Question

In the class lecture notes, the result of the first-order perturbation calculation for the energy of lithium was given as E(1) = 2J, +J Lists – K,625 - This energy is added to the E© value obtained by treating each electron as if it were in an independent hydrogen-like orbital in order to estimate the total energy of the lithium atom. 1s 2s The calculation to determine the E9) energy correction involves solving many integrals because the wavefunction for lithium has 6 terms, the Hamiltonian has 3 pairwise electron-electron potential terms and E" = (410°4). Two of the many integrals are shown below. Show all of the work necessary to evaluate these integrals in terms of Jij and Kij. + 1. $1$(1)a(1)15*(2)$(2)2*(3)a^(3) * * * (1)a(1)14(2)(2)2-(3)a(3)dt 2. ſ 1s (1)c*(1)is* (2) ** (2)2s* (3)aº(3), * * 25(1)(1)1s(2)6(2)1=(3)a(3)dt ri2 '13

Answer 1

some Too electrons must have different quantum numbee re either spin palts are with orbital part dituerent or Orbital but different spin pert for J- integral to be nonzero some we must have their spin part some (4,41 or (B, B) for K- integral For 12 Palt with wave function 25C124(125C2 )BL)5(3*(3) on teem total o parts of the complete of anti- symmetrized Waller function ist palt 92 we have J2 518 -(K2515 - 0) 212 b) It is =) we have Jasis & kasis ri 13 -) we () have Tesis (kisis =o) 72 3 Add at btc. ť Jisis + k2s18 Wave function has 4/5 V31 numberator which cancel's others 2 3 2515 + Jisist k2sis Els Jos fs Cisor and Ę teems in CS Scanned with CamScanner