now104 = 24 nm : ...
View
Full
Document
This note was uploaded on 01/22/2012 for the course PHYS 2101 taught by Professor Grouptest during the Fall '07 term at LSU.
 Fall '07
 GROUPTEST

Click to edit the document details