The Quantum Mechanical Treatment of the Hydrogen Atom

Motivation:

Hydrogen atom orbitals and energy levels form the basis of an

understanding of larger molecules.

covalent bonding molecular orbitals hybridization

In fact, the results of the quantum mechanical treatment of the

hydrogen atom permeate through our modern description of

chemistry.

For example, we often speak of the

p-bond in a C-C double bond as a

result of the interaction between two p

orbitals on each carbon.

The orbitals we so often speak of come from the quantum

mechanical treatment of the hydrogen atom.

H atom ground state

electron density

total electron

density of

benzene

3-D plot of the p-bonding orbital in ethylene from a quantum

chemical calculation.

The Hydrogen Atom – The Physical Model

The hydrogen atom is a two-particle system

consisting of a proton and an electron.

We will treat the proton and electron as point masses that interact

via Coulomb’s Law given by (in SI units):

2 1( )

4 o

eV r

rp

r is the distance between the electron and nucleus.

e is the charge of the electron:

o is the vacuum permittivity constant and in SI units has a value of:

191 602 10e . x C

12 2 2 1 38.854188x10 o C s kg m

p+

e-

How does this

potential compare

to harmonic

oscillator?

r

The quantum mechanical Hamiltonian for this system is:

p+

e-2 2222

2

1

2ˆ

4 ope

ep

mH

rm

e

p

This is a two-particle system, and just like the Harmonic Oscillator and

Rigid Rotor problem, we can separate the Hamiltonian into a center of

mass coordinate and a relative coordinate system:

translational motion of the center of mass

22ˆ

2CM CM

p e

Hm m

relative or internal motion

2 22

int

1ˆ2 4 o

eH

r p

where is the reduced mass

e p

e p

m m

m m

intˆ ˆ ˆ

total CMH H H

2 2 22 2 1ˆ

2 42total CM

op e

eH

rm m p

intˆ ˆ

CMH H

This Hamiltonian is separable, with the solution of the center of mass

Hamiltonian being that of the free particle, we have previously

examined.

intĤ E

We want to find the wave functions that govern the internal motion of

the system.

These solutions are the hydrogen atom wave functions that we are

interested in.

2 22

int

1ˆ ˆ2 4 o

eH H

r p

The Hamiltonian operator for the internal energy of the hydrogen atom is:

kinetic energy of

internal motionpotential energy,

Coulomb’s law.

In many textbooks, you will often see the hydrogen-atom Hamiltonian

written with me, the mass of the electron, instead of .2 2

2 1ˆ2 4e o

eH

m rp

Who is correct? Are these other textbooks

incorrect?

e p

e p

m m

m m

This problem has the same spherical symmetry as the 3-D rigid rotor,

so this suggests that it would be easier to solve our problem in polar

spherical coordinates.

x

y

z

r

r0

p 0

2π0

What do we need to do to convert

our Hamiltonian into polar

spherical coordinates?

2 22 1ˆ

2 4 o

eH

r p

( , , )r

Schrödinger’s equations for the H atom is Best Solvedin Polar Spherical Coordinates

Polar Spherical Coordinates

The Laplacian operator in polar spherical coordinates is:

22 2

2 2 2 2 2

1 1 1sin

sin sinr

r rr r r

(there is no need to memorize the above, but you

should know how to work with it and what it means)

The Integration Element

In Cartesian coordinates the integration over all

space involves the following integration element:

d dx dy dz The equivalent integration element in polar

spherical coordinates is given by:

2 sin d r dr d d dr

rd

In polar spherical coordinates the Schrodinger equation becomes:

Unlike the rigid rotor, r is not fixed because the electron is free to movecloser or farther away from the nucleus.

ˆ ( , , ) ( , , )H r E r

2

2

2

2

22

2 2 2

1 1sin

sin s2 i

1

n

1

4 o

er

r r rr rE

r

p

Therefore, we cannot simplify the Laplacian further. This is unfortunately

the Schrödinger equation we have to work with!

The Hamiltonian is not separable, but if we multiply both sides of the

SE by 2r2, it becomes pseudo-separable.

2

2 2

2 22 2 22 1 2

1 1sin

sin si 4n o

r er r E

r r r

p

The above is pseudo-separable, so we can collect angular components

( and ) on one side and radial components on the other side.

depends only on and

22 2

2 2 22

2 2

2 12

4

1 1sin

sin sino

r er r E

r r r

p

angular dependence

(depends on and )radial dependence

(depends on r)

Collecting sides gives:

This suggests that we can write our total hydrogen atom wave function

as a product of a radial and angular functions.

The Hydrogen Atom Wave Function can be Written as a

Product of Radial and Angular Components

Using this form of the wave function and substituting into the last

equation on the previous slide and rearranging gives:

2 2 22

2 2

22

21 2 1 24

1 1sin

sin sino

r er r E R

R r r rA

A

p

2 22 2 21 2 1 2 ( )

( ) 4 o

r er r E R r lhs

R r r r r

p

Because the lhs depends only on the variable r, and the rhs depends only

on the angular variables and , each side must be equal to a constant.

2 2

2 2

1 1sin ( , )

( , ) sin sinrhs A

A

),(),,( AR(r)r

The Angular Part of the Hydrogen Atom Wave Function

Let’s first examine the angular part of the hydrogen atom Schrödinger

equation:2 2

2 2

1 1sin ( , )

( , ) sin sinA

A

Multiplying both sides by A gives:

22

2 2

1 1sin ( , ) ( , )

sin sinA A

The above is simply an eigenvalue equation, where the operator is the L2

operator.

2L̂

2ˆ ( , ) ( , )L A A

Haven’t we already solved these already? What is A?

Spherical Harmonics Make up the Angular Component of

the Total Hydrogen Atom Wave Function

The spherical harmonics are eigenfunctions of the L2 operator, recall:

22ˆ (( , ) ( , )1)m mY YL ...2,1,0

,...2,1,0m

2 ( , )ˆ ( , )AL A

Comparing the above and our angular Schrödinger equation

tells us that:

( , ) ( , )mA Y 2 ( 1)

So the angular component of the total wave function of the hydrogen

atom are the spherical harmonics:

),()(),,( , mYrRr

200Y

210Y

21 1Y

xy

z

220Y

This is why the angular plots of the spherical

harmonics looked like our familiar s, p and d

orbitals.

Some of them don’t, like the donut below (to be

discussed shortly).

We have solved for the angular part of

the total hydrogen atom wave function,

now we need to solve for the radial

component.

Solving for the Radial Component of the H-Atom Wave Function

From a few slides back, we had an expression for the radial wave

function R(r),2 2

2 2 21 2 1 2 ( )( ) 4 o

r er r E R r

R r r r r

p

From the solution of the angular part, we know that rearranging

2 ( 1) equation and dividing through by 2r2 gives:

2 2 22

2 2

1 ( 1) 1( ) ( )

4 22 o

er R r ER r

r r rr rp

This is known as the radial Schrödinger equation for the hydrogen

atom. This is the only new equation we have to solve for to get our total

wave function for the hydrogen atom.

Again, we will not explicitly solve this equation. The solutions will simply

be presented and analyzed.

Radial Wave Functions of the Hydrogen Atom

)(rRnnormalization

constant

exponential decay

function:polynomial

in ronare

/

The solution of the radial Schrödinger equation gives what the radial

wave function that depends on a new quantum number, n.

)(rRn...3,2,1 n

0, 1, 2... 1n

Notice the new restriction on the quantum number ℓ (why?) and that that

radial wave function does not depend on the quantum number m.

The general form of the radial wave function is:

All hydrogen atom wave functions decay exponentially away from the

nucleus.

3/ 2

10

1 /2 o

o

r aR e

a

3/ 2

20

1 1 / 2(1 )

22o

o o

r r aR e

a a

3/ 2 2

30 2

2 1 2 2 / 31

33 3 27

o

o o o

r r r aR e

a a a

5/ 2

21

1 1 / 2

2 6o

o

r aR re

a

7 / 2

232

4 1 / 3

81 30o

o

r aR r e

a

3/ 2 2

31 2

8 1 / 3

27 6 6

o

o o o

r r r aR e

a a a

Normalized Radial Functions of the Hydrogen Atom

)(rRn...3,2,1 n

0, 1, 2... 1n n = 1

n = 2

n = 3

Bohr radius ao = 0h2/pe2

= 0.0529 nm

r (ao)

r (ao)

r (ao)

The radial wave functions are plotted as

a function of r in units of

Bohr radii ao = (oh2)/(pe2)

3/ 2

10

1 /2 o

o

r aR e

a

5/ 2

21

1 1 / 2

2 6o

o

r aR re

a

r (ao)

Plots of Radial Wave Functions

The Energy of the Hydrogen Atom

2 2 22

2 2

1 ( 1) 1( ) ( ) 0

4 22 o

er R r E R r

r r rr rp

The energy of the hydrogen atom can be derived by solving the radial

Schrödinger equation. It is given by:

4 2

2 2 2 2 2

1 1

832 o oo

e eE

an n

pp - ...3,2,1 n

The total energy of the hydrogen atom depends only on the quantum

number n, and not on the other two quantum numbers, ℓ and m.

NOTE that E is the total energy (kinetic + electrical potential energy)of the hydrogen atom, not just the ‘radial energy’.

The energy levels are essentially

in exact agreement with the

discrete line spectra of atomic

hydrogen.1 2

2

2 22 1

1 1

8n ,n

o o

eΔE

a n np

One of the most powerful experimental verifications of the quantum

mechanical description of the hydrogen atom involves the emission

spectra of hydrogen atoms. Selection rule for H-atom electronic

transitions:

∆n unrestricted

Lyman Balmer Paschen

Ultraviolet Visible Infrared

656.5

nm

Experimental Verification of the Hydrogen-Atom Solution

The numerical value of the ground state energy

(13.59670 eV) is given by:

The Ground State Energy and the Ionization Energy

2

1 2

11311.9 kJ/mol

8 1o o

eE

ap -

The ionization energy is the energy required to

remove an electron from the ground state of an

atom.

Experimentally, the hydrogen atom ionization energy

is 1311.9 kJ/mol.

This is another powerful way the quantum

mechanical description of the hydrogen atom can be

confirmed experimentally.

n=1

n=2

n=3

n=4n=0

energ

y

1311.9 kJ/mol

2

2

1

8 o o

eE

a np - ...3,2,1 n

The hydrogen atom energy levels are depicted to the

right.

n = 1

n = 2

0

energ

y

n = 3

n =4

Notice that the energies are all negative, and as

n increases, the energy tends to zero.

Does this make sense?

Lagoon Nebula using the VLT (very large telescope) in Chile

Trifid Nebula

Red: H alpha

(n = 3 to n = 2)

Blue: scattered

light

Dark brown:

dust

Rosette Nebula using the VLT

),),,( ((r)YRr mnmn

n = 1, 2, 3…

ℓ = 0, 1, 2…(n 1)m = 0, ±1, ±2…±ℓ

Total Wave Functions for “Hydrogen-Like” Atoms(One-Electron Systems: H, He+, Li2+, Be3+, …)

The total wave function of the hydrogen atom is the product of the

radial wave function R(r) and the spherical harmonic Ynl(,):

Notice that there are three distinct quantum numbers that characterize

the state of a hydrogen atom.

We, therefore label the total wave function with these three quantum

numbers, n, ℓ and m, usually in this order.

0 0, 1, 100

mn 1 2, 3, 132 mn

Considering all possible combinations of ℓ and m

consistent with n, we find that the degree of

degeneracy, g, of hydrogen atoms is:

Degeneracies of the Hydrogen-Like Atom States

H(Z = 1), He+ (Z = 2), Li2+(Z = 3), Be3+(Z = 4), …

ℓ = 0, 1, 2…(n-1)m = 0, ±1, ±2…±ℓ

The energy levels are highly degenerate because the

energy depends only on the quantum number n.

100

200 210 211 21-1

300 310 311 31-1 320 321 32-1 322 32-2

Energ

y

g = n2

2

00

2

222

0

4 1

8

1

8)1,atom H(

na

e

nh

eZEn

2

00

22

222

0

42 1

8

1

8)(

na

eZ

nh

eZZEn

n = 1, 2, 3, …

3/ 2

100

1 1 / o

o

r ae

a

p

3/ 2

200

1 1 / 22

4 2o

o o

r r ae

a a

p

5/ 2

21 1

1 1 / 2sin

8o

o

r a ire e

a

p

3/ 2 2

300

1 1 18 / 3(27 2 )

81 3o

o o o

r r r ae

a a a

p

3/ 2 2

31 1

1 1 6 / 3( ) sin

81o

o o o

r r r a ie ea a a

p

7 / 2

232 1

1 1 / 3sin cos

81o

o

r a ir e e

a

p

7 / 2

2 232 2

1 1 / 3 2sin

162o

o

r a ir e e

a

p

n = 2, ℓ = 0

n = 1, ℓ = 0

n = 2, ℓ = 15/ 2

210

1 1 / 2cos

4 2o

o

r are

a

p

n = 3, ℓ = 0

n = 3, ℓ = 1

n = 3, ℓ = 27 / 2

2 2320

1 1 / 3(3cos 1)

81 6o

o

r ar e

a

p

Complete Hydrogen Atom Wave Functions

wave functions Ynlm(r,,)for n = 1, 2, and 3

3/ 2 21/ 2

310

1 2 1 6 / 3( ) cos

81o

o o o

r r r ae

a a a

p

Probability distribution functions

Y*nlm(r,,) Ynlm(r,,) r2sindrdd

for n = 1, 2, 3

Show that the hydrogen atom wave function Y100 is normalized.

o

2/3

o

100

/11 are

a

p

Show that the hydrogen atom wave function Y100 is orthogonal to

the Y211 wave function.

p

ie

arre

a

sin

2/1

8

1o

2/5

o

211

The One-Electron Wave Function

The hydrogen atom wave functions were derived by considering the

two-body (proton + electron) system as an effective “one-body” system.

Due to the large mass difference of the proton and electron, the

hydrogen atom Hamiltonian and wave functions essentially describe the

motion of the electron.Why?

A One-Electron Wave Function is Called an Orbital

The hydrogen atom wave functions are often called orbitals, a term

frequently used in chemistry. Formally an orbital has the following

definition:

An orbital is the wave function describing a

single electron. In other words, an orbital is

a one-electron wave function.

What do Hydrogen Atom Wave Functions ‘Look’ Like?

The hydrogen atom wave functions are 3-dimensional functions of space.

In other words, associated with every point in space, is a value of that

wave function. There are many different ways to visualize wave

functions.

3D-Isosurface Plots

When a chemist thinks of a 3-D representation of a

orbital, such as a 210 or 2pz orbital, they usuallyhave a 3-D isosurface plot in mind.

An isosurface plot is 3-dimensional surface of which

the function has the same value. Thus, associated

with each isosurface plot, should be a value of that

surface. r = 0.02 a.u.

Shown to the right is the isosurface plot of the

density of the 210 orbital:

2210

How ‘big’ the orbitals are when we visualize them depends on what

isosurface value we choose.

Due to the exponential decay

of all H-atom wave functions,

the larger the isosurface value,

the ‘smaller’ the orbital is

depicted in an isosurface plot.

The size of an Orbital in an Isosurface plot is Arbitrary

0 1.5 2.0 2.5

4.0

2.0

1.0 3.00.5

r/a0

|100|2

large isosurface

value

small isosurface

value

When wave functions are plotted, the color

represents the sign of the wave function.

We will discuss complex-valued wave functions

later.

23 zdΨ

2200

2300

2320

2210 2

21 1

2310

231 1

232 1

232 2

Isosurface plots of the densities

of the n = 1, 2 and 3 orbitals

2100

z

Spectroscopic Designation of States

What about our familiar s, p and d orbitals? There is a relation between

the quantum numbers ℓ and the s, p, d labels we give the atomicorbitals.

s p d f g h

ℓ = 0 1 2 3 4 5

Ignoring the quantum number m for the time being, we have:

1s = 10 n = 1, ℓ = 0

2s = 20 n = 2, ℓ = 0

2p = 21 n = 2, ℓ = 1

3s = 30 n = 3, ℓ = 0

3p = 31 n = 3, ℓ = 1

3d = 32 n = 3, ℓ = 2

Notice that we are not specifying the x, y, z , x2-y2 etc. states yet.

Real and Directed Orbitals

In chemistry, we often speak about the s, px, py, pz and the five

d orbitals, dxy, dyz, dxz, dz2 and dx2-y2.

How are these orbitals, that we often use to explain chemical bonding,

related to the atomic hydrogen orbitals? To illustrate this point, consider

the n = 2 states, where we would expect to have:

2s 2 zp 2 yp

2 xp

200 210 211 21 1

Are they

Equivalent?

Consider the 200 wave function:3/ 2

200

1 1 / 22

4 2o

o o

r r ae

a a

p

There is no angular dependence, meaning that

this is a perfectly spherical function.

x

y

z

r

r0

p 02π0

Indeed, the n = 2, ℓ = 0, m = 0 wave function is

the familiar 2s orbital.

200 2s

What about the 210 wave function which is

given as:5/ 2

210

1 1 / 2cos

4 2o

o

r are

a

p

An isosurface plot of the 210 wave function shown tothe right shows that this is indeed our familiar 2pzorbital. (Here, positive is “red”, negative is “blue”.)

z

5/ 2

2

1 1 / 2cos

4 2zo

po

r are

a

p

cosz r

What about the 21+1 wave function? Is this one of the 2px or 2pyorbitals?

This is a complex function with both real and imaginary components,

making it a bit difficult to plot.

5/ 2

21 1

1 1 / 2sin

8o

o

r a ire e

a

p

We know however that the square of the wave function is always

positive, so let’s try plotting this:

221 1

This means that:5

2 2 2211

1 1 /sin

64o

o

r ar e

a

p

We will get the same result for the 21-1 orbital!

52 2 2

21 1

1 1 /sin

64o

o

r ar e

a

p

When we plot this we get a donut isosurface plot!

221 1

So what are all of the 2px, 2py, and other

orbitals we are familiar with?

Are they donuts?

Are they complex?

Real and Directed Orbitals are Linear Combinations of

the Complex Set We have Derived

The orbitals we use in chemistry are purely real functions that are

constructed as a linear combination of the wave functions we have

derived.

12121122

1-p ΨΨΨ x

2 211 21 11

2yp -Ψ Ψ Ψ

i

Show this for the 2px orbital.

Show that the 2px orbital as defined above is normalized.

The Energies of the Real and Directed Orbitals

Do the real and directed orbitals still satisfy the Schrödinger equation

for the hydrogen atom and what are their energies?

Consider the 2py orbital which is a linear combination of the 211

and 21-1 orbitals.

Indeed, any linear combination of degenerate wave functions is also a

solution to the Schrödinger equation with the same energy.

2 211 21 11

2yp -Ψ Ψ Ψ

i

200 210 211Ψ AΨ BΨ CΨ

Show this for the following linear combination:

2px orbital directed along the x axis:

p

p

p

cossin1

4

1

)sincossin(cossin1

8

1

esinesin1

28

1

)(2

1

0

0

0

2/

2/5

0

2/

2/5

0

2/

2/5

0

121211p2

ar

ar

iiar

x

rea

iirea

rea

2py orbital directed along the y axis:

p

p

p

sinsin1

4

1

)sincossin(cossin1

28

1

esinesin1

28

1

)(2

1

0

0

0

2/

2/5

0

2/

2/5

0

2/

2/5

0

121211p2

ar

ar

iiar

y

rea

iireai

reai

i

Is the 2px orbital normalized?

1

)11(2

1

d2

1

d2

1

d2

1

d22

d

*

121

*

121211

*

211

*

121

*

121211

*

121121

*

211211

*

211

121211

*

121211

121211

*

121211

p2

*

p2

xx

Any linear combination of degenerate wave functions

has the same energy.

2

2112102002

211221022002

211210200

211210200

211210200

)(

ˆˆˆ

ˆˆˆ

)(ˆˆ

E

CBAE

CEBEAE

HCHBHA

CHBHAH

CBAHH

The Real and Directed d-Orbitals

The d-orbitals that we often use in transition metal chemistry are also

normalized linear combinations of the complex functions we have

derived.

What about the n = 3, ℓ = 2, m = 0 hydrogen atom orbital which is a

purely real function?7 / 2

2 2 2320

1 1 / 3(3 cos )

81 6o

o

r ae r r

a

p

z = r cos 2 23z r

So our familiar dz2 orbital is shorthand for a 3z2 r2 orbital.

The other d orbitals we are familiar with are linear combinations of the

remaining four m = ±1 and m = ±2 functions.

Linear combinations used to construct nd orbitals:

p

p

p

p

p

2sinsin16

15)()(

2

1d

2cossin16

15)()(

2

1d

sincossin4

15)()(

2

1d

coscossin4

15)()(

2

1d

)1cos3(16

5)(d

2

22222

2

22222)(

21221

21221

2

220

22

2

rRi

rR

rRi

rR

rR

nnnnxy

nnnyxn

nnnnyz

nnnnxz

nnnz

The familiar d-orbitals are the ‘real and directed’ d-orbitals.

Here, “positive” is blue and “negative” is yellow.

dx2-y2 dz2

dxy dxz dyz

The real and directed 4f orbitals

Angular Momentum of the Hydrogen Atom

The hydrogen atom energy levels are

n2-fold degenerate.

g = n2En

erg

y

1, 0, 0n m

2, 0,1, 1,0,1n m

What is different about the states

if the energies are the same?

Each of the states has a different angular momentum. In other words,

these states each have different values for L2 and Lz.

Recall that the total hydrogen atom wave function is a product of the

radial wave function and the spherical harmonics.

),),,( ((r)YRr mnmn

The spherical harmonics are eigenfunctions of the L2 and Lz operators:22ˆ (( , ) ( , )1)m mY YL

, ) ( , )ˆ (m mzY YL m

ˆzL i

22 2

2 2

1 1ˆ sinsin sin

L

Because the L2 and Lz operators act only on and ,

We know that the hydrogen atom wave functions are eigenfunctions of

the L2 and Lz operators. Also, we know that each of the hydrogen atom

states has a definite square of the angular momentum and z-component of the angular momentum.

Show this

The Quantum Numbers ℓ and m Define the Angular Momentum of the Hydrogen Atom State

The quantum number ℓ defines the square of the magnitude of theangular momentum vector of the state:

2 2 ( 1)L

The quantum number m defines the z-component of the angularmomentum vector of the state:

zL m

Because the angular momentum of the hydrogen atom is due to the

angular motion of the electron, it is often referred to as the orbital

angular momentum.

Sketch the angular momentum vectors of the

hydrogen atom in a 1s and 2pz orbitals.

The H-Atom Quantum Numbers have Common Names

The principle quantum number, specifies the energy of the

hydrogen atom.n

The angular momentum quantum number, (sometimes called

the azimuthal quantum number), specifies the magnitude of the

orbital angular momentum of the state.

ℓ

2 2 ( 1)L 0,1,2... 1n

2

2

1

8 o o

eE

a np - ...3,2,1 n

The magnetic momentum quantum number, specifies the

z-component of the angular momentumm

zL m 0, 1, 2...m

The Spatial and Nodal Structure of the Hydrogen Atom Orbitals

When using atomic orbitals to discuss molecular bonding, the spatial

and nodal structure of the wave functions is important

Motivation

For example, the overlap between orbitals on different atoms, either

constructive, destructive or zero overlap, is highly dependent upon the

nodal structure of the orbitals involved.

We will discuss this in terms of the purely real and directed orbitals.

Recall both sets of orbitals are equivalent, and can be obtained from

one another by rotation.

Nodes of the Wave Function

Recall that nodes are where the wave function changes sign and has

zero value.

+

-+ -

Radial and Angular Nodes

The number of nodes is determined by the quantum numbers n and ℓ.The total number of nodes is given by:

total number of nodes = n 1

So the ground state hydrogen atom wave function 100 has no nodes.

Check this!

The nodes can be categorized as radial and angular nodes arising from

each component of the total hydrogen atom wave function, R and Y.

number of radial nodes = n – ℓ 1

number of angular nodes = ℓ

Angular Nodes

The real and directed orbitals have nodal planes (or equivalent) where

the wave function changes sign. The number of angular nodes is given

by the quantum number ℓ.

ℓ = 0, the ‘s’ orbitals

The s orbitals have no angular dependence, and are spherical wave

functions and therefore have no angular nodes.

ℓ = 1, the ‘p’ orbitals

The p orbitals have one angular plane, each.

+

-x

y

z

+- x

y

z

x

y

z

+ -

dxz dxy dyz

x

y

z

+

-

-

+

x

y

z

-++

-

x

y

z

++-

-

+ x

y

z

+

-

-

dx2-y2 dz2

x

y

z

+

+-

ℓ = 2, the ‘d’ orbitals

The d orbitals have two angular planes each.

Where are the two angular nodes of the dz2 orbital?

2

7 / 2

2 2320 3

1 1 / 3(3cos 1)

81 6o

dzo

r ar e

a

p

Radial Nodes

Plots of the radial wave function show that the number of radial nodes is

given by:number of radial nodes = n ℓ 1

Below are plotted the 1s, 2s, 3s and 4s radial wave functions.

r(ao)r(ao) r(ao) r(ao)

The radial nodes give isosurface plots of the electron density that are

layered like an onion.

22s

23s

Explain

The nodes of the radial wave functions give the higher angular

momentum wave functions a similar structure.

2

2 zp

2

3 zp

Sketched below are 2-D slices of the isosurface plots of the wave

function.

Notice the alternation of the sign of the layers and the larger spatial

extent as n varies from 2 to 4.

2 zp 3 zp 4 zp

+

-

+

-

+

-

-

++-

The same layered structure exists for the 4d and 5d orbitals!

The reason we don’t sketch or represent the inner orbitals when

discussing bonding is because it is only the outer ‘layer’ that participates

in the interaction.

4 xzd

+

+ --

-

-

+

--

++

+

-

-

24 zd