Download Helium Atom

Lecture 11 Title : Helium Atom
Page-1
In this lecture we will extend the problem to more than one electron atoms. We will start with the
simple one such as He atom.
The quantum mechanical calculations of the energy levels in different approximations are
discussed.
From the ionization energy i.e the energy required to remove one electron from Helium atom, the
ground state energy of Helium is calculated to be -79.0eV.
We use this information to check the different method approximation for calculating the Helium
atom energy levels.
Page -2
The emission spectra of He consists of a number of series in the visible region of the spectrum as
well as in the near & far UV regions.
There are twice as many line series as for the alkalis; two principal series in the visible and near
UV, as well as two diffuse, two sharp and two fundamental series.
This observation cannot be described by simple concept from hydrogen atom.
When more than one electron is present, the Hamiltonian for an atom in free space becomes
H =−
2
2me
∑ ∇i2 − ∑
i
i
Z e2
e2
+∑
ri
i > j rij
=TE + VEN + VEE
……..11.1
VEN is the Coulomb interaction between the nucleus and the electrons. VEE is the
electrostatic repulsion of the free electrons, summed over all pairs i and j at separation rij . This
equation (1) does not contain spin part, we assume that such effects can be treated separately.
2
Equation (1) contains terms e
rij
for interelectronic repulsions that are not present in the
Hamiltonian for hydrogen like atoms.
Because of these terms, the motion of each electron affects the motion of every other
electrons, and the wave equation for polyelectronic atoms cannot be solved exactly.
Various approximate methods of solution exist. Most methods start by assuming that
complex atoms can be “built up” by feeding electrons successively into atomic orbitals (AOs).
That is, every electron in an atom is assigned a wave function φi , defined by a set of quantum
numbers.
The wave function for the whole atom is written in terms of these electron wave
functions φi . The best sets of wave functions φi are those for which the residual difference
between the calculated energy of the whole atom and the actual energy is as small as possible.
Page-3
Independent Particle Model :
One-Electronic Wave Functions
A crude method, zeroth order approximation to the φi is obtained by ignoring altogether
the terms VEE in (1), giving the Hamiltonian
2 N 2 N Z e2
∑ ∇i − ∑
2me i 1 =i 1 ri
=
H =−
…………………11.2
Then, in the approximation in which electron spin is neglected, the eigen functions of this
Hamiltonian are products of the occupied single electron orbital wave functions φi for the N
electrons,
Φ (1, 2,...., N ) =
φ1 ( r1 θ1 ϕ1 ) φ2 ( r2 θ 2 ϕ2 ) .... φN ( rN θ N ϕ N )
……….11.3
Since interactions between the electrons are ignored, each electron moves independently of the
others in the electrostatic field of the nucleus alone. The φi is therefore the hydrogen like
wavefunctions, the eigen values of Hamiltonian are the sum or the individual energies Ei for the
electrons in hydrogen like orbitals:
E = E1 + E2 + .... + EN
`
…………11.4
Page-4
Let us take the example of Helium (Z = 2) that consists of two electrons as shown in the figure11.1
In this approximation, Hamiltonian ,
H =−
 2 2 Z e2  2 2 Z e2
∇1 −
−
∇2 −
2m
r1
2m
r2
e−
……………11.5
e−
r12
a
b

r1

r2
2e
Figure-11.1
Schrödinger equation,

2 2 2 2 Z e2 Z e2 
∇1 −
∇2 −
−
 Φ ( r1 , r2 ) = E Φ ( r1 , r2 )
2m
r1
r2 
 2m
………….11.6
So,  −
Here we take φa ( r1 ) and φb ( r2 ) are the one electron wavefunction as in hydrogen atom.
Φ ( r1 , r2 ) =
φa ( r1 ) φb ( r2 )
 2 2 Z e2 
∇1 −
−
 φa ( r1 ) =Ea φa ( r1 )
r1 
 2m
 2 2 Z e2 
∇2 −
−
 φb ( r2 ) =Eb φb ( r2 )
r2 
 2m
=
E Ea + Eb
………..11.7
Page-5
Two electrons “a” and “b” move independently of each other and the total energy of the system is
simply sum of the energies of non-interacting particle.
φa ( r1 ) = Rn l ( r1 ) Yl m (θ1 , φ1 )
11
1 1
φb ( r2 ) = Rn l ( r2 ) Yl m (θ 2 , φ2 )
2 2
2
2
Z 2 m e4
Z2
En =
− 2
=
−
EH
n 2 2
n2
m e4
=
where, EH =
13.605 eV
2 2
The ground state energy of Helium (when both the electrons are in 1s level)
4
4
E=
Ea + Eb =
− 2 EH − 2 EH =
−8 EH =
−108.8 eV
I
I
The experimental value of the ionization energy i.e. the energy required to remove one electron
from Helium atom = 24.63 eV
When one electron is removed then the energy of ionized Helium ion (He+)
= -Z2×13.6 eV =-54.4 eV
So the ground state energy of Helium = -54.5 eV - 24.63 eV = -79.03 eV
Thus, Atomic energies calculated in this manner are the poor approximation to the actual eigen
values.
However, the calculated energies and wavefunctions can be improved by introducing the
interelectronic repulsion terms VEE as a perturbation.
Page-6
Considering electron repulsion:
H =−
2 2 2 2 Z e2 Z e2 e2
∇1 −
∇2 −
−
+
2m
2m
r1
r2
r12
………..11.8
We will consider the first order perturbation theory. In this approximation, we take unperturbed
Hamiltonian as
H 0 =−
2 2 2 2 Z e2 Z e2
∇1 −
∇2 −
−
2m
2m
r1
r2
And consider the perturbed Hamilrtonian as
H′ =
e2
r12
Substituting in the Schrodinger Equation
E φa ( r1 ) φb ( r2 )
( H 0 + H ′ ) φa ( r1 ) φb ( r2 ) =
We are interested to calculate the ground state energy. So both the electrons are in 1s level.
So, n = 1, l = 0 and m = 0 for both the electrons. Let us take the ground state wavefuntion of the
hydrogen atom for each electron.
3
=
φnl m φ=
100 ( r )
1  Z  2 − Zr a0
  2e
4π  a0 
3
1  Z  2 − Zr1 a0
φ100 ( r1 ) =
  2e
4π  a0 
3
1  Z  2 − Zr2 a0
φ100 ( r2 ) =
  2e
4π  a0 
Page-7
So,
1
φ100 ( r1 ) φ100 ( r2 ) =
4π
3
− Z ( r1 + r2 )
Z
a0
  4e
 a0 
3
 Z ( r1 + r2 ) 

a0

1  Z  −
=   e 
π  a0 
Evaluating,
e2
e2
*
*
r
r
= ∫∫ φ100
φ
φ100 ( r1 ) φ100 ( r2 ) dτ 1dτ 2
( 1 ) 100 ( 2 )
r12
r1 − r2
=
−
Z 6 e2
dτ 1 e
2 6 ∫
π a0
2 Zr1
a0
∫ dτ 2 e
 1  r 
 ∑  2  P ( cos θ )
 r1   r1 
1
=
r1 − r2  1  r 
1
 ∑   P ( cos θ )
 r2   r2 
θ ⇒ between r1 & r2
dτ 2 = r dr2 sin θ dθ dφ
2
2
=
ρ1
−
2 Zr2
a0
1
r1 − r2
for r1 ≥ r2
Z
for r2 ≥ r1
r1
θ
r2
27 r1
27 r2
=
ρ2
a0
a0
Figure-11.2
Page -8
e2
r12

 1 ρ1

 ρ2 
1
− ρ2
2
 ∫ ρ 2 d ρ1 e × ∑   ∫∫ ρ ( cos θ ) sin θ dθ dφ 
ρ1   ρ1 
 ρ1 0

Z e2
=
ρ12 d ρ1 sin θ1 dθ1 dφ1 e − ρ1 

2
∫∫∫

∞
32π a0
 ρ1 
1


− ρ2
2
 + ∫ ρ 2 d ρ 2 e × ρ ∑  ρ  ∫∫ ρ ( cos θ ) sin θ dθ dφ 
2   2 
 ρ1

∞
 1 ρ1 2 − ρ

−ρ
∫∫∫ ρ d ρ1 sin θ1 dθ1 dφ1 e  ρ1 ∫0 ρ2 e 2 d ρ2 + ρ∫ ρ2 e 2 d ρ2 


1
e2
5 e 2 5Z
=
= 13.605 eV
Z=
EH
EH
where, =
8 a0
4
2a0
Z e2
= 2
8π a0
2
1
− ρ1
According to first order perturbation theory the ground state energy of He,
5Z
10
EH =
−8 EH + EH
4
4
=
−108.84 + 34.01 =
−74.83 eV
Experimental = −79.0 eV
Error = 4.2 eV
Eground =
−8 EH +
Page-9
Variational Method
Trial wavefunction,
Φ ( r1 , r2 ) =
φa ( r1 ) φb ( r2 )
 Z 3  − Zr1 a0 − Zr2 a0
=
e
e
3 
 π a0 
2
The effect of the repulsive term e
r12
, is to reduce the nuclear charge on each electron. One way
to accommodate this is to assume hydrogen like wavefunction corresponding to effective nuclear
charge Z ′ , which will be assumed to be the variable parameter. We assume the (trial) normalized
wavefunction,
 Z ′3  − Z ′r1 a0 − Z ′r2 a0
e
e
Φ ( r1 , r2 ) =

3 
 π a0 
Z = 2;
φ is normalized, so we have to minimize the integral
E ( Z ′ ) = ∫∫ φ * H φ dτ 1dτ 2
 2
1  e2 
2
2
2 1
= ∫∫ φ  −
( ∇1 + ∇2 ) − Z e  r + r  + r  φ dτ1dτ 2
 1 2  12 
 2m
*
Page-10
Since Φ ( r1 , r2 ) is the solution of
 2 2 Z ′ e2 
2
∇1 −
−
 Φ ( r1 , r2 ) =− Z ′ EH Φ ( r1 , r2 )
r1 
 2m
Z ′2 e 2 Z ′2 e 2 Z e 2 Z e 2 e 2 
+
−
−
+  φ dτ 1dτ 2
r1
r2
r1
r2
r12 


1  e2 
*
2
2 1
= ∫∫ φ  −2 Z ′ EH + ( Z ′ − Z ) e  +  +  φ dτ 1dτ 2
 r1 r2  r12 

……..11.9
E ( Z ′) =

*
2
2
∫∫ φ −Z ′ EH − Z ′ EH +
1st integration:
∫∫ φ ( −2Z ′ E ) φ dτ dτ
*
2
H
1
2
=
−2 Z ′2 EH
2nd integration:
1
 r1
*
2
∫∫ φ ( Z ′ − Z ) e  +
1
 φ dτ 1dτ 2
r2 
1 1
=
( Z ′ − Z ) e2 ∫∫ φ *  +  φ dτ1dτ 2
 r1 r2 
( Z ′ − Z ) e 2∫∫
=
2
=
( Z ′ − Z ) e2
φ *φ
r1
dτ 1dτ 2
2Z ′
=
4 Z ′ ( Z ′ − Z ) EH
a0
e2
where, EH =
2a0
3rd Integration:
*
∫∫ φ
e2
5
φ dτ 1dτ 2 = Z ′EH
r12
4
Page-11
Substituting in equation-11.9,
5
−2 Z ′2 EH + 4 Z ′ ( Z ′ − Z ) EH + Z ′EH
E ( Z ′) =
4
∂E ( Z ′ )
5
⇒
=
−4 Z ′EH + 8Z ′EH − 4 ZEH + EH
0=
∂Z ′
4
5

⇒  −4 Z ′ + 8Z ′ − 4 Z +  EH =
0
4

5
0
⇒ 4Z ′ − 4Z + =
4
5
⇒ Z ′ =Z −
16
So, the ground state energy,
E = −2 Z ′2 EH
2
5

=
−2  Z −  EH
16 

2
5

=
−2  Z −  13.605
16 

= −77.48 eV
Experimental = −79.0 eV
Error = 0.02 %
Other trial function yield values within 0.01% of the observed one.
Page-12
Recap
What we observed is that the perturbation method fails to calculate the ground state energy of
Helium. The basic difficulty in solving the Schrödinger equation with this Hamiltonian stems
2
from the fact that the e
r12
is too large an effect to be treated as a perturbation. On the other
hand the variational principle is an effective approximation and is able to produce the
experimental observation.
Now the question is that,
“In this Helium problem can we use perturbation method?.”
The answer is : Yes, but we have to use “Central Field Approximation. In the next lecture, we
elaborate the central field approximation.