Transcript CH107 Special Topics

CH107 Special Topics
Part A: The Bohr Model of the
Hydrogen Atom (First steps in
Quantization of the Atom)
Part B: Waves and Wave Equations
(the Electron as a Wave Form)
Part C: Particle (such as an Electron)
in a Box (Square Well) and Similar
Situations
1
A. The Bohr Model of the
Hydrogen Atom
n3
n2
n1
n3
Photon
2
Setting a Goal for Part A
• You will learn how Bohr imposed Planck’s
hypothesis on a classical description of
the Rutherford model of the hydrogen
atom and used this model to explain the
emission and absorption spectra of
hydrogen (Z = 1) and other single electron
atomic species (Z = 2, 3, etc).
3
Objective for Part A
• Describe how Bohr applied the quantum
hypothesis of Planck and classical
physics to build a model for the hydrogen
atom (and other single electron species)
and how this can be used to explain and
predict atomic line spectra.
4
Setting the Scene
At the turn of the 19th/20th centuries, classical
physics (Newtonian mechanics
and Maxwellian wave theory) were unable
explain a number of observations relating to
atomic phenomena:
• Line spectra of atoms (absorption or emission)
• Black body radiation
• The photoelectric effect
• The stability of the Rutherford (nuclear) atom
5
The Beginnings of Bohr’s Model
In 1913, Niels Bohr, a student of Ernest
Rutherford, put forward the idea of
superimposing the quantum principal of Max
Planck (1901) on the nuclear model of the H
atom.
He believed that exchange of energy in
quanta (E = hn) could explain the lines in the
absorption and emission spectra of H and
other elements.
6
Basic Assumptions (Postulates) of
Bohr’s Original Model of the H Atom
1.
2.
3.
4.
The electron moves in a circular path around the
nucleus.
The energy of the electron can assume only
certain quantized values.
Only orbits of angular momentum equal to
integral multiples of h/2p are allowed (meru = nħ,
which is equation (1); n = 1, 2, 3, …)
The atom can absorb or emit electromagnetic
radiation (hn) only when the electron transfers
between stable orbits.
7
First Steps: Balance of Forces
The centripetal force of circular
motion balances the
electrostatic force of attraction:
e- (me)
r
+
(Ze+)
u
meu2
Ze2
=
r
4p r 2
0
or
2
meu =
Ze2
4p 0 r
(2)
0 is known as the permitivity of vacuum
8
Determination of Potential Energy (PE
or V) and Kinetic Energy (KE)
The PE arises from the attractive electrostatic force
between the electron and the nucleus.
In general, V(r) =
_
r
F(r) dr
0
_
2
Ze
For the Bohr H atom, V(r) =
4p0 r
The KE exists by virtue of the electron's (circular) motion.
1
meu2
KE =
2
9
Determination of Total Energy and
Calculation of u
2
Hence the total energy is
Ze
1
2 _
meu
2
4p0 r
2
Ze
1
_ m u2 (since
=
=
e
2
4p0 r
meu2 )
(3)
Imposition of the quantum restriction meru = nh (1)
and substituting for r in equation (2) gives
2
Ze
u =
2h0n
(4)
10
Determination of Total Energy
Substitution of the value of u from equation (4)
into equation (2) gives
En =
_ e4me
Z2
802h2 n2
(5)
This is the Bohr equation for quantized electron
energy levels: the integer n, introduced in
equation (1) with the quantum hypothesis, defines
the energy for each stable orbit, known as energy
levels.
11
Determination of Total Energy Continued
The expression inside the brackets of
equation (5) is a constant, equivalent
to 2.18 x 10-18 J.
This is a convenient unit of electronic
energy and is called the rydber g.
Equation (5) thus reduces to
Z2 (rydberg)
_
En =
n2
(6)
where n = 1, 2, 3,...
and for H, Z = 1
12
Energy Level Diagram for the Bohr H
Atom
n
En
0.00 J
_ 1 rydberg
)
16
_
( 1 rydberg )
9
4
-1.36 x 10-19 J (
3
-2.42 x 10-19 J
2
-5.45 x 10-19 J (
1
_
-2.18 x 10-18 J ( 1.00 rydberg)
_ 1 rydberg
)
4
13
Determination of Orbit Radius
Elimination of u between equations (1) and (4),
and solving for r, gives
rn =
0h2
pmee2
n2 (6)
Z
This equation describes the radii of the allowed orbits
corresponding to different values of n, and with energies
given by equation (5).
As before, Z = 1 for the H atom and n = 1, 2, 3,...
14
Determination of Orbit Radius Continued
All the parameters in equation (6),
apart from Z and n2,
are constants, which is equivalent to
5.29 x 10-11 m (0.529 A).
This is known as the Bohr r adius (a0),
and is a convenient measure of
electronic distance in atoms. Thus
equation (6) reduces to
n2a
rn =
0
(7)
Z
15
Electronic Radius Diagram of Bohr H
Atom
Each
electronic
radius
corresponds
to an energy
level with
the same
quantum
number
r 3 =9a0
r 1 = a0
n =3
r 2 =4a0
n =2
n =1
r 4 =16a0
n =4
16
The Line Spectra of Hydrogen
The major triumph of the Bohr model of the
H atom was its ability to explain and predict
the wavelength of the lines in the absorption
and emission spectra of H, for the first time.
Bohr postulated that the H spectra are
obtained from transitions of the electron
between stable energy levels, by absorbing or
emitting a quantum of radiation.
17
Electronic Transitions and Spectra in
the Bohr H atom
Electron relaxation
n3
n2
Emission
n1
n3
Electron promotion
Photon
Absorption
18
Qualitative Explanation of Emission Spectra of H
in Different Regions of the Electromagnetic
Spectrum
7
6
5
4
Brackett series
(mid infrared)
Energy
3
2
Paschen series
(near infrared)
Balmer series
(visible)
n = 1 Lyman series
(ultraviolet)
19
Quantitative Explanation of Line Spectra of
Hydrogen and Hydrogen-Like Species
When H or H-like species (a one-electron ion)
undergoes a transition between two energy
_ hn
levels Ei and Ef, E = +
ni
E
nf
hn
nf
Emission
Z2e4me _ Z2e4me
hn =
802h2 nf 2 802h2 ni 2
ni > nf
hn =
hn
ni
Absorption
Z2e4me _ Z2e4me
802h2 ni 2
802h2 nf 2
nf > ni
20
…….Continued
From the previous slide,
n =
Z2e4me
2 3
80 h
n =
Z2e4me
802h3
1 _ 1
(Emission)
nf 2 ni 2
(8)
1 _ 1
(Absorption) (9)
2
2
ni
nf
4
The constant e me is equivalent to Rydberg's
802h3
empirical constant (3.29 x 1015 s-1), hence
1 _ 1
(Emission) (10)
n = (3.29 x 1015 s-1) Z2
2
2
nf
ni
n = (3.29 x 1015 s-1) Z2
1 _ 1
2
2 (Absorption) (11)
ni
nf
21
Conclusion
For the H atom (Z = 1), the predicted
emission spectrum associated with nf = 1
corresponds to the Lyman series of lines in
the ultraviolet region.
That associated with nf = 2 corresponds to
the Balmer series in the visible region, and
so on.
Likewise, the lines in all the absorption
spectra can be predicted by Bohr’s equations.
22
Aftermath – the Successes and Failures of
Bohr’s Model
• For the first time, Bohr was able to give a
theoretical explanation of the stability of the
Rutherford H atom, and of the line spectra of
hydrogen and other single electron species
(e.g. He+, Li2+, etc).
• However, Bohr’s theory failed totally with two-and
many-electron atoms, even after several drastic
modifications. Also, the imposition of quantization
on an otherwise classical description was uneasy.
• Clearly a new theory was needed!
23
B. Waves and Wave Equations
24
Setting a Goal for Part B
• You will learn how to express equations
for wave motions in both sine/cosine
terms and second order derivative terms.
• You will learn how de Broglie’s matter
wave hypothesis can be incorporated into
a wave equation to give Schrödinger-type
equations.
• You will learn qualitatively how the
Schrödinger equation can be solved for
the H atom and what the solutions mean.
25
Objective for Part B
•Describe wave forms in general and
matter (or particle) waves in particular,
and how the Schrödinger equation for a 1dimensional particle can be constructed.
•Describe how the Schrödinger equation
can be applied to the H atom, and the
meaning of the sensible solutions to this
equation.
26
The Basic Wave Form
Wave
amplitude
y(x)
Basic wave properties
c = n (1)
A = amplitude
 = wavelength
 = 
n = frequency
c = velocity
 = wavenumber
E = hn
= hc/
= hc (2)
27
Basic Wave Equations
For a travelling wave,
_ 2pn t (3)
p
2
x
y (x ) = Asin

For a standing wave,
p
2
x
sin
(
)
A
yx =

(4)
28
Characteristics of a Travelling Wave
29
Characteristics of a Standing Wave
30
Wave Form Related to Vibration and
Circular Motion
y
x
Looking
along x
axis
The wave form
appears as a
vibration
or oscillation
This can be
resolved into
two opposing
circular motions
31
The Dual Nature of Matter – de
Broglie’s Matter Waves
• We have seen that Bohr’s model of the H
atom could not be used on multi-electron
atoms. Also, the theory was an
uncomfortable mixture of classical and
modern ideas.
• These (and other) problems forced
scientists to look for alternative theories.
• The most important of the new theories
was that of Louis de Broglie, who
suggested all matter had wave-like
character.
32
de Broglie’s Matter Waves
• De Broglie suggested that the wave-like character
of matter could be expressed by the equation (5),
for any object of mass m, moving with velocity v.
 =
h
(5)
mv
• Since kinetic energy (Ek = 1/2mv2) can be written
as
2
Ek =
(mv)
2m
• De Broglie’s matter wave expression can thus be
written
h
 =
(6)
2mEk
33
de Broglie’s Matter Waves, Continued
•
Since h is very small, the de Broglie wavelength will
be too small to measure for high mass, fast
objects, but not for very light objects. Thus the
wave character is significant only for atomic
particles such as electrons, neutrons and protons.
•
De Broglie’s equation (5) can be derived from
(1) equations representing the energy of photons
(from Einstein – E = mc2 – and Planck – E = hc/)
and also
(2) equations representing the electron in the Bohr H
atom as a standing wave (mevr = nh/2p; n = 2pr)
34
The Electron in an Atom as a Standing
Wave
An important
suggestion of
de Broglie was
that the
electron in the
Bohr H atom
could be
considered as
a circular
standing wave
35
Differential Form of Wave Equations
• Consider a one-dimensional standing wave. If we
suppose that the value y(x) of the wave form at
any point x to be the wave function y(x), then we
have, according to equation (4)
y(x) = A sin
2px

or B cos
2px

(7)
• Of particular interest is the curvature of the wave
function; the way that the gradient of the gradient
of the plot of y versus x varies. This is the
second derivative of y with respect to x.
36
Differential Form of Wave Equations,
Continued
Thus
2
d2y(x) = _ A 2p sin 2px


dx2
2y
d
(x)
_
or
=
dx2
(8)
2p 2 y (x) (9)

Equation (9) is a second order differential equation
whose solutions are of the form given by equation
(7).
37
Differential Wave Equation for a OneDimensional de Broglie Particle Wave
We now consider the differential wave equation for
a one-dimensional particle with both kinetic energy
(Ek) and potential energy (V(x)).
h
d2y
2pmv 2 y
_
(x)
If  =
, then
=
2
mv
dx
h
Manipulating the above equation to get a kinetic
energy term, since Ek = (mv)2/2m
2
d2y
_ h
8p2m dx2
=
h2
d2y
8p2m dx2
=
_
(mv)2
y(x)
2m
Ek y(x)
or,
(10)
38
The Schrödinger Equation for a
Particle Moving in One Dimension
• Equation (10) shows the relationship between the
second derivative of a wave function and the kinetic
energy of the particle it represents.
• If external forces are present (e.g. due to the
presence of fixed charges, as in an atom), then a
potential energy term V(x) must be added.
• Since E(total) = Ek + V(x), substituting for Ek in
Equation (10) gives
2
2
y
h
d
_
V(x) = E y(x)
+
(11)
2
2
8p m dx
This is the Schrödinger equation for a particle
moving in one dimension.
39
The Schrödinger Equation for the
Hydrogen Atom
Erwin Schrödinger (1926) was the first to act upon de
Broglie’s idea of the electron in a hydrogen atom
behaving as a standing wave. The resulting equation
(12) is analogous to equation (11);
It represents the wave form in three dimensions and is
thus a second-order partial differential equation.
2
_ h
2
8p 2 m  x
or
2
+
_ h2
2
y
2
+
2
2
z
2
y +
+ V(x,y,z) y (x,y,z) = Ey (x,y,z)
(12)
Vy
= Ey
2m
40
The Schrödinger Equation, Continued
• The general solution of equations like equation
(12) had been determined in the 19th century (by
Laguerre and Legendre).
• The equations are more easily solved if
expressed in terms of spherical polar coordinates
(r,q,f), rather than in cartesian coordinates (x,y,z), in
which case,
2
1  r2 
r
r2  r
in equation (12) becomes


q
sin
+
q
r2sinq  q
1
2
+
2
2q  f2
r sin
1
41
Erwin
Schrödinger
Students:
I hope you are staying
awake
while the professor
talks about my work!
Love,
Erwin
42
Spherical Polar Coordinates
z
The electron position with
respect to the nucleus
in spherical polar coordinates
is r,q ,f (in cartesian coordinates
this is x,y,z)
x
P
q
r
0
f
q is the angle between OP
the z axis.
f (the azimuthal angle) is
y the angle between the
projection of OP onto
the xy plane.
x = rsinqcosf
y = rsinqsinf
z = rcosq
r 2 = x2 + y2 + z2
43
Solutions of the Schrödinger Equation
for the Hydrogen Atom
• The number of solutions to the Schrödinger
equation is infinite.
• By assuming certain properties of y (the wave
function) - boundary conditions relevant to the
physical nature of the H atom - only solutions
meaningful to the H atom are selected.
• These sensible solutions for y (originally called
specific quantum states, now orbitals) can be
expressed as the product of a radial function
[R(r)] and an angular function [Y(q,f)], both of
which include integers, known as quantum
numbers; n, l and m (or ml ).
44
Solutions of the Schrödinger Equation
for the Hydrogen Atom, Continued
Y(r,q,f) = Rnl(r)Ylm(q,f)
(13)
• The radial function R is a polynomial in r of
degree n – 1 (highest power r(n-1), called a
Laguerre
polynomial)
multiplied
by
an
exponential function of the type e(-r/na0) or e(-/n),
where a0 is the Bohr radius.
• The angular function Y consists of products of
polynomials in sinq and cosq (called Legendre
polynomials) multiplied by a complex exponential
function of the type e(imf).
45
Solutions of the Schrödinger Equation
for the Hydrogen Atom, Continued
• The principal quantum number is n (like the Bohr
quantum number = 1, 2, 3,…), whereas the other
two quantum numbers both depend on n.
• l = 0 to n - 1 (in integral values).
• m = -l through 0 to +l (again in integral values).
• The energies of the specific quantum states (or
orbitals) depend only on n for the H atom (but not
for many-electron atoms) and are numerically the
same as those for the Bohr H atom.
46
Orbitals
• Orbitals where l = 0 are called ‘s orbitals’; those
with l = 1 are known as ‘p orbitals’; and those with
l = 2 are known as ‘d orbitals’.
• When n = 1, l = m = 0 only; there is only one 1s
orbital.
• When n = 2, l can be 0 again (one 2s orbital), but l
can also be 1, in which case m = -1, 0 or +1
(corresponding to three p orbitals).
• When n = 3, l can be 0 (one 3s orbital) and 1 (three
3p orbitals) again, but can also be 2, whence m
can be –2, -1, 0, +1 or +2 (corresponding to five d
orbitals).
47
Energy Levels of the H atom
48
The 1s Wave Function of H and Corresponding
Pictorial Representation
The 1s wave function is the solution of the Schrödinger
equation when n = 1; l = 0; m = 0. It represents the orbital
with the lowest energy.
Y100 = 2
a0-3/2
e-r/a0
2
1
p
R10 is a function
Y is a constant for
of r0 and of e-r/a0 00
all cases where l = m = 0
(s orbitals), indicating
spherical symmetry
The 3D boundary
surface of the 1s
orbital
49
The 2pz Wave Function of H and
Corresponding Pictorial Representation
The 2pz wave function is the solution of the Schrödinger
equation when n = 2; l = 1; m = 0. It represents the p orbital
pointing along the z axis.
Y210 =
1
2 6
R21 a
function of
r, as well as
of e-r/2a0
a0-5/2 r e-r/2a 0
2
3
p
cosq
Y10 is a function
of q only, indicating
directionality along
the z axis
The 3D boundary
surface of the 2pz
orbital
50
C. Particle in a
One-Dimensional Box
oo
oo
V(x)
0
0
x
a
51
Setting a Goal for Part C
• You will learn how the Schrödinger
equation can be applied to one of the
simplest problems; a particle in a onedimensional box or energy well.
• You will learn how to calculate the
energies of various quantum states
associated with this system.
• You will learn how extend these ideas to
three dimensions.
52
Objective for Part C
• Describe how the Schrödinger equation
can be applied to a particle in a onedimensional box (and similar situations)
and how the energies of specific quantum
states can be calculated.
53
Particle in a
One-Dimensional Box
• The simplest model to which the
Schrödinger equation can be applied is
the particle (such as a ‘1-D electron’) in a
one-dimensional box or potential energy
well.
• The potential energy of the particle is 0
when it is in the box and  beyond the
boundaries of the box; clearly the particle
is totally confined to the box.
• All its energy will thus be kinetic energy.
54
Defining the Problem
V(x) = 0 (0 < x < L)
V(x) = oo (x < 0 or x > L)
oo
oo
0
L
V(x)
0
x
55
Setting up the Schrödinger Equation
The Schrödinger equation for this model
is derived from equation (11) in Part B, with
V(x) = 0 for 0 < x < L
_ h2
d2y
8p2m dx2
=
Ey
(1)
Separating variables in the Schrödinger
equation gives equation (2)
d2y
dx2
2
_ 8p mE y
=
h2
-k2y
(2)
56
Solution of the Schrödinger Equation
and use of the First Boundary
Condition
The solution to this equation is
y(x) = A sin(kx) or B cos(kx),
but after imposition of the boundar y
condit ion that y (x) = 0 at x = 0,
the solution must be
y(x) = A sin(kx)
Substituting this into equation 1 gives
_
2
A k sin(kx) =
_ 8p2mE A sin(kx)
h2
(3)
57
Evaluation of the Constant k and Use
of the 2nd Boundary Condition
From equation (3),
k =
8p2mE
1
2
and hence
h2
y(x) = A sin
8p2mE
1
2
x
(4)
h2
Imposition of the 2nd boundary condit ion,
that y(x) = 0 at x = L, implies equation (5)
8
p2
mE
h2
1
2
L =
np (5) ,
since sin(x) is zero only when x = np
58
Determination of the Energy Levels
This boundary condition is
satisfied if E is restricted to
values that satisfy equation
(5): that is if En is the value
of energy that satisfies
equation (5) for given allowed
value of n, then
En
=
n2h2
8mL2
(6)
(n = 1, 2, 3,....)
This gives rise to the set
of energy levels opposite,
showing the corresponding
wave forms
The spacing of energy levels;
E = En-1 _En =
h2
8 mL
2
(2n + 1)
59
Determination of the Constant A
From equations (4) and (5), it can be seen that
y(x) = A sin npx
(7)
L
A is called the nor malization const ant . To evaluate A,
we refer to the well-behaved nat ur e of y that the
integral of y2 over all space (here between x = 0 and
x = L) must be 1, since the particle is somewhere in
the box.
a
a
i.e
npx dx
y2dx =
A2sin2
=1
L
0
0
L
2
or A2 2 = 1 from which, A =
L
npx
2
( n = 1, 2, 3,....) (8)
Hence y(x) =
sin
L
L
60
A Particle in a Three-Dimensional Box
• The arguments in the previous slides can be
extended to a particle confined in a 3D box of
lengths Lx, Ly and Lz.
• Within the box, V(x,y,z) = 0; outside the cube it is

• A quantum number is needed for each dimension
and the Schrödinger equation includes
derivatives with respect to each coordinate.
• The allowed energies for the particle are given by
h2 nx2 ny2 nz2
En xnynz =
2+
2+
Ly
Lz2
8m Lx
(9)
61
Calculation of Energies of a Particle
in a 3-D Box
Comparative energies of a particle confined
to a box of sides 2L, L, L.
Enxnyn =
z
h2
nx2
8m
(2L)2
+
ny2
L2
+
nz2
L2
For the ground state, nx = ny = nz = 1
Hence E111 =
h2
8mL2
1
+1 + 1
4
=
9h2
32mL2
62
….Continued
For the excited states where one of nx, ny or nz is 2,
the others are 1, we have
E211 =
h2
8mL
E121 = E112 =
2
4 + 1 + 1
4
h2
8mL
1
2
4
=
+ 4 + 1
3h2
12h2
8mL2
32mL2
=
21h2
32mL2
The last two are degenerate (of the same energy).
If all the box sides had been L, then all three of the
above excited state energies would be degenerate.
Note that there are many other excited states; E311,
E321, etc.
63
Calculation of Energy Spacing in
Different Situations
Consider two situations:
(1) an electron in a one-dimensional box of length 1.0 A.
(2) an electron in a cube of lengths 10 cm on an edge.
Calculate the energy difference between the ground state
and the first excited state.
(1) E =
3h2
8mL
2
=
3 (6.626 x 10-34 Js) 2
8 (9.11 x 10-31 kg)(1.0 x 10-10 m) 2
= 1.807 x 10-17 J ( 10,880 kJ/mol)
64
…Continued
(2) For the cube, where Lx = Ly = Lz = L = 10 cm (0.1 m)
E111 (ground state) =
E211 = E121 = E112 =
h2
8mL2
h
12 + 12 + 12
E =
2
8mL2
22 + 12 + 12
3 h2
8mL2
(First excited state)
E =
3 (6.626 x 10-34 Js) 2
8 (9.11 x 10-31 kg)(0.1 m)2
= 1.807 x 10-35 J
This is equivalent to 1.088 x 10-14 kJ/mol. The energy levels
in general are so close together that they appear continuous;
quantum effects are minimal, quite unlike the case with the
electron in a 1-dimensional box of atom-size length 1.0 A.
65