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Unformatted text preview: 62 (c) From Eq. (6.25) 1E' = (A+1)2
VEcos(}s + VE(A2 Binary Nuclear Reactions
 Chap. 6 1 + cos2 (}s) + A(A + l)Q} 2 ..j8 cos(45)+ ../8(144 1 + cos2(45))+ 12(13)! 4.439)} 2 = 3.22 MeV. Derive Eq. (6.21) from Eq. (6.11). Solution: Begin with the general result of Eq. (6.11), namely VIF y = E Yr~;~~:~~~~~;~xmYEx ( my + my ) 2 COS (}y
[ :!: mxmyEx (my+my)2 COB y + 2 {} my mx (my+my) E+ x myQ (my+my). (P6.1) For a heavy particle scattering elastically from an electron at rest, we identify particles X and y in Eq. (P6.1) as the electron, so mx = my = me, Ey = Ee (the recoil electron energy), Os= Oe,mx = my = M (the mass of the heavy particle), and Ex = EM (the kinetic energyof the incident heavy particle). For this scattering process,there is no change in the rest massesof the reactants, i.e., Q = 0, and only the + sign of the ::f: choice in the above equation has physical significance. Substitution of these variables into Eq. (P6.1) gives V Ee V~:+
r;;""I~~~ M)""2COS e+ V~ O I MmeEM M)""2 COS e 0 O+ ~~2l1 In Squaring this result and using M > > me, we find that the recoil energy of the electron is July 24, 2002 3. ...
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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).
 Spring '06
 CADY

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