Chapter 11 - Ground State Energies of Lithium and Beryllium
by Reinaldo Baretti Machín
e mail : reibaretti2004@yahoo.com
Lectures on Quantum Mechanics -Introduction
Chapter 1 - Some experimental facts
Chapter 2 - The Old Quantum Theory
Chapter 3 - The Postulates of Quantum Mechanics
Chapter 6 - Periodic Potentials
Chapter 8 - The Hydrogen AtomChapter 10 -The two electron atom
Chapter 11- Ground state energies of lithium and beryllium
Chapter 12 - Self Consistent Field Calculations
Chapter 13- Monte Carlo methods
Chapter *** -The H2 molecule
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References:
1. Principles of quantum mechanics; nonrelativistic wave mechanics with illustrative applications.
2. Introduction to Quantum Mechanics with Applications to Chemistry by Linus Pauling and E. Bright Wilson Jr.
3. Quantum Mechanics: Non-Relativistic Theory, Volume 3, Third Edition (Quantum Mechanics) by L. D. Landau and L. M. Lifshitz
4. Applied mathematics for engineers and physicists , by Louis Albert Pipes
5. Theoretical physics by A. S Kompaneets
6. The helium atom ,http://farside.ph.utexas.edu/teaching/qmech/lectures/node128.html
7.The Hartree-Fock Method for Atoms: A Numerical Approach by Charlotte Froese Fischer
8. Quantum Mechanics of One- and Two-Electron Atoms by Hans A. Bethe and Edwin E. Salpeter
9. http://en.wikipedia.org/wiki/File:IonizationEnergyAtomicWeight.PNG
Two examples will be given where the variation method with one or two parameters a1 , a2 , is applied to a many electron system like lithium and beryllium. By using hydrogenic wave functions we avoid the question of self consistency. That is the hydrogenic wave functions are solutions to the one electron problem , not the many electron problem.
The self consistent field method (SCFM) requires that the i-th wave function be found from one electron equations where the wave function of the other j electrons appears in the potential V(r). For example the n-th iteration of φi is labeled φi (n) and depends on the (n-1) th iteration of the wave functions through the one electron equation ,
-(1/2)∆ φi (n) -(Z/r) φi (n) + [ ∫ ∑ j≠i (/r - r' / )-1 {φj (n-1) }2 dτ'] φi (n) = ε i φi (n) ,
or -(1/2)∆ φi (n) -(Z/r) φi (n) + V(r) φi (n) = ε i φi (n) .
The SCFM is the subject of chapter 12.
Lithium ground state electronic configuration 1s2 2s , 2S
There are anti parallel spin in the 1s orbitals but the remaining electron in the 2s orbital has an unpaired spin. There are two coulombic terms of the type J(1s,2s) and one J(1s,1s) . The sole exchange term is K(1s,2s) . The total energy is
E(a) = 2 ( a2 /2 - z a) + (1/4) ( a2 /2 - z a) + J(1s,1s) + 2 J(1,2s) + K(1s,2s) . (1)
Using Table 1, from chapter 10, we get
E(a) = (9/4) ( a2 /2 - z a) +(5/8)a + 2 (17/81)a -(16/729) a . (2)
Fig 1. E vs a for lithium . E minimum ≈ -7.29 au . E Hartree - Fock = -7.43 au .
The results of eq (2) can be improved if two variation parameters are introduced one a1 for φ1s and a parameter a2 for φ2s .
Using Sage code we found the modified J(1s,2s) to be
J(1s,2s) = (8.*a1**4*a2 + 20.*a1**3*a2**2 + 12.*a1**2*a2**3
+ 10.*a1*a2**4 +a2**5)*a1**3/(32.*a1**7 + 80.*a1**6*a2 +
80.*a1**5*a2**2 + 40.*a1**4*a2**3 +10.*a1**3*a2**4 +a1**2*a2**5)
. (3)
We retain the previous K(1s,2s) = -(16/729) a1 . Now is a function E (a1 , a2) but we fix a1 = Z-5/16 and only a2 .
Then
E(a1 = Z-5/16 , a2 ) = 2 ( a12 /2 - z a1) + (1/4)( a22 /2 - z a2) + (5/8)a1 + J1s2s(a1 ,a2 ) -(16/729) a1
Fig 2. E vs a2 for lithium , a1 = Z-5/16 . E minimum ≈ -7.45 au . E exact = -7.48 au (see page 183 Ref. 7 )
Beryllium ground state electronic configuration 1s2 2s2 , 1S
The coulombic terms are J(1s,1s) , 4 J(1s,2s) , J(2s,2s) and there are two identical exchange terms 2 K(1s,2s) .
a) the one parameter variation ( a1 = a2 =a)
E(a) = 2(a2/a - Za) + 2(1/4)( a2 /2 - Za) + J(1s,1s) + 4 J(1s,2s) + J(2s,2s) +2K(1s,2s)
= (5/2)( a2 /2 - Za) +(5/8)a +4(17/81)a +(77/512)a - 2(16/729)a . (4)
The plot of E(a) is given in Fig. 3.
Fig 3. E vs a for beryllium ( . E minimum ≈ -14.2 au . E Hartree-Fock = -14.6 au (see page 28, Ref. 7 )
We introduce again two parameters a1 = Z-5/16 and the variable parameter a2 . Employ J(1s,2s) as given in (3) and retain the approximation K(1s,2s) = -(16/729) a1 . The total energy is
E(a1 , a2) = 2(a1 2/a - Za1) + 2(1/4)( a2 2 /2 - Za2) + (5/8)a1 + 4 J(1s,2s,a1 , a2 ) + (77/512)a2 +2{-16/729}a1 . (5)
Fig 4. E vs a2 for beryllium (a1 = Z-5/16) . E minimum ≈ -14.64 au . E exact = -14.66 au (see page 191, Ref. 7 )
Ionization potentials
The ionization energy equals the difference between the singly ionized atom ,say E (A+) and the total energy of its ground state E(A),
I = E(A+) - E(A) . (6)
Fig 5. The ionization energies .
For helium we had , with the single variation parameter that E (1s2 ) ≈ - 2.848 au (see eq.(44) , Chapter 10 ).
The energy of singly ionized helium is E(1s) = - Z2 /2 = -2.00 au , thus the ionizing energy is
I(helium) = .85 au = .85(27.2eV/au) = 23 eV . (7)
In the case of lithium the energy of the singly ionized state is approximately -(Z-5/16)2 = -7.22 au.
The plot in Fig 2 shows the ground state to be -7.45 au , , hence I = 0.23 au = 6.3 eV .
To calculate the energy of Be+ we modify the energy formula (5) to read , with one less electron,
E(a1 , a2) = 2(a1 2/a - Za1) + (1)(1/4)( a2 2 /2 - Za2) + (5/8)a1 + 2*J(1s,2s,a1 , a2 ) + (1){-16/729}a1 . (8)
The plot of E gives E(Be+) = -14.29 au .
Fig 6. Plot of E (Be+ ) vs a2 .
From Fig (4) we have E (Be) = -14.64 au . Then I = -14.29 + 14.64 = 0.35 au = 9.5 eV .
In conclusion we have obtained
I (helium) = 23 eV
I (lithium) = 6.3 eV
I (beryllium) = 9.5 eV
which compare very well with the experimental values given in Fig 5.
The pioneer experiments in measuring energy levels and ionization potentials with electrons shot through gases is that of
Franck and Hertz in 1914 (see http://www.dartmouth.edu/~physics/labs/writeups/franck.hertz.pdf ) .
END OF CHAPTER
