Question

Calculate the binding energy per nucleon (in ) for Li and Li. The atomic masses are 6.015122 u for Li, and 7.016004 u for Li. (Enter unrounded values. Assume that the mass of H 1.007275 u, m 1.007825 u, 1.008666 u, and m 0.000549 respectively.) 6L x J/nucleon 4.80-12 7 J/nucleon

Answer 1

Step 1 frist we cal culate dhe mass defect 3 pratong P)/ 3 electrong Ce). hag 3 heutang (G amd Thue, due total mags + 3me 3 mp t 3m M= laus oCon sesvatren o mass) Coy OPuen mp maas Paoton opeutron oo 866bu Mag electan me= mass 0 Ooo 549u we haue va uag eqn ) Thug ubstreating de 3 (t.004295 u) +3 C008666) +3 Co-0005) 549 6.04947u Caiuen) 6. 015122u GLi Afomic mass of Li thug Mags defect Md Total mas A fomf maos 79 6.04947 6 o15122u 0.034348 -

Sep2 diffesence 03 mass yoous we know dat deet s atame to getes bfndi ng eneaay Comueeted to ehaoay to bind de 42 3933 xio amu Ne knos uat O 03434 8 x4923933 xio Thug O 0343 4 8 Omu G.051 26 0구251 X10T Contang 3 paotong avd 3 heutang total 양 Since 6uclkons amd hence pes nucleon of Bibduig enery 6 O51 26 07 251 X100J hucleong 13 Pckon nuclkon 8-543 45 418 xio

13 mag defect Calculate dha he fist protone CP 3 electong e) and has heutrong Cn Thug due total mags oL? is : -Ca) M- law af Consee valfon op maes) CBy gfuen mp Mag of paton mass o0 8666u mMe mags of electan o O0o 549 u we haue valuag eqn 2 Thug, Substituting the CI.008666)+3 Co.co051) 007295) M- 2 4 7 0581 36 u Carueu) t01 6004 u L Afom?c ass Thug May deect, Md lotal mass AtamPe mass 7 O58136 u 7.016004u = 0. 042132 u

0. 04 2132 amu Step 2 Conues tee defeet to bindimg enaseng mags we knon fhat lamu 4923983 xioT Thug, O 0 421 32 amu O 042132 x 1.4923933 xio J 0. o628 7751 46 xioT da s 8ihce Licontains 3 protong amd 4 heutrong total f huckong hence Bindug enear pea hacleon Bindng nucle ong DO O 06287F 51 F5 Xio" hucleong 3uckn 8 98250207 xio nucle on 8.18 xi3 J