Transcript Atomic Structure

Chapter 41
Atomic Structure
PowerPoint® Lectures for
University Physics, Thirteenth Edition
– Hugh D. Young and Roger A. Freedman
Copyright © 2012 Pearson Education Inc.
Goals for Chapter 41
•
To see how to write the Schrödinger equation for a three-dimensional
problem
•
To learn how to find the wave functions and energies for a threedimensional box
•
To examine the full quantum-mechanical description of the hydrogen
atom
•
To study how an external magnetic field affects the orbital motion of
an atom’s electrons
•
To learn about the intrinsic angular momentum (spin) of the electron
•
To understand how the exclusion principle affects the structure of
many-electron atoms
•
To study how the x-ray spectra of atoms indicate the structure of these
atoms
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The Schrödinger equation in 3-D
• Electrons in an atom can move in all three dimensions of space. If
a particle of mass m moves in the presence of a potential energy
function U(x, y, z), the Schrödinger equation for the particle’s
wave function (x, y, z, t) is
  2   x , y , z , t   2   x , y , z , t   2   x, y , z , t  





2
2
2
2m 
x
y
z

  x, y, z, t 
 U  x, y , z    x, y , z , t   i
t
2
• This is a direct extension of the one-dimensional Schrödinger
equation from Chapter 40.
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The Schrödinger equation in 3-D: Stationary states
•
If a particle of mass m has a definite energy E, its wave function (x, y, z, t) is a
product of a time-independent wave function
(x, y, z) and a factor that depends on time but not position. Then the
probability distribution function |(x, y, z, t)|2 = |(x, y, z)|2 does not depend on
time (stationary states).
  x, y, z, t     x, y, z  eiEt /
•
The function (x, y, z) obeys the time-independent Schrödinger equation in
three dimensions:
  2  x, y, z   2  x, y, z   2  x, y, z  





2m 
x 2
y 2
z 2

 U  x, y, z   x, y, z   E  x, y, z 
2
•
The normalization of the three-dimensional wave function probability
distribution function is
   x, y, z  dV  1
2
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Particle in a three-dimensional box
•
In analogy with our infinite square-well potential (U(x) = 0 inside, U(x) = ∞
outside), let us consider a three-dimensional region in space (box) of equal
sides of length L, with the same potential (U(x) = 0 inside, U(x) = ∞ outside).
•
We will consider the wave function as separable, that is, can be written as a
product of the three independent dimensions x, y and z:
  x, y, z   X ( x)Y ( y)Z ( z)
•
The Schrödinger equation inside the box becomes

2 X  x
 2Y  y 
2Z  z  

 X ( x) Z ( z )
 X ( x)Y ( y )
 Y ( y)Z ( z)
  EX ( x)Y ( y ) Z ( z )
2m 
x 2
y 2
z 2 
2
•
Or, dividing by X(x)Y(y)Z(z) we have:
2
2
 1 2 X  x
1  Y  y
1  Z z 




E
2
2
2
2m  X ( x) x
Y ( y ) y
Z ( z ) z 
2
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Particle in a three-dimensional box
•
This says that the total energy is contributed to by three terms on the left, each
depending separately on x, y and z. Let us write E = Ex + Ey + Ez. Then this
equation can be separated into three equations:
2 X  x

 EX  x 
2
2m x
2
 2Y  y 

 EY  y 
2
2m y
2
2Z  z 

 EZ  z 
2
2m z
2
•
These obviously have the same solutions separately as our original particle in
an infinite square well, and corresponding energies:
n x
X nX ( x)  C X sin X
(nX  1, 2,3...)
L
n y
YnY ( y )  CY sin Y
(nY  1, 2,3...)
L
n z
Z nZ ( z )  CZ sin Z
(nZ  1, 2,3...)
L
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nX 2 2 2
EX 
(nX  1, 2,3...)
2mL2
nY 2 2 2
EY 
(nY  1, 2,3...)
2mL2
nZ 2 2 2
EZ 
(nZ  1, 2,3...)
2
2mL
Particle in a three-dimensional box
•
A particle’s wave-function is the product of these three solutions,
  x, y, z   X ( x)Y ( y ) Z ( z )  C sin
nX  x
n y
n z
sin Y sin Z
L
L
L
•
We can use the three quantum numbers nX, nY, and nZ to label the stationary
states (states of definite energy). Here is an example of a particle in three
possible states (nX, nY, nZ) = (2, 1, 1), (1, 2, 1) or (1, 1, 2).
•
The three states shown here are degenerate: Although they have different
values of nX, nY, and nZ, they have the same total energy E.
4 2 2  2 2  2 2 3 2 2
E  EX  EY  EZ 



2
2
2
2mL
2mL 2mL
mL2
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Particle in a three-dimensional box
•
Example 41.1: (a) Find the value of the normalization constant C. (b) Find the
probability that the particle will be found somewhere in the region 0 ≤ x ≤ L/4,
for the cases (nX, nY, nZ) = (1, 2, 1), (2, 1, 1) and (3, 1, 1).
(a) The integral    x, y, z  dV  1 can be written as the product of three
integrals   x, y, z  2 dV  C 2 L sin 2 nX  x dx L sin 2 nY  y dy L sin 2 nZ  z dz  1

0
0
0
L
L
L
2
1
Recall that sin   2 (1  cos 2 ) so that
L
L
L  nX  x 1
L
 2n X  x  
2 nX  x
sin
dx


sin



0
L
2nX   L
2  L   0 2
2
The other terms give the same result, so |C|2 (L/2)3 = 1, and C = (2/L)3/2.
(b) For the first case, (nX, nY, nZ) = (1, 2, 1), we now want to find
3
L
L
2
 2  L /4 2  x
2 2 y
2 z

x
,
y
,
z
dV

sin
dx
sin
dy
sin
dz


  0

0
0
L
L
L
L
Only the first integral differs from L/2, due to its different high limit

L /4
0
2 x
L /4
L   x 1  2 x  
L L
sin
dx 

sin




L
2  L 2  L   0
8 4
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3
 2   L L  L  L 
so    
      0.091
 L   8 4   2   2 
Energy Degeneracy
•
For a particle in a three-dimensional box, the allowed energy levels are
surprisingly complex. To find them, just count up the different possible states.
•
Here are the first 6 for an equal-sided box:
3 2 2
E1,1,1 
2mL2
6-fold degenerate
Not degenerate
3-fold degenerate
3-fold degenerate
3-fold degenerate
If length of sides of box are different:
 nX 2 nY 2 nZ 2   2 2
E  2  2  2 
LY
LZ  2m
 LX
(breaks the degeneracy)
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Not degenerate
The hydrogen atom: Quantum numbers
•
For a hydrogen atom, the three-dimensional
potential energy depends only on the electron’s
distance from the proton:
1 e2
U (r )  
4 0 r
•
As before, we will seek separable wave functions,
but this time, due to the spherical symmetry we will
use spherical coordinates: (r, , ) rather than (x, y,
z). Then   r, ,   R(r )( )( )
•
Following the same procedure as before, we can
separate the problem into three separate equations

d  2 dR (r )   2l (l  1)

r


U
(
r
)
 R (r )  ER  r 

 
2  dr 
dr   2  r 2

2
ml 2
1 d 
d ( )  
 sin 
   l (l  1)  2
sin  d 
d  
sin 

 ( )  E( )

d 2  ( )
 ml 2  ( )  E ( )
2
d
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•
Energies are
En  
•
 e4
1
 4 0 
2
2n 2
2

13.6 eV
n2
Same as for Bohr
model! Quantum
numbers are n, l and ml.
Description of solution
•
We will not go through the mathematics of the solution, but note that we can
only accept solutions for which the wave function is normalizable (does not
blow up). Some radial solutions blow up at r = 0 or r = ∞, and so must be
discarded. As in the harmonic oscillator problem, the radial solutions R(r)
turn out to be an exponential function ear, multiplied by a polynomial in r.
The angular solutions () are polynomials containing powers of sin and
cos. The azimuthal solutions () have to be periodic (have the same value
at () and (+2, etc. They turn out to depend on eiml, where ml is an
integer, either positive, negative or zero.
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Quantization of Angular Momentum
•
In addition to quantized energy (specified by
principle quantum number n), the solutions subject to
physical boundary conditions also have quantized
orbital angular momentum L. The magnitude of the
vector L is required to obey
L  l (l  1)
(l = 0, 1, 2, ..., n  1)
where l is the orbital quantum number.
•
Recall that the Bohr model of the hydrogen atom
also had quantized angular moment L = nħ, but the
lowest energy state n = 1 would have L = ħ. In
contrast, the Schrödinger equation shows that the
lowest state has L = 0. This lowest energy-state
wave function is a perfectly symmetric sphere. For
higher energy states, the vector L has in addition
only certain allowed directions, such that the zcomponent is quantized as
Lz  ml
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(ml = 0,  1,  2, ...,  l )
The hydrogen atom: Degeneracy
•
States with different quantum numbers l and n
are often referred to with letters as follows:
l value
letter
0
s
1
p
2
d
3
f
4
g
5
h
n value
shell
1
K
2
L
3
M
4
N
•
Hydrogen atom states with the same value of n
but different values of l and ml are degenerate
(have the same energy).
•
Figure at right shows radial probability
distribution for states with l = 0, 1, 2, 3 and
different values of n = 1, 2, 3, 4.
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The hydrogen atom: Quantum states
•
Table 41.1 (below) summarizes the quantum states of the hydrogen atom. For
each value of the quantum number n, there are n possible values of the quantum
number l. For each value of l, there are 2l + 1 values of the quantum number ml.
•
Example 41.2: How many distinct states of the hydrogen atom (n, l, ml) are
there for the n = 3 state? What are their energies?
The n = 3 state has possible l values 0, 1, or 2. Each l value has ml possible
values of (0), (-1, 0, 1), or (-2, -1, 0, 1, 2). The total number of states is then
1 + 3 + 5 = 9. We will see later that there is another quantum number s, for
electron spin (±½), so there are actually 18 possible states for n = 3.
Each of these states have the same n, so they all have the same energy.
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The hydrogen atom: Probability distributions I
• States of the hydrogen atom with l = 0 (zero orbital angular
momentum) have spherically symmetric wave functions that
depend on r but not on  or . These are called s states. Figure 41.9
(below) shows the electron probability distributions for three of
these states.
• Follow Example 41.4.
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The hydrogen atom: Probability distributions II
• States of the hydrogen atom with nonzero orbital angular
momentum, such as p states (l = 1) and d states (l = 2), have wave
functions that are not spherically symmetric. Figure 41.10 (below)
shows the electron probability distributions for several of these
states, as well as for two spherically symmetric s states.
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Magnetic moments and the Zeeman effect
•
Electron states with nonzero orbital angular momentum (l = 1, 2, 3, …) have a
magnetic dipole moment due to the electron motion. Hence these states are
affected if the atom is placed in a magnetic field. The result, called the Zeeman
effect, is a shift in the energy of states with nonzero ml. This is shown in Figure
(below).
•
The potential energy associated with a magnetic dipole moment  in a magnetic
field of strength B is U = B , and the magnetic dipole moment due to the
orbital angular momentum of the electron is in units of the Bohr magneton,
B 
e
2m
U  ml B B (orbital magnetic interaction energy)
Lower
energy
•
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Same
energy
Higher
energy
Think of magnets,
which like to be antialigned. This is a lower
energy state.
Zeeman Effect (Sunspots)
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Zeeman Effect (Sunspots)
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The Zeeman effect and selection rules
•
Figure 41.14 (lower left) shows the shift in energy of the five l = 2 states
(each with a different value of ml) as the magnetic field strength is
increased.
•
An atom in a magnetic field can make transitions between different
states by emitting or absorbing a photon A transition is allowed if l
changes by 1 and ml changes by 0, 1, or –1. (This is because a photon
itself carries angular momentum.) A transition is forbidden if it violates
these selection rules. See Figure 41.15 (lower right).
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The anomalous Zeeman effect and electron spin
• For certain atoms the Zeeman effect does not follow the simple
pattern that we have described (see Figure 41.16 below). This is
because an electron also has an intrinsic angular momentum,
called spin angular momentum.
S
1
2
( 12  1)

3
4
S z  ms
(ms =  12 )
U  2.00232 ms  B B (e  spin magnetic interaction energy)
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Electron spin and the Stern-Gerlach experiment
•
The experiment of Stern and Gerlach demonstrated the existence of
electron spin (see Figure 41.17 below). The z-component of the spin
angular momentum has only two possible values (corresponding to
ms = +1/2 and ms = –1/2).
B 
e
2m
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U  2.00232 ms  B B (e  spin magnetic interaction energy)
Quantum states and the Pauli exclusion principle
•
The allowed quantum numbers for an atomic electron (see Table
41.2 below) are n ≥ 1; 0 ≤ l ≤ n – 1; –l ≤ ml ≤ l; and ms = ±1/2.
•
The Pauli exclusion principle states that if an atom has more than
one electron, no two electrons can have the same set of quantum
numbers.
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A multielectron atom
•
Figure 41.21 (at right) is a
sketch of a lithium atom, which
has 3 electrons. The allowed
electron states are naturally
arranged in shells of different
size centered on the nucleus.
The n = 1 states make up the K
shell, the n = 2 states make up
the L shell, and so on.
•
Due to the Pauli exclusion
principle, the 1s subshell of the
K shell (n = 1, l = 0, ml = 0) can
accommodate only two
electrons (one with ms = + 1/2,
one with ms = –1/2). Hence the
third electron goes into the 2s
subshell of the L shell (n = 2,
l = 0, ml = 0).
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Ground-state electron configurations
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Screening in multielectron atoms
•
An atom of atomic number Z has a nucleus of charge +Ze and Z electrons of
charge –e each. Electrons in outer shells “see” a nucleus of charge +Zeffe, where
Zeff < Z, because the nuclear charge is partially “screened” by electrons in the
inner shells.
•
Example 41.8: Determine Zeff experimentally. The measured energy of a 3s
state of a sodium atom is -5.138 eV. Calculate the value of Zeff.
From the table, Na (Z=11) has 1s22s22p63s. In analogy with hydrogen, which
has a ground state of -13.6 eV, we can define an effective (reduced) Z by
knowing the energy of a state, i.e.
Z eff 2
En   2 (13.6 eV)
n
implies
Zeff  n En / (13.6 eV)  3 5.138 /13.6  1.84
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X-ray spectroscopy
•
When atoms are bombarded
with high-energy electrons,
x rays are emitted. There is a
continuous spectrum of x rays
(described in Chapter 38) as
well as strong characteristic
x-ray emission at certain
definite wavelengths (see the
peaks labeled Ka and K in
Figure 41.23 at right).
•
Atoms of different elements
emit characteristic x rays at
different frequencies and
wavelengths. Hence the
characteristic x-ray spectrum
of a sample can be used to
determine the atomic
composition of the sample.
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X-ray spectroscopy: Moseley’s law
•
Moseley showed that the square root of the x-ray frequency in Ka
emission is proportional to Z – 1, where Z is the atomic number of the
atom (see Figure 41.24 below). Larger Z means a higher frequency and
more energetic emitted x-ray photons. Follow Example 41.10.
•
This is consistent with our model of multielectron atoms. Bombarding
an atom with a high-energy electron can knock an atomic electron out
of the innermost K shell. Ka x rays are produced when an electron from
the L shell falls into the K-shell vacancy. The energy of an electron in
each shell depends on Z, so the x-ray energy released does as well.
•
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The same principle applies to
K emission (in which an
electron falls from the M shell
to the K shell) and K emission
(in which an electron falls
from the N shell to the K
shell).