Problem Set 9 Solutions

10
6. b.) Continued
From the information given, we know that ε 1 = −0.619 a.u. We also can determine from above that
J 11 = K 11 = φ 1 (1) φ 1 (2)
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1
r 12
φ 1 (1)φ 1 (2) = 0.566 a.u.
Substituting, the electronic energy of H 2 becomes
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( ) = 2ε 1 − ( 2J 11 − K 11 )
E el H 2
= 2ε 1 − J 11
= 2 −0.619 a.u.
( ) − ( 0.566 a.u. )
( ) = −1.804 a.u.
E el H 2
c.) Determine the total € energy from the SCF calculation for H 2 at an internuclear
separation of 1.40 a.u. The total energy includes the electronic energy plus the
nuclear-nuclear repulsion energy.
The total energy of the system is just the electronic energy plus the nuclear-nuclear repulsion,
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E TOT = E el + V nn ,
where the nuclear-nuclear repulsion V nn in atomic units is
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V nn = Z α Z β
r αβ
.
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For H 2 ,
Z α = Z β = 1. Substituting, the total energy becomes
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E TOT = E el +
1
r HH
= −1.804 a.u. +
1
1.40 a.u.
= −1.804 + 0.714 a.u.
E TOT = −1.090 a.u.
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