hw12_solutions - Physics 325 Homework #12 Solutions 1)...

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Unformatted text preview: Physics 325 Homework #12 Solutions 1) Using ( E,cp x ,cp y ,cp z ) as the Energy-momentum four vector, the dot product gives us the following invariant: E 2- c 2 p 2 = ( mc 2 ) 2 . (I note that the four-vector can be written with the energy term first or last.) Therefore the momentum can be written as: pc = q E 2- ( mc 2 ) 2 a) E = 5 mc 2 ,pc = 5 2- 1 2 ( mc 2 ) (5 mc, 24 mc, , 0) b) E = mc 2 ,p = 0, ( mc 2 , , , 0) c) pc = 3 mc 2 ,E = q p 2 c 2 + ( mc 2 ) 2 = 3 2 + 1 2 mc 2 ( 10 mc 2 , , , 3 mc 2 ) d) E = mc 2 + T = 5 mc 2 ,pc = mc 2 5 2- 1 2 (5 mc 2 , ,- 24 mc 2 , 0) e) E = 10 mc 2 ,pc = 10 2- 1 2 mc 2 = 99 mc 2 p 2 = p 2 x + p 2 y + p 2 z = (1 p x ) 2 + (2 p x ) 2 + (3 p x ) 2 = 99( mc ) 2 p x = q 99 14 mc (10 mc 2 , q 99 14 mc 2 , 2 q 99 14 mc 2 , 3 q 99 14 mc 2 ) 2) The total kinetic energy of the particles produced in the neutron decay is given by T = m n c 2- m p c 2- m e c 2 = 939 . 565 MeV- 938 . 272 MeV- . 511 MeV = 0 . 782 MeV a) The neutrino is assumed massless here, so its momentum is just p = E /c As the four-vector of neutrino is ( E /c,E /c, , 0) and that of the proton is ( m p c, , , 0), then the four-vector of electron is given by ( E e /c,- E /c, , 0), because the momenta of the electron and the neutrino must be equal and opposite. Now E e + E = m e c 2 + T = 1 . 293 MeV, where the kinetic energy, T, is that of the neutrino and electron combined. Therefore, E e = 1 . 293 MeV- E ....
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This note was uploaded on 02/23/2009 for the course PHYS 325 taught by Professor Lamb during the Fall '09 term at University of Illinois at Urbana–Champaign.

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hw12_solutions - Physics 325 Homework #12 Solutions 1)...

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