# Li first energy (Homework 6-325)

1. Use the variational principle with a trial wave function  $\psi(1, 2) = \psi(r1) \psi(r2) \chi0$, where  $\psi(r) = (\zeta( 3/ )1/\pi)^{1/2}e^{-ζr}$ (in atomic units), to estimate the ground state energy of a Li+ ion. Express your result in both Hartrees and eV. Compare the “effective charge” $\zeta$  with that

for a He atom.

Li has Z=3 and $Li^+$ has just 2 electrons on the 1s2 shell. Therefore the wave function is

$Ψ(1,2)=√(ξ^3/π)*exp⁡[-ξ(r_1+r_2 ) ]$

For Li the Hamiltonian is

$H=-ℏ^2/2m*(∇_1^2+∇_2^2 )-e^2/(4πϵ_0 )*[3/r_1 +3/r_2 -1/(|(r_1 ⃗)-(r_2 ⃗)|)$

This Hamiltonian can be rearranged as

$H={-ℏ^2/2m*(∇_!^2+∇_2^2 )-e^2/(4πϵ_0 ) [ξ/r_1 +ξ/r_2 ] }+$

$+{e^2/(4πϵ_0 ) [(ξ-3)/r_1 +(ξ-3)/r_2 ] }+{e^2/(4πϵ_0 )*1/(|(r_1 ⃗) -(r_2 ⃗))}$

Therefore

$<H> =2ξ^2 E_1+2(ξ-3)*e^2/(4πϵ_0 )*〈1/r〉+ <V_ee>$

where $V_ee$  is the exchange interaction given by the last ${…}$

The unknown values of  $<1/r>$  and $<V_ee>$ are computed for the effective charge $ξ (=Z/a)$   so that

$<1/r> =ξ$    and  $<V_ee>=-5ξ/4*E_1$
So that finally

$<H> =[2ξ^2-4ξ(ξ-3)-5ξ/4]*E_1=[-2ξ^2+43/4 ξ]*E_1$   with $E_1=-13.6 eV$

The variational principle implies

$(d<H>)/dξ=0 so that-4ξ+43/4=0 or ξ=2.6875$

$<H>=-196.46 eV=-7.22 Ha$

The effective charge of nucleus in Li is $ξ=2.6875$ (screened down by the electrons from the theoretical value of Z=3). For comparison He has an an effective charge of $ξ=1.69$ (down from Z=2). Since He has a smaller nucleus positive charge its charge is more screened by the two outside electrons than in Li+.