Thanks to Miranda's diligence and intelligence we have identified thee errors in the solutions to Part B. I do apologize, but I did warn you that there were no guarantees due to the hour at which I posted these. [For complete corrected Part B solutions see later post of 1/25]
22) I forgot to divide by 2 when answering part A therefore the correct answers are:
vLi = -1/6 vp = -.5E6; KEi = 8.7E-15 KEf = 6.49E-13; He4 has 3/4 v of He 3 so it has 3/7 of total KE so He4=sqrt(3/7 x 2x6.49E-13/(6.68E-27)) = 9.1 E6 m/s
24) I changed the problem and then compared the solution to the old problem instead of the new one ( it was very late at night) so here are the corrected answers:
it should be T ends up with vTy' = 2E6 m/s in y direction so for (a)
vDy'= -3E6m/s; vDx' = -1.5E6 m/s
(b)DpTx = +1E6(5E-27)=+5E-21 pods DpTy =-1E-20 pods
(c)J= (sqrt(DpTxsqrd+DpTysqrd)= 1.12 E-20 pods so F = J/E-10 = 1.12E-10 N; r = sqrt( kQTQD/F) = 1.42 E-9 m
26) 20 keV = 3.2 x10^ -15J v=sqrt(2x3.2x10^ -15/5x10^-27)=1.13x10^6; vperp 2/3 x v ; Mvperp^2/r =QvB cancel one vperp so QB = Mvperp/r so B = Mvperp/Qr = 5E-27x 7.5E5/(1.6E-19 x .02) =1.17 T
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