# ch44 - 1 Conservation of momentum requires that the gamma...

• Notes
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1. Conservation of momentum requires that the gamma ray particles move in opposite directions with momenta of the same magnitude. Since the magnitude p of the momentum of a gamma ray particle is related to its energy by p = E / c , the particles have the same energy E . Conservation of energy yields m π c 2 = 2 E , where m π is the mass of a neutral pion. The rest energy of a neutral pion is m π c 2 = 135.0 MeV, according to Table 44-4. Hence, E = (135.0 MeV)/2 = 67.5 MeV. We use the result of Problem 83 of Chapter 38 to obtain the wavelength of the gamma rays: 5 6 1240 eV nm 1.84 10 nm 18.4 fm. 67.5 10 eV λ = = × = ×

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2. We establish a ratio, using Eq. 22-4 and Eq. 14-1: ( )( ) ( )( ) 2 11 2 2 31 2 2 2 gravity 2 2 2 2 9 2 2 19 electric 43 6.67 10 N m C 9.11 10 kg 4 9.0 10 N m C 1.60 10 C 2.4 10 . e e F Gm r Gm F ke r e ε 0 × × π = = = × × = × Since F F gravity electric , << we can neglect the gravitational force acting between particles in a bubble chamber.
3. Since the density of water is ρ = 1000 kg/m 3 = 1 kg/L, then the total mass of the pool is ρ ν = 4.32 × 10 5 kg, where ν is the given volume. Now, the fraction of that mass made up by the protons is 10/18 (by counting the protons versus total nucleons in a water molecule). Consequently, if we ignore the effects of neutron decay (neutrons can beta decay into protons) in the interest of making an order-of-magnitude calculation, then the number of particles susceptible to decay via this T 1/2 = 10 32 y half-life is N M m p = = × × = × 10 18 10 18 5 27 32 4 32 10 167 10 144 10 pool kg kg . . . . c h Using Eq. 42-20, we obtain R N T = = × ln . ln . / 2 144 10 2 10 1 1 2 32 32 c h y decay y

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4. By charge conservation, it is clear that reversing the sign of the pion means we must reverse the sign of the muon. In effect, we are replacing the charged particles by their antiparticles. Less obvious is the fact that we should now put a “bar” over the neutrino (something we should also have done for some of the reactions and decays discussed in the previous two chapters, except that we had not yet learned about antiparticles). To understand the “bar” we refer the reader to the discussion in §44-4. The decay of the negative pion is π + µ v . A subscript can be added to the antineutrino to clarify what “type” it is, as discussed in §44-4.
5. From Eq. 37-45, the Lorentz factor would be γ = = × = E mc 2 6 15 10 75000 . . eV 20 eV Solving Eq. 37-8 for the speed, we find γ γ = ¡ = 1 1 1 1 2 2 ( / ) v c v c which implies that the difference between v and c is 2 2 1 1 1 1 1 1 2 c v c c § · § · § · = + ¨ ¸ ¨ ¸ ¨ ¸ ¨ ¸ γ γ © ¹ © ¹ © ¹ " where we use the binomial expansion (see Appendix E) in the last step. Therefore, 2 2 1 1 (299792458m s) 0.0266m s 2.7cm s 2 2(75000) c v c § · § · = = ¨ ¸ ¨ ¸ γ © ¹ © ¹ .

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6. (a) In SI units, K = (2200 MeV)(1.6 × 10 –13 J/MeV) = 3.52 × 10 –10 J.
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