HW2-3 - 2-3 3 In fission reactors one deals with neutrons...

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Unformatted text preview: 2-3 3. In fission reactors one deals with neutrons having kinetic energiesas high as 10 MeV. How much error is incurred in computing the speed of 10-MeV neutrons by using the classical expressionrather than the relativistic expression for kinetic energy? Solution: A neutron with rest mass mn = 1.6749288x 10-27 kg has a kinetic energy T = (107 eV)(1.602177 x 10-19 JjeV) = 1.602177x 10-12 J. For the neutron mnc2 = 939.5656MeV. Classically: = 4.373993X 107m/s. Relativistically: From the text we have T = mc2 -mOc2 = -7i~g~~ VI -v2/& -moc2, Solving this equation for v yields the relativistic speed Vr Vr = c Substitution then gives 1 -moc [ 2 2) 1/2 I T+moc2 I (P2.4) 4. What speed (m s-1) and kinetic energy (MeV) would a neutron have if its relativistic mass were 10% greater than its rest mass? Solution: We are given (m -mo)/m ==f = 0..1. From Problem 2-2 "2 = V 1 -U2 ~ = V 1 -~ v 11} /. 1 = 0.4167. Thus the neutron's speed is v = 0.4167c= 1.25 X 108 mfg. The kinetic energy can be calculated from July 24, 2002 ...
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