How to solve this problem?

3.2 b[12 points] Using the SEMF calculate the neutron and proton separation energy of 65Cu, explain the difference using the the SEMF (use the neutron and proton mess excess from the appendix). 1 u = 931.502 MeV/c2 1 eV = 1.602 * 10-19 J C = 2.99792458 * 10 m/s av = 15.5MeV as = 16.8MeV ac = 0.72MeV asym = 23MeV ap = 34.0MeV m(H) = 1.007825u m(n) = 1.008665u m(12C) = 12.0000u 12 C abundence = 98.93% 13 C Vabundence = 1.07% Mass Excess (MeV) n 8.0713181 1 H 7.28897106 2 H 13.135722895 HE 14.94981090 4 He 2.42491587 12 C 0 13 C 3.12500933 160 -4.7370022 170 -0.8087642 52 Cr -55.41951 56 Fe -60.6072 87 Rb -84.597802 88 Sr -87.921629 89 Y -87.7112 90 -86.4969 90 71 -88.77255 91 Nb -86.638 92 Mo -86.80859 93 Nb -87.2128 204 Pb -25.1098 205 Pb -23.7702 206 Pb -23.7855 207 Pb -22.4520 208Pb -21.7485