586 Nuclear StructureAdditional Problems*P44.54(a)With mnand vnas the mass and speed of the neutrons, Eq. 9.23 of the text becomes, aftermaking appropriate notational changes, for the two collisions vmmmvnnn112=+FHGIKJ,vmvnnn222=+FHGIKJ∴+=+=∴−=−∴=−−v mmv mvmv v mv mvmmv mvvvnnnnnn22112111222bg(b)mn=×−××=13 30 10144 70 104 70 103 30 101167667u msmsmsuafej...P44.55(a)QMMM mcn=+−−94 122BeHeCQ− −=9 012 1824 002 60312 000 0001 008 665931 55 70..uuuuMeV uMeV(b)Mmn=−−223HHeQ−=2 2 0143 016 029 1 008 6653 27..102u 931.5 MeV uMeV exothermicbgafP44.56(a)At threshold, the particles have no kinetic energy relative to each other. That is, they movelike two particles that have suffered a perfectly inelastic collision. Therefore, in order tocalculate the reaction threshold energy, we can use the results of a perfectly inelasticcollision. Initially, the projectile Mamoves with velocity vawhile the target MXis at rest.We have from momentum conservation for the projectile-target system:MvMM vaaaX c.The initial energy is:
This is the end of the preview. Sign up
access the rest of the document.
This note was uploaded on 12/15/2011 for the course PHY 203 taught by Professor Staff during the Fall '11 term at Indiana State University .