HW2-1 - to give . (b) For an electron with T = 109 eV =...

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PROBLEMS An accelerator increases the total energy of electrons uniformly to 10 GeV over a 3000 m path. That means that at 30 m, 300 m, and 3000 m, the kinetic energy is 108, 109, and 1010 eV, respectively. At each of these distances, compute the velocity, relative to light (v/c), and the mass in atomic mass units. Solution: From Eq. (2.10) in the text T = mc2 -moc2 we obtain m = Tlc2 + mo. (P2.1) From Eq. (2.5) in the text m = mol VI -v2 I c2, which can be solved for v I c
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Unformatted text preview: to give . (b) For an electron with T = 109 eV = 1000 MeV, we similarly obtain m = 1.0741 u and vjc = 0.99999987. (c) For an electron with T = 1010 eV = 104 MeV, we similarly obtain m = 10.736 u and vjc = 0.9999999987. Alternative solution: Use Eq. (P2.4) developed in Problem 2-3, namely 1/2 { [ 2 ] 2 } v-I meC ~ --T+mec2 2-1 July 24, 2002...
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This note was uploaded on 02/01/2010 for the course ECE 4130 taught by Professor Cady during the Spring '06 term at Cornell University (Engineering School).

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