CHM 4412 Chapter 13 - School of Chemical Sciences

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Lecture 18
Hydrogen’s wave functions and energies
(c) So Hirata, Department of Chemistry, University of Illinois at Urbana-Champaign. This material has
been developed and made available online by work supported jointly by University of Illinois, the
National Science Foundation under Grant CHE-1118616 (CAREER), and the Camille & Henry Dreyfus
Foundation, Inc. through the Camille Dreyfus Teacher-Scholar program. Any opinions, findings, and
conclusions or recommendations expressed in this material are those of the author(s) and do not
necessarily reflect the views of the sponsoring agencies.
The energy expression
The nuclear charge
Z me e
E=2 2 2 2
32p e 0 n
2
Discrete energies
are negative
4
This explains the
experiment. Note that the
angular momentum
quantum numbers do not
enter the energy
expression.
Homework challenge #4

The special theory of relativity states that a
mass cannot travel faster than the speed of
light. By assuming that the energy of the
ground-state hydrogenic atom is equal to the
negative of the classical kinetic energy (cf.
the virial theorem) of the electron and using
the above speed limit, can we find an upper
limit of the atomic number Z? Does this
explain the fact that there are only 120 or so
atom types in nature and not so many more?
The energy levels
When an electron is
given an energy greater
than that required to
excited into the highest
state, it escapes from the
Coulomb force of the
nucleus – ionization into
an unbound, continuum
state
Z me e
E=2 2 2 2
32p e 0 n
2
4
+
H®H +e
-
There are an infinite number
of bound states with a
negative energy
Atomic orbitals

AO has the form:
Y nlm sm = N nl Rnl (r)Ylm (f ,q )s sm
l


s
l
s
With three orbital quantum numbers: n, l, ml.
n = 1, 2, 3,… Principal quantum number
l = 0, 1, 2,…, n–1
ml = –l, –(l–1),…, (l–1), l
Also spin quantum numbers: s = ½, ms =
±½.  1 1   ,  1 1   They are orthonormal
2
, 2
2
, 2
functions
Shells





The orbitals are classified
by their principal quantum
number n.
n = 1, K shell.
n = 2, L shell.
n = 3, M shell, etc.
Because energies are
determined by n, the
orbitals in the same shell
have the same energy.
Subshells




For each value of n, we
classify the orbitals by
l.
l = 0, s subshell
(1 orbital).
l = 1, p subshell
(3 orbitals because
2l +1-fold degeneracy:
ml = – 1,0,1).
l = 2, d subshell
(5 orbitals).
Atomic orbitals
Principal quantum
number n
3d+2
Angular momentum
quantum number l
Angular momentum
quantum number ml
The s orbitals
The higher the quantum number n, the more
diffuse the orbitals are
6
2.0
1.0
1s
0.8
1.5
2s
5
3s
4
0.6
1.0
0.4
3
2
0.5
1
0.2
2
4
6
8
10
12
14
2
0
2
4
6
8
10
r / a0
12
14
0.5
r / a0
1
4
6
8
r / a0
The higher the quantum number n, the higher the
energy and the more (n–1) nodes
The orbital has a kink
10
12
14
Homework challenge #5


Given the fact that the electron in the
hydrogen atom can exist exactly on the
nucleus, why is it that the energy of the atom
is not −∞ (stability of matter of the first kind)?
Given the fact that the particles in a chemical
system interact through two-body Coulomb
forces, why is it that the energy of the system
grows only asymptotically linearly with the
number of particles (not quadratically as the
number of particle-particle pairs does)
(stability of matter of the second kind)?
The p orbitals: radial part
The higher the quantum number n, the more
diffuse the orbitals are
1.4
5
2p
1.2
4
1.0
3p
3
0.8
2
0.6
1
0.4
5
0.2
10
15
20
25
30
1
5
10
15
20
r / a0
l
æ rö
Rnl (r) = N nl ç ÷ Lnl ( r )e- r /2n
è nø
25
30
2
r / a0
The p orbitals are zero and kinked
at the nucleus.
The number of nodes is n–2.
The pz orbital
Angular part
3
l = 1, m = 1, Ylm =
sin q eij
8p
Radial part
1 æZö
n = 2, l = 1, Rnl =
ça ÷
4 6è 0ø
3/2
re- r /4 l = 1, m = 0, Ylm =
3
cosq
4p
3
l = 1, m = -1, Ylm = sin q e- ij
8p
The product of n = 2 radial function
and l = 1 and ml = 0 angular function
gives rise to r cosθ = z. It is a product
of a spherical s-type function times z.
x  r sin  cos
y  r sin  sin 
z  r cos
The pz orbital
e
- r /4
r cosq = z
z / a0
re- r /4 cosq
z / a0
The px and py orbitals
Angular part
3
l = 1, m = 1, Ylm =
sin q eij
8p
Radial part
1 æZö
n = 2, l = 1, Rnl =
ça ÷
4 6è 0ø
3/2
re- r /4 l = 1, m = 0, Ylm =
3
cosq
4p
3
l = 1, m = -1, Ylm = sin q e- ij
8p
The l = 1 and ml = ±1 angular
functions do not lend themselves to
such simple interpretation or
visualization; they are complex.
x  r sin  cos
y  r sin  sin 
z  r cos
The px and py orbitals
Radial part
1 æZö
n = 2, l = 1, Rnl =
ç ÷
4 6 è a0 ø
Angular part
3/2
re
- r /4
However, we can take the
linear combination of these
to make them align with x
and y axes.
r sinq (eij + e-ij ) = 2r sinq cosj = 2x
3
l = 1, m = 1, Ylm =
sin q eij
8p
3
l = 1, m = 0, Ylm =
cosq
4p
3
l = 1, m = -1, Ylm = sin q e- ij
8p
r sinq (eij - e-ij ) = 2ir sinq sinj = 2iy
We are entitled to take any linear combination of
degenerate eigenfunctions (with the same n and l) to
form another eigenfunction (with the same energy and
total angular momentum but no well-defined ml).
x  r sin  cos
y  r sin  sin 
z  r cos
The px and py orbitals
The d orbitals

The linear combination of d+2, d+1, d0, d–1, d–2 can give
rise to dxy, dyz, dzx, dx –y , d3z –r .
They are degenerate (the same n and l = 2) and have
the same energy and same total angular momentum.
They no longer have a well-defined ml except for ml = 0
(d3z –r ).
2


2
2
2
2
2
Size of the hydrogen atom

Calculate the average radius of the hydrogen
atom in the ground state (the electron is in the
1s orbital).
1 -r/a
Y ( r,q ,j ) =
e
3
p a0
0
ˆ dt =
W = ò Y WY
ò
*
¥
0


0
p
2p
0
0
ò ò
n  ax
xe
ˆ 2 sinq dr dq dj
Y*WYr
n!
dx  n 1
a
The solution
r = ò Y rY d t = ò r Y d t
*
2
1 ¥ p 2p -2r /a0 2
= 3 ò ò ò re
r sinq drdq df
p a0 0 0 0
4 ¥ 3 -2r /a0
3
= 3ò re
dr = a0
2
a0 0
Radial distribution functions

Because the wave function is the product of
radial (R) and angular (Y) parts, the
probability density is also the product …
2
1 = òòò Y n,l ,m r 2 sin q dr dq df
l
2
2
= òòò Rn,l (r) Yl ,m (f ,q ) r 2 sin q dr dq df
l
2
= ò Rn,l (r) r dr
2
2
1 = òò Yl,m (f ,q ) sinq dq df
Radial distribution function
(probability of finding an electron in
the shell of radius r and thickness dr)
l
Size of the hydrogen atom

Calculate the most probable radius of the
hydrogen atom in the ground state (the electron
is in the 1s orbital).
R(r ) =
2
a
3
0
4
e
R(r )
3
2
- r/a0
2
R(r ) r 2
2
1
0
1
2
3
r / a0
4
5
6
The solution
0=
(
2
¶ R r2
¶r
) = 4 æç 2re
-2r /a0
a è
3
0
2r -2r /a0 ö
e
÷
a0
ø
2
r = a0
Most probable
point
Average
radius
Most probable
radius
4
R(r )
3
2
2
R(r ) r 2
2
1
0
1
2
3
r / a0
4
5
6
Summary

Examining the solutions of the hydrogenic
Schrödinger equation, we have learned the
quantum-mechanical explanations of
chemistry concepts such as





discrete energies of the hydrogenic atom
ionization and continuum states
atomic shell structures
s, p, and d-type atomic orbitals
atomic size and radial distribution function

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