hw3_solutions

# hw3_solutions - PHY 251 Fall 2009 homework problem set 3...

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PHY 251 Fall 2009: homework problem set 3, due PHY 251 drop box in room A-129 by noon on Friday, Sep. 25. 1. A proton with a kinetic energy of 10 GeV collides with an antiproton at rest. They annihilate each other, and produce two identical-energy gamma rays (very energetic photons) that come out at angles θ above and below, respectively, the direction of the proton’s motion. Calculate the energy E and momentum p of the gamma rays, and the angle θ . Answer: Conservation of energy gives ( E K + m p c 2 ) + m p c 2 = 2 E γ because we go from a proton with kinetic energy E K = 10 GeV and the mass-energy of the proton plus the antiproton (same mass as proton), to two photons of identical energy E γ . Since m p = 938 . 3 MeV/ c 2 we have E γ = (10 + 0 . 9383 + 0 . 9383) / 2 = 5 . 94 GeV. Note for the proton we have E K = ( γ 1) m p c 2 γ = 1 + E k m p c 2 = 1 + 10 0 . 9383 = 11 . 66 from which we can find β = r 1 1 2 = 0 . 996 . At the same time, conservation of momentum gives p p = 2 p γ cos θ γm p v = 2 E γ c cos θ γβm p c 2 = 2 E γ cos θ cos θ = γβm p c 2 2 E γ θ = cos 1 p γβm p c 2 2 E γ P = cos 1 p 11 . 66 · 0 . 996 · 0 . 9383 2 · 5 . 94 P = cos 1 (0 . 703) = 23 . 5 2. A atom of the radioactive isotope of carbon 14 C (mass 14 . 003 242 amu, where 1 amu=931.494 MeV/ c 2 ) decays into an electron (mass 0 . 000 549 amu) and a singly ionized 14 N atom (mass 14 . 003 074 0 . 000 549 amu). If the 14 C atom were at rest, and no other particles were in- volved, calculate the total energy released by this decay, and the energy and momentum of the electron and 14 N atoms after the decay. Answer: Let’s have m C = 14 . 003 242 amu represent the mass of the 14 C atom before the decay, m N = 14 . 003 074 0 . 000 549 am represent the mass of the 14 N atom minus an elec- tron after the decay, and m e = 0 . 000 549 amu represent the mass of the electron. The total energy released in the decay is given by Δ E c 2 = Δ m = [ m C m N m e ] = [14 . 003 242 (14 . 003 074 0 . 000 549) 0 . 000 549] amu = [14 . 00 3242 14 . 003 074] amu = 0 . 000 168 amu or Δ E = (0 . 000 168 amu ) · ( 931 . 494 MeV /c 2 amu ) · c 2 = 0 . 156 MeV 1

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or 156 keV. Now even if all of that energy went to the electron it would have a value of
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hw3_solutions - PHY 251 Fall 2009 homework problem set 3...

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