Chapter-44 - Chapter 44 1 The total rest energy of the...

Info icon This preview shows pages 1–3. Sign up to view the full content.

View Full Document Right Arrow Icon
1655 Chapter 44 1. The total rest energy of the electron-positron pair is 2 2 2 2 2(0.511 MeV) 1.022 MeV e e e E m c m c m c = + = = = . With two gamma-ray photons produced in the annihilation process, the wavelength of each photon is (using 1240 eV nm hc = ) 3 6 1240 eV nm 2.43 10 nm 2.43 pm. / 2 0.511 10 eV hc E λ - = = = × = × 2. 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 1240 eV nm hc = 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 = = × = × 3. We establish a ratio, using Eq. 22-4 and Eq. 14-1: ( ( ( 29( 29 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. 4. Since the density of water is ρ = 1000 kg/m 3 = 1 kg/L, then the total mass of the pool is ρ ς = 4.32 × 10 5 kg, where ς 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
Image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
CHAPTER 44 1656 ( 5 pool 32 27 (10/18) 4.32 10 kg (10/18) 1.44 10 . 1.67 10 kg p M N m - × = = = × × 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 5. 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. 6. (a) In SI units, the kinetic energy of the positive tau particle is K = (2200 MeV)(1.6 × 10 –13 J/MeV) = 3.52 × 10 –10 J. Similarly, mc 2 = 2.85 × 10 –10 J for the positive tau. Eq. 37-54 leads to the relativistic momentum: ( 29 ( 29( 29 2 2 2 10 10 10 8 1 1 2 3.52 10 J 2 3.52 10 J 2.85 10 J 2.998 10 m/s p K Kmc c - - - = + = × + × × × which yields p = 1.90 × 10 –18 kg·m/s.
Image of page 2
Image of page 3
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

What students are saying

  • Left Quote Icon

    As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

    Student Picture

    Kiran Temple University Fox School of Business ‘17, Course Hero Intern

  • Left Quote Icon

    I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

    Student Picture

    Dana University of Pennsylvania ‘17, Course Hero Intern

  • Left Quote Icon

    The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

    Student Picture

    Jill Tulane University ‘16, Course Hero Intern