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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.
© 2010 Pearson Education South Asia Pte Ltd
Physical Chemistry 2nd EditionChapter 20: 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
© 2010 Pearson Education South Asia Pte Ltd
Physical Chemistry 2nd EditionChapter 20: The Hydrogen 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
© 2010 Pearson Education South Asia Pte Ltd
Physical Chemistry 2nd EditionChapter 20: The Hydrogen Atom
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
© 2010 Pearson Education South Asia Pte Ltd
Physical Chemistry 2nd EditionChapter 20: The Hydrogen Atom
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
© 2010 Pearson Education South Asia Pte Ltd
Physical Chemistry 2nd EditionChapter 20: The Hydrogen Atom
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.
© 2010 Pearson Education South Asia Pte Ltd
Physical Chemistry 2nd EditionChapter 20: The Hydrogen Atom
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
© 2010 Pearson Education South Asia Pte Ltd
Physical Chemistry 2nd EditionChapter 20: The Hydrogen Atom
20.3 Eigenvalues and Eigenfunctions20.3 Eigenvalues and Eigenfunctions for the Total Energy for the Total Energy
• 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
© 2010 Pearson Education South Asia Pte Ltd
Physical Chemistry 2nd EditionChapter 20: The Hydrogen Atom
20.3 Eigenvalues and Eigenfunctions20.3 Eigenvalues and Eigenfunctions for the Total Energy for the Total Energy
• The quantum numbers associated with the wave functions are
© 2010 Pearson Education South Asia Pte Ltd
Physical Chemistry 2nd EditionChapter 20: The Hydrogen Atom
20.3 Eigenvalues and Eigenfunctions20.3 Eigenvalues and Eigenfunctions for the Total Energy for the Total Energy
• The quantum numbers associated with the wave functions are
© 2010 Pearson Education South Asia Pte Ltd
Physical Chemistry 2nd EditionChapter 20: The Hydrogen Atom
Example 20.1Example 20.1
Normalize the functions in three-dimensional spherical coordinates.
iarar eeare sin/ and 00 20
2
© 2010 Pearson Education South Asia Pte Ltd
Physical Chemistry 2nd EditionChapter 20: The Hydrogen Atom
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
© 2010 Pearson Education South Asia Pte Ltd
Physical Chemistry 2nd EditionChapter 20: The Hydrogen Atom
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
© 2010 Pearson Education South Asia Pte Ltd
Physical Chemistry 2nd EditionChapter 20: The Hydrogen Atom
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
© 2010 Pearson Education South Asia Pte Ltd
Physical Chemistry 2nd EditionChapter 20: The Hydrogen Atom
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
© 2010 Pearson Education South Asia Pte Ltd
Physical Chemistry 2nd EditionChapter 20: The Hydrogen Atom
20.3 Eigenvalues and Eigenfunctions20.3 Eigenvalues and Eigenfunctions for the Total Energy for the Total Energy
• The angular part of each hydrogen atom total energy eigenfunctions is a spherical harmonic function.
© 2010 Pearson Education South Asia Pte Ltd
Physical Chemistry 2nd EditionChapter 20: The Hydrogen Atom
Example 20.2Example 20.2
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
© 2010 Pearson Education South Asia Pte Ltd
Physical Chemistry 2nd EditionChapter 20: The Hydrogen Atom
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.
© 2010 Pearson Education South Asia Pte Ltd
Physical Chemistry 2nd EditionChapter 20: The Hydrogen Atom
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
© 2010 Pearson Education South Asia Pte Ltd
Physical Chemistry 2nd EditionChapter 20: The Hydrogen Atom
SolutionSolution
We use the standard integral,
Using the relationship
20
2
1
0
8/
:/!
ahE
andxex
kinetic
naxn
2for ,32 00
2
nEa
eEkinetic
© 2010 Pearson Education South Asia Pte Ltd
Physical Chemistry 2nd EditionChapter 20: The Hydrogen Atom
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
© 2010 Pearson Education South Asia Pte Ltd
Physical Chemistry 2nd EditionChapter 20: The Hydrogen Atom
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
© 2010 Pearson Education South Asia Pte Ltd
Physical Chemistry 2nd EditionChapter 20: The Hydrogen Atom
20.3 Eigenvalues and Eigenfunctions20.3 Eigenvalues and Eigenfunctions for the Total Energy for the Total Energy
• 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
© 2010 Pearson Education South Asia Pte Ltd
Physical Chemistry 2nd EditionChapter 20: The Hydrogen Atom
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
© 2010 Pearson Education South Asia Pte Ltd
Physical Chemistry 2nd EditionChapter 20: The Hydrogen Atom
Example 20.3Example 20.3
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
© 2010 Pearson Education South Asia Pte Ltd
Physical Chemistry 2nd EditionChapter 20: The Hydrogen Atom
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.
© 2010 Pearson Education South Asia Pte Ltd
Physical Chemistry 2nd EditionChapter 20: The Hydrogen Atom
Example 20.4Example 20.4
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.
© 2010 Pearson Education South Asia Pte Ltd
Physical Chemistry 2nd EditionChapter 20: The Hydrogen Atom
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*
© 2010 Pearson Education South Asia Pte Ltd
Physical Chemistry 2nd EditionChapter 20: The Hydrogen Atom
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
© 2010 Pearson Education South Asia Pte Ltd
Physical Chemistry 2nd EditionChapter 20: The Hydrogen Atom
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
© 2010 Pearson Education South Asia Pte Ltd
Physical Chemistry 2nd EditionChapter 20: The Hydrogen Atom
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.
© 2010 Pearson Education South Asia Pte Ltd
Physical Chemistry 2nd EditionChapter 20: The Hydrogen Atom
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
© 2010 Pearson Education South Asia Pte Ltd
Physical Chemistry 2nd EditionChapter 20: The Hydrogen Atom
Example 20.6Example 20.6
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?
© 2010 Pearson Education South Asia Pte Ltd
Physical Chemistry 2nd EditionChapter 20: The Hydrogen Atom
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
© 2010 Pearson Education South Asia Pte Ltd
Physical Chemistry 2nd EditionChapter 20: The Hydrogen Atom
SolutionSolution
These functions are plotted as a function of in the following figure:
0/ ar
© 2010 Pearson Education South Asia Pte Ltd
Physical Chemistry 2nd EditionChapter 20: The Hydrogen Atom
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
© 2010 Pearson Education South Asia Pte Ltd
Physical Chemistry 2nd EditionChapter 20: The Hydrogen Atom
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.