Chemistry 431
Lecture 7
The Schrödinger
g equation
q
for hydrogen
y g
The electronic structure of atoms
Absorption
p
spectra
p
of atoms
Many-electron atoms
NC State University
The Solar Spectrum
I
I
There are gaps in the solar emission
called Frauenhofer lines.
The gaps arise from specific atoms in
the sun that absorb radiation.
Experimental observation of
hydrogen atom
Hydrogen
y g atom emission is “quantized”.
q
It
occurs at discrete wavelengths (and therefore
at discrete energies).
I The Balmer series results from four visible
lines at 410 nm,
nm, 434 nm,
nm, 496 nm and 656 nm.
nm.
I The relationship between these lines was
shown to follow the Rydberg relation.
I
Atomic spectra
Atomic spectra consist of series of narrow lines.
I Empirically it has been shown that the
wavenumber of the spectral lines can be fit by
I
⎛
⎜
⎜
⎜
⎜
⎜
⎜
⎝
⎞
⎟
⎟
⎟
⎟
⎟
⎟
⎠
ν~ = 1 = R 12 − 12 (n2 > n1)
λ
n1
n2
where R is the Rydberg constant
constant, and n1 and n2
are integers.
El t i St
Electronic
Structure
t
off Hydrogen
H d
I The
Schrödinger equation for hydrogen
I Separation of variables
Radial and angular parts
I Hydrogen atom wavefunctions
I Expectation values
I Spectroscopy of atomic hydrogen
Schrödinger equation for hydrogen:
The kinetic energy operator
The Schrödinger equation in three dimensions is:
2
2
h
–
∇ Ψ + VΨ = EΨ
2μ
The
Th operator
t ddeldell-squaredd is:
i
2
2
2
∂
∂
∂
∇ = 2+ 2+ 2
∂x
∂y
∂z
2
The pprocedure uses a spherical
p
ppolar coordinate system.
y
Instead of x, y and z the coordiantes are θ, φ and r.
Schrödinger
g equation
q
for hydrogen:
y g
The form of the potential
I The
Coulomb potential between the electron and
the proton is V = -Ze2/4
/4πε
πε0r
I The hamiltonian for both the proton and electron
2
2
is:
2
2
h
h
H=–
∇ N–
∇e+V
2m N
2m e
I Separation of nuclear and electronic variables
results in an electronic equation in the centercenter-ofofmass coordinates: H = -h2/2
/2μ∇
μ∇2 - Ze2/4
/4πε
πε0r
(1/μ
(1/
μ=1/me+1/mN).
Separation of variables
ariables
It would be impossible to solve the equation
with all three variables simultaneously.
y
Instead a procedure known as separation of variables
is used.
The steps are:
1.
1 Substitute in Ψ(r,
(r θ,φ) = R(r)Y(θ
(r,θ
R(r)Y(θ,φ)
2. Divide both sides by R(r)Y(θ
R(r)Y(θ,φ)
3. Multiply
p y both sides byy 2μ
2μr2
Separation of variables
It is not possible to solve the equation
for all three variables simultaneously.
y
The steps for the separation of variables are:
1. Substitute in Ψ(r,
(r,θ
θ,φ) = R(r)Y(θ
R(r)Y(θ,φ)
2 Di
2.
Divide
id b
both
th sides
id b
by R(
R(r)Y(
R(r)Y(θ
)Y(θ,φ)
)Y(θ
3. Multiply both sides by 2μ
2μr2
2
2
2
2μr
e
h
d
2 dR
–
r
+ 2
+ E R(r)
4πε
r
R(r) dr dr
h
0
2
2
h
d
Y
1
d
dY
1
–
sinθ
+
dθ
Y(θ,φ) sinθ dθ
sin 2θ dφ 2
=0
Radial and angular
g
equations
q
Separate into two equations by setting each
equation equal to a separation constant, β.
The radial equation is:
2
2
2μr
h
e
d
dR
–
r2
+ 2
+ E R(r)
( ) = –β
β
R(r) dr dr
h 4πε 0r
2
The angular equation is:
2
–
2
h
1 d sinθ dY + 1 d
=β
2
2
Y(θ,φ)
( ,φ) sinθ dθ
dθ
sin θ dφ
Separation of angular variables
The sperical harmonics arise from the product
( ) (φ) after substituting
g Y(θ,φ)
( ,φ) = Θ(θ)Φ(φ)
( ) (φ)
of Θ(θ)Φ(φ)
1 ∂ Φ + sin2θβ+ sinθ ∂ sinθ ∂Θ = 0
Φ(φ) ∂ϕ 2
Θ(θ) ∂θ
∂θ
2
The θ and φ equations are obtained with separation
constant m2.
sin2θβ+ sinθ ∂ sinθ ∂Θ = m 2
Θ(θ) ∂θ
∂θ
2
∂
Φ
1
Φ(φ) ∂ϕ 2
= – m2
Solution of the φ equation
The φ equation has a simple form:
∂ Φ (φ)
2
=
–
m
Φ(φ)
2
∂
∂ϕ
2
The solution of this equation is:
Φ= e
±imφ
This is an oscillatory function since:
e
±ix
= cos (x) ± i sin(x)
Solutions of the angular equation
I The
wavefunction solutions of the angular
equation are spherical harmonics Ylm (θ,φ).
I These functions describe the angular distribution
of atomic orbitals and are the wavefunctions for
the rigid rotor of polyatomic molecules.
I The
degeneracy of a given orbital is 2l+1 and
the angular momentum of the electron is
√l(l + 1)h
1)h.
MAPLE
worksheet
k h
on spherical
h i lh
harmonics
i
The form of the spherical harmonics Ynl (θ,φ) is quite
familiar. The shape
p of the ss--orbital resembles the first
spherical harmonic Y00.
Attached to this lecture are three MAPLE worksheets that
illustrate the s, p and d orbitals respectively. The idea is to
obtain an interactive picture of the mathematical form and
the plots of the functions.
Y0 or the
th ss--orbital
bit l
Y00
Y1 or the p
p--orbitals
Y11
Y10
Y2 or the
th dd-orbitals
bit l
Y22
Y21
Y20
The effective potential: result of
the
th solution
l ti off the
th angular
l partt
The solutions for the angular part result in a term in
ppotential energy
gy equal
q to:
V’ = h2l(l+1)/2
+1)/2μ
μr2
This term contains the contributions to the energy from
angular terms.
Together with the Coulomb potential the effective
potential energy is:
Veffff = - Ze2/4
/4πε
πε0r + h2l(l+1)/2
+1)/2μ
μr 2
The radial equation for hydrogen
Making the above approximations we have an
radial hamiltonian (energy operator)
2
h l (l +1)
2
2
h
Ze
–
∇ R–
R–
R = ER
2
2μ
4πε 0r
2μr
2
l -ρ/2 L (r)
The solutions have the form: Rn,
n,ll = Nn,
n,ll ρ e
n,ll
n,
where ρ = 2Zr/na0 and a0 = 4πε
4πε0h2/me2
Nn,
n,ll is the normalization constant.
Ln,
polynomial
n,ll(r) is an associated Laguerre polynomial.
The Bohr radius
The quantity a0 = 4πε0h2/me
/ 2 is known as the Bohr
radius. The Bohr radius is a0 = 0.529 Å. Since it
emerges from the calculation of the wave functions and
energies of the hydrogen atom it is a fundamental unit.
In soso-called atomic units the unit of length
g is the Bohr
radius. So 1 Å is approximately 2 Bohr radii.
You
Y should
h ld do
d a di
dimensional
i l ((unit)
it) analysis
l i andd verify
if
that a0 has units of length!
Solutions of the radial equation
q
The normalization constant depends on n and l.
1/2
n –l – 1 !
N n,l = –
2n n + 1 !
3
2 l
na 0
+ 3/2
The associated Laguerre polynomials are:
n=1
n=2
n=2
n=3
n=3
n=3
l
l
l
l
l
l
= 0 L 11(x)
( )=–1
= 0 L 12(x) = – 2!(2 – x)
These are given
for completeness.
= 1 L 33(x) = – 3!
= 0 L 13(x) = – 3!(3 – 3x + 1/2x 2)
= 1 L 34(x) = – 4!(4 – x)
= 2 L 55(x) = – 5!
Normalization of the radial functions
Each of the radial equation solutions is a polynomial
multiplying an exponential. The normalization is
obtained from the integral:
∞
0
*
nl
2
R R nl r dr = 1
The volume element here is r2dr which is the “r”
r
part of the spherical coordinate volume element
r2sin
sinθ
θdrd
drdθ
θdφ.
MAPLE worksheet
Normalization of the radial functions
A MAPLE worksheet attached to this lecture illustrates
the normalization of the first three radial functions. The
worksheet includes plots of the functions.
When examining
Wh
i i a plot
l t kkeep in
i mind
i d th
thatt you can plot
l t
the wave function or the square of the wave function.
We often plot the square of the wave function
function, because
the integral of the square of the wave function gives the
pprobability.
y
Energy levels of hydrogen atom
I The
energy levels of the hydrogen atom are
specified by the principal quantum number n:
I All
μe 4
1
E=–
2 2 2
2
32π ε 0 h n
states with the same quantum number n
have the same energy.
I All states of negative energy are bound states,
states of ppositive energy
gy are unbound and are
part of the continuum.
The Rydberg Constant
I The
energy levels calculated using the
Schrödinger equation permit calculation of the
Rydberg constant.
constant
I One major issue is units. Spectroscopists often
use units of wavenumber or cm-1. At first this
~
seems odd, but hν
hν = hc/λ
hc/λ = hcν
hcν where ν is ~the
value of the transition in wavenumbers.
wavenumbers
4
μe
1
R=
hc 32π 2ε 0 2h 2
in cm-11
Shells and subshells
I All
of the orbitals of a given value of n for a shell.
shell
I n = 1, 2, 3, 4 .. correspond to shells K, L, M, N…
I Orbitals with the same value of n and different
values of l form subshells.
I l = 0, 1, 2, ... correspond to subshells s, p, d …
I Using
the quantum numbers that emerge from
solution of the Schrödinger equation the
subshells can be described as orbitals.
orbitals
Hydrogenic
y g
orbitals
Is
orbitals are spherically symmetrical. The 1s
wavefunction decays exponentially from a
maximum value of (1/π
(1/πa03)1/2 at the nucleus.
I p orbitals have zero amplitude at r=0, and the
electron possesses an angular momentum of
h√l(l +1
+1).). The orbital with m = 0 has zero angular
momentum about the z axis
axis. The angular
variation is cosθ
cosθ which can be written as z/r
leading to the name pz orbital.
orbital
Hydrogen 1s radial wavefunction
The 1s orbital has no
nodes and decays
exponentially
exponentially.
I R1s = 2(1/a0)3/2e-ρ/2
2.0
I
I
n = 1 and l = 0 are
the quantum numbers
for this orbital.
orbital
R 1 ,0
1.5
Ψ1s
1.0
0.5
0
2
4
ρ
6
8
10
The Radial Distribution in
Hydrogen 2s and 2p orbitals
0.6
n ,l
Ψ2s
Ψ2pp
R
0.4
0.2
0.0
0
2
4
ρ
6
8
10
Expectation values
The expectation
p
((or average)
g ) value of an observable <r> is ggiven
by the general formula.
∞
Ψ r Ψ dτ
*
<r>=
0
As written the above integral describes the expectation value
off the
th mean value
l off the
th radius,
di r. Integration
I t
ti over the
th angular
l
part gives 1 because the spherical harmonics are normalized.
The volume element can be written dτ
dτ = r2dr. The mean value is:
∞
Ψ r Ψ r dr =
*
<r>=
0
∞
Ψ Ψ r 3dr
*
2
0
MAPLE worksheet on
Expectation (Average) values
There is a MAPLE worksheet attached to this lecture
that illustrates the use of expectation values. The
example of the position <r> is given for the 1s, 2s and
3s radial wave functions. The expectation value or
average value of r gives the average distance of an
electron from the nucleus in a particular orbital. Since
the 2s orbital has one radial node and the 3s orbital has
two radial nodes the average distance of the electron
from the nucleus is shown to increase as:
<r> for 3s > <r> for 2s > <r> for 1s
Spectroscopy of atomic hydrogen
Spectra reported in wavenumbers,
1
ν=
I Rydberg
y
g fit all of the series of hydrogen
y g spectra
p λ
with a single equation,
ν=R 1 – 1
n
n
I Absorption or emission of a photon of frequency
ν occurs in resonance with an energy change,
ΔE = hν
hν (Bohr frequency condition).
I Solutions of Schrödinger equation result in
further selection rules.
rules
I
2
1
2
2
Spectroscopic transitions
IA
transition requires a transfer from one state
with its qquantum numbers ((n1, l1, m1) to another
state (n2, l2, m2).
I Not
all transitions are possible: there are
selection rules, Δl = ± 1, m = 0, ± 1
I These
rules demand conservation of angular
momentum. Since a photon carries an intrinsic
angular momentum of 1.
Many-electron atoms:
ManyThe orbital approximation
The orbital approximation to the total many
many-electron wavefunction Ψ(r1, r2, ...) is to rewrite
it as a p
product of “one“one-electron” wavewavefunctions ψ(r1)ψ(r2)… so that
Ψ((r1, r2, ...)) = ψ((r1)ψ((r2))…
I The configuration of an atom is the list of
occupied orbitals.
I The Pauli exclusion principle states that the
spins must be paired if two electrons are to
occupy one orbital and no more than two
electrons may occupy an orbital.
I
Penetration and shielding
g
In many
many--electron atoms the s, p, d etc. orbitals
of each shell are not degenerate.
I An electron distance r from the nucleus
experiences the nuclear charge Z shielded by
all of the other electrons. The effective charge
is Zeff.
I Zeff = Z - σ where σ is the shielding parameter.
I An s electron has a greater penetration thro
through
gh
inner shells than a p electron of the same shell
because the p electron has a node at the
nucleus.
I
The Aufbau principle
I Aufbau
means “building“building-up” in German.
I The configurations of atoms are built up by
population
p
p
of the hydrogenic
y g
orbitals.
I Imagine a bare nucleus with charge Z and
then add Z electrons to the orbitals in the
following order:
1s 2s 2p
p 3s 3p
p 4s 3d 4p
p 5s 4d 5p
p 6s etc.
I Hund’s rule an atom in its ground state
adopts a configuration with the greatest
number of unpaired spins.
Why is 4s lower than 3d?
I Shielding
of 3d electrons is less effective
than 4s. Thus, an occupied 3d wave
function is pulled closer to the nucleus than
a 4s wave function.
I The more confined volume for 3d electrons
means that they have greater electronelectronelectron repulsion than 4s.
I To relieve electronelectron-electron repulsion there
is a general trend that (n+1)s is occuplied
before nd in the periodic table (n > 2).
The problem of multiple electrons
I
The central difficulty with application of the
Schrödinger equation is the presence of
electron--electron interaction terms in the
electron
potential energy:
Σ
I
V=–
i=1
I,J
2
Ze
e
+ Σ
4πε 0r i
i j = 1 4πε 0r ij
i,
2
electron – nuclear
electron – electron
No analytical solutions
solutions.
I Hartree
Hartree--Fock procedure. Find solutions that
optimize the electron in each orbital in the
presence of the field of all of the other orbitals.
I
Hartree--Fock procedure
Hartree
As an example,
e ample for He we
e can write
rite the ttwo
o electron
wave function as a product of orbitals.
Ψ(r1,rr2) = φ(r1)φ(r2)
The probability distribution, φ*(r2)φ(r2)dr2 for electron
2 corresponds to a charge density in classical
physics. Therefore, we can say that the effective
potential felt by
p
y electron 1 is:
V
eff
1
(r1) =
*
1
φ (r 2) r φ (r 2)dr 2
12
*
The antianti-symmetry
y
y
requirement
I The
Th
wave function
f
ti mustt be
b antianti
ti-symmetric
ti
with respect of electron exchange. This
means that is must change sign if any two
electrons change places.
I Slater came up with a general way to do this
using a determinant. Here, we show it for
two electrons:
Ψ r 1,r 2
φa(r 1) φb(r 1)
1
=
= 1 φa(r 1)φb(r 2) – φa(r 2)φb(r 1)
2 φa(r 2) φb(r 2)
2
The Slater determinant
I The
two--electron case is pretty easy. The
two
general case for N orbitals is:
φa(r
( 1)
φa(r 2)
φa(r 3)
1
Ψ r 1,rr 2,rr 3,rr 4.... =
N! φa(r 4)
...
...
IA
φb(r
( 1) φc(r
( 1) φd(r
( 1)...
)
φb(r 2) φc(r 2) φd(r 2)...
φb(r 3) φc(r 3) φd(r 3)...
φb(r 4) φc(r 4) φd(r 4)...
N x N determinant is painful to do by
hand but is very easy to automate for use
hand,
in computer programs.
The selfself-consistent field method
The effective or average potential can be used in a
one electron hamiltonian operator for electron 1
eff
2
1
2
H (r 1) = – ∇ 1 – r + V1 (r 1)
2
eff
1
The Schrödinger equation is solved for electron 1
H φ(r 1) = ε 1φ(r 1)
eff
1
Start with a trial function φ(r2) and solve for φ(r1).
)
Using φ(r1) calculate an effective potential for 2
and solve for φ(
φ(r2)). Continue until convergence
g
is
reached.
Atomic term symbols
The letter indicates the total orbital angular
momentum quantum number of all electrons
in an atom L = l1 + l2 , l1 + l2 - 1,
1 ….|| l1 - l2 |
I The left superscript gives the multiplicity
2S+1,
2S+1 where S = s1 + s2 + s3 + ….
I The right subscript gives the total angular
momentum J = L + S
S, L + S - 1,
1 … |L - S|
I The term symbol is
2S+1
J
I
X
L
Hund’s rules determine the
ground state term symbol
Each state is designated by a term symbol corresponds
to a wave function that is an eigenfunction of L2 and S2
with
ith unique
i
energy.
I The state with the larges value of S is the most stable.
I For states with the same value of S, the state with the
largest value of L is the most stable.
I If the
th states
t t have
h
the
th same value
l off L andd S
- subshell less than half filled, smallest J is most stable
- subshell more than half filled,
filled the state with the largest
value of J is the most stable.
I
Explicit Rules for Determining
Term Symbols
I
I
I
I
I
I
To determine the states(terms) of a given atom or ion:
Write down the electronic configuration (ignore closed
subshell electrons)
Determine the number of distinct microstates that can
represent that configuration. If you have e electrons in a
single open subshell of 2l+1 orbitals
orbitals,, this value is
#microstates = (2(2l+1))!/ e!(2(2l+1)
!(2(2l+1)-- e)!
Tabulate the number of microstates that have a given ML
and MS
Decompose your table into terms by elimination
Test the total degeneracy of the resultant terms to account
f allll th
for
the microstates
i
t t counted
t d in
i parts
t 2 and
d3
Determine the lowest term using Hunds Rules.
Examples
p
I
H(1s1)
ground state term symbol is 2S1/2
I
He(1s2)
Ground state term symbol is 1S0
I
He(1s12s1) An Excited State Configuration
Terms: 1S0 , 3S1 {There is no 3S0 or 3S-1 term }
I
I
B(1s22s22p1)
T
Term
symbol:
b l 2P1/2, 2P3/2
2
The spin orbit splitting is regular so P1/2. is the ground state.
Examples
C(1s22s22p2)
Calculate the number of possible electron
arrangements in the given configuration
I There
Th
are 6!/4!2! = 15 microstates
i
t t expected
t d
I Write down all these possibilities
I Tabulate
T b l t the
th total
t t l numbers
b
by
b ML and
d MS
I Decompose this table into terms
I
Examples
I
This listing of configurations is
d i db
derived
by systematically
t
ti ll
considering every possible
arrangement of 2 electrons
in six possible positions.
3 orbitals + 2 spins
Degeneracy =
(2l + 1)(2s + 1) = 2(2l + 1)
Examples
I
This is an example of a
t bl off microstates.
table
i
t t
I
These are the terms