Nuclear stopping power: Coulomb scattering(assumed elastic Incident ion interacts with nucleus of stationary ion b= impact parameter 1.0 0.8 dE/E 04 2 0.0 2468 Energy lost by incoming ion(microscopic) Mo/M △E=E1{1- cos esin+ cos p sin e The angles depend on masses and on b 4M,M, △E=E Max energy loss is when b=0, 0=0 (M,+M 3.155J/6.152J.2003Nuclear stopping power: Coulomb scattering (assumed elastic) Incident ion interacts with b q f b E1, M1 M2 Energy lost by incoming ion (microscopic) DE = E1 1 - sin2 f cosq sinf + cosf sinq Ï Ì Ó ¸ ˝ ˛ 0 2 4 6 8 10 M2/M1 1.0 0.8 dE/E 0.4 0.2 0.0 nucleus of stationary ion = impact parameter The angles depend on masses and on b. Max. energy loss is when b = 0, f = 0: 4 M1M2 DE = E1 (M1 + M2 )2 3.155J/6.152J, 2003 9