Physical Chemistry 2 nd Edition Thomas Engel, Philip Reid Chapter 20 The Hydrogen Atom

Physical Chemistry 2Physical Chemistry 2ndnd Edition EditionThomas Engel, Philip Reid

Chapter 20 Chapter 20 The Hydrogen Atom

© 2010 Pearson Education South Asia Pte Ltd

Physical Chemistry 2nd EditionChapter 20: The Hydrogen Atom

ObjectivesObjectives

• Solve the Schrödinger equation for the motion of an electron in a spherically symmetric Coulomb potential.

• Emphasize the similarities and differences between quantum mechanical and classical models.

• Comparison is made between the quantum mechanical picture of the hydrogen atom.



OutlineOutline

1. Formulating the Schrödinger Equation2. Solving the SchrödingerEquation for the

Hydrogen Atom3. Eigenvalues and Eigenfunctions for the

Total Energy4. The Hydrogen Atom Orbitals5. The Radial Probability Distribution

Function6. The Validity of the Shell Model of an

Atom



20.1 Formulating the Schrödinger Equation20.1 Formulating the Schrödinger Equation

• Hydrogen atom as made up of an electron moving about a proton located at the origin of the coordinate system.

• The two particles attract one another and the interaction potential is given by a simple Coulomb potential:

where e = electron charge me = electron mass

ε0 = permittivity of free space

r

e

r

erV

0

2

0

2

44



20.1 Formulating the Schrödinger 20.1 Formulating the Schrödinger EquationEquation

• As the potential is spherically symmetrical, we choose spherical polar coordinates to formulate the Schrödinger equation.

,,,,4

,,

sin

1

,,sin

sin

1,,1

2

0

2

2

2

2

22

22

rErr

e

r

r

r

rr

rr

rr

m

h

e



20.2 Solving the Schrödinger Equation20.2 Solving the Schrödinger Equation for the Hydrogen Atom for the Hydrogen Atom

• Separation of variables

• Thus the differential equation for R(r) is obtained.

• The second term can be viewed as an effective potential.

rERrRr

e

rm

llh

dr

rdRr

dr

d

rm

h

ee

0

2

2

22

2

2

42

1

2

r

e

rm

llhrV

eeff

0

2

2

2

42

1



20.2 Solving the Schrödinger Equation20.2 Solving the Schrödinger Equation for the Hydrogen Atom for the Hydrogen Atom

• Each of the terms that contribute to Veff (r) and their sums can be graphed as a function of distance.



20.3 Eigenvalues and Eigenfunctions20.3 Eigenvalues and Eigenfunctions for the Total Energy for the Total Energy

• Note that the energy, E, only appears in the radial equation and not in the angular equation.

• R(r) can be well behaved at large values of r [R(r) → 0 as r → ∞].

where (Bohr radius) • The definition leads to

.1,2,3,4,.. for , 8 222

0

4

nnh

emE en

529.02

20

0 em

ha

e

,...4,3,2,1 60.1310179.2

8 22

18

200

2

nn

eV

n

J

na

eEn




• The other two quantum numbers are l and ml, which arise from the angular coordinates.

• Their relationship is given by

lm

nl

n

l

,...3,2,1,0

1,...,4,3,2,1,0

,...4,3,2,1




• The quantum numbers associated with the wave functions are




• The quantum numbers associated with the wave functions are



Example 20.1Example 20.1

Normalize the functions in three-dimensional spherical coordinates.

iarar eeare sin/ and 00 20

2



SolutionSolution

In general, a wave function is normalized by multiplying it by a constant N defined by . In three-dimensional spherical coordinates, it isThe normalization integral

For the first function,

t

ddrdrd sin2

1*2 dN

1,,,,*sin 2

0

2

00

2

drrrrddN

1sin0

22/2/2

00

2 00

drreeddN arar



SolutionSolution

We use the standard integral

Integrating over the angles , we obtain

Evaluating the integral over r,

140

22/2/2 00

drreeN arar

10

!

naxn

a

ndxex

23

030

2 1

22

1or 1

1

!24

aN

aN

and



SolutionSolution

For the second function,

This simplifies to

Integrating over the angles using the result , we obtain

13

8 2/

2

0 0

2 0

drrea

rN ar

1sinsinsin 2/

00

/

0

2

00

2 00

drreea

ree

a

rddN iariar

and

1sin 2/

2

0 0

2

00

32 0

drrea

rddN ar



SolutionSolution

Using the same standard integral as in the first part of the problem,

2/3

050

20

2 1

8

1or 1

/1

!41

3

8

a

Naa

N




• The angular part of each hydrogen atom total energy eigenfunctions is a spherical harmonic function.




a. Consider an excited state of the H atom with the electron in the 2s orbital.Is the wave function that describes this state,an eigenfunction of the kinetic energy? Of the potential energy?

b. Calculate the average values of the kinetic and potential energies for an atom described by this wave function.

02/

0

2/3

0200 2

1

32

1 area

r

ar



SolutionSolution

a. We know that this function is an eigenfunction of the

total energy operator because it is a solution of the

Schrödinger equation. You can convince yourself that

the total energy operator does not commute with either

the kinetic energy operator or the potential energy

operator by extending the discussion of Example

Problem 20.1. Therefore, this wave function cannot

be an eigenfunction of either of these operators.



SolutionSolution

b. The average value of the kinetic energy is given by

0

/440

/330

/

0

/220

30

0

220

203

0

2/2/

030

0

22/

0

22

2/

00

2

030

0000

0

0

00

4

1389

8

1

2

10164

28

1

2

21

2sin32

1

2

ˆ*

arararar

arar

arar

kinetickinetic

era

era

rea

eraa

h

drrrraara

ee

a

r

a

h

drrea

r

dr

dr

dr

d

re

a

rdd

a

h

dEE



SolutionSolution

We use the standard integral,

Using the relationship

20

2

1

0

8/

:/!

ahE

andxex

kinetic

naxn

2for ,32 00

2

nEa

eEkinetic



SolutionSolution

The average potential energy is given by

2for 216

28

1

4

144

8

1

4

21

2sin32

1

4

ˆ*

00

2

203

00

2

0 0 0

/320

/2

0

/300

2

0

22/

0

2/

00

2

0300

2

000

00

nEa

e

aa

e

drera

drera

drea

e

drea

r

re

a

rdd

a

e

dEE

n

ararar

arar

potentialkinetic



SolutionSolution

We see that

The relationship of the kinetic and potential energies is a specific example of the virial theorem and holds for any system in which the potential is Coulombic.

kineticpotentialtotalpotential EEEE 2 and 2




• The radial distribution function is used to extract information from the H atom orbitals.

• We first look at the ground-state (lowest energy state) wave function for the hydrogen atom,

• We need a four-dimensional space to plot as a function of all its variables.

0/

2/3

0100

11 area

r



20.4 The Hydrogen Atom Orbitals20.4 The Hydrogen Atom Orbitals

• Since such a space is not readily available, the number of variables is reduced.

• It is reduced by evaluating in one of the x–y, x–z, or y–z planes by setting the third coordinate equal to zero.

• r are spherical nodal surfaces rather than nodal points (one-dimensional) potentials.

222 zyxr




Locate the nodal surfaces in

Solution:The radial part of the equations is zero for finite values of for .

This occurs at .

cos612

81

1,, 03/

20

2

0

2/3

0

2/1

310are

a

r

a

r

ar

0/ ar 0//6 20

20 arar

06 and 0 arr



20.5 The Radial Probability Distribution 20.5 The Radial Probability Distribution FunctionFunction

• 3D perspective plots of the square of the wave functions for the orbitals is indicated.




a. At what point does the probability density for the electron in a 2s orbital have its maximum value?

b. Assume that the nuclear diameter for H is 2 × 10-15 m. Using this assumption, calculate the total probability of finding the electron in the nucleus if it occupies the 2s orbital.



SolutionSolution

a. The point at which and , therefore, has its greatest value is found from the wave function:

which has its maximum value at r=0, or at the nucleus

02/

0

2/3

0200 2

1

32

1 area

r

ar

d*



SolutionSolution

b. The result obtained in part (a) seems unphysical, but is a consequence of wave-particle duality in describing electrons. It is really only a problem if the total probability of finding the electron within the nucleus is significant. This probability is given by

drea

rrdd

aP ar

rnucleus0/

2

00

2

0

2

00

2sin1

32

1



SolutionSolution

Because , we can evaluate the integrand by assuming that is constant over the interval

0arnucleus

nucleusrr 0

15

0

/0

3/

2

0

3

0

0

2/

2

0

3

0

100.96

1

,1 and 2/2 Because

3

42

1

32

1

241

32

1

0

0

0

a

rP

ear

rea

r

a

drrea

r

aP

nucleus

arnucleus

nucleusarnucleus

rarnucleus

nucleus

nucleus

nucleus

nucleus

0/20/2 arear



SolutionSolution

Because this probability is vanishingly small, even though the wave function has its maximum amplitude at the nucleus, the probability of finding the electron in the nucleus is essentially zero.



20.5 The Radial Probability Distribution 20.5 The Radial Probability Distribution FunctionFunction

• It is most meaningful for the s orbitals whose amplitudes are independent of the angular coordinates.

• The radial distribution P(r) is the probability function of choice to determine the most likely radius to find the electron for a given orbital drrRrdrrP 22




Calculate the maxima in the radial probability distribution for the 2s orbital. What is the most probable distance from the nucleus for an electron in this orbital? Are there subsidiary maxima?



SolutionSolution

The radial distribution function is

To find the maxima, we plot P(r) and

versus and look for the nodes in this function.

0/3220

306

0

16832

arerraaa

r

dr

rdP

0/

2

0

2

3

0

22 21

32

1 area

rr

arRrrP

0/ ar



SolutionSolution

These functions are plotted as a function of in the following figure:

0/ ar



SolutionSolution

The resulting radial distribution function only depends on r, and not on . Therefore, we can display P(r)dr versus r in a graph as shown

and



20.6 The Validity of the Shell Model 20.6 The Validity of the Shell Model of an Atom of an Atom

• The idea of wave-particle duality is that waves are not sharply localized.

Documents

Physical Chemistry 2 nd Edition Thomas Engel, Philip Reid Chapter 20 The Hydrogen Atom