Unformatted text preview: 23 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 10MeV neutrons by using the classical expressionrather than the relativistic expression for kinetic energy? Solution:
A neutron with rest mass mn = 1.6749288x 1027 kg has a kinetic energy T = (107 eV)(1.602177 x 1019 JjeV) = 1.602177x 1012 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 s1) 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 22 "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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 Spring '06
 CADY
 Kinetic Energy, Mass, Special Relativity

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