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## Atomic Physics

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### Atomic Physics Step Potential Consider a particle of energy E moving in region in which the potential energy is the step function U(x) = 0, x0 What ... – PowerPoint PPT presentation

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Title: Atomic Physics

1
Atomic Physics

2
Step Potential
• Consider a particle of energy E moving in region
in which the potential energy is the step
function
• U(x) 0, xlt0
• U(x) V0, xgt0
• What happened when
• a particle moving from
• left to right encounters
• the step?
• simple to the left of the
• step, the particle moves
• with a speed v v2E/m

3
Step Potential
At x 0, an impulsive force act on the particle.
If the initial energy E is less than V0, the
particle will be turned around and will then move
to the left at its original speed that is, the
particle will be reflected by the step. If E is
greater than V0, the particle will continue to
move to the right but with reduced speed given by
v v2(E U0)/m

4
Step Potential
• We can picture this classical problem as a ball
rolling along a level surface and coming to a
steep hill of height h given by mghV0.
• If the initial kinetic energy of the ball is
less than mgh, the ball will roll part way up the
hill and then back down and to the left along the
lower surface at it original speed. If E is
greater than mgh, the ball will roll up the hill
and proceed to the right at a lesser speed.

5
• The quantum mechanical result is similar when E
is less than V0. If EltV0 the wave function does
not go to zero at x0 but rather decays
exponentially. The wave penetrates slightly into
the classically forbidden region xgt0, but it is
eventually completely reflected.

6
Step Potential
• This problem is somewhat similar to that of
total internal reflection in optics.
• For EgtV0, the quantum mechanical result differs
from the classical result. At x0, the wavelength
changes from
• ?1h/p1 h/v2mE
• to
• ?2h/p2 h/v2m(E-V0).
• When the wavelength changes suddenly, part of
• the wave is reflected and part of the wave is
transmitted.

7
Reflection Coefficient
• Since a motion of an electron (or other
particle) is governed by a wave equation, the
electron sometimes will be transmitted and
sometimes will be reflected.
• The probabilities of reflection and
transmission can be calculated by solving the
Schrödinger equation in each region of space and
comparing the amplitudes of transmitted waves and
reflected waves with that of the incident wave.

8
Reflection Coefficient
• This calculation and its result are similar to
finding the fraction of light reflected from the
air-glass interface. If R is the probability of
reflection, called the reflection coefficient,
this calculation gives
• where k1 is the wave number for the incident
wave and k2 is the wave number for the
transmitted wave.

9
Transmission Coefficient
• The result is the same as the result in optics
for the reflection of light at normal incidence
from the boundary between two media having
different indexes of refraction n.
• The probability of transmission T, called the
transmission coefficient, can be calculated from
the reflection coefficient, since the probability
of transmission plus the probability of
reflection must equal 1
• T R 1
• In the quantum mechanics, a localized particle
is represented by the wave packet, which has a
maximum at the most probable position of the
particle.

10
• Time development of a one dimensional wave
packet representing a particle incident on a step
potential for EgtV0. The position of a classical
particle is indicated by the dot. Note that part
of the packet is transmitted and part is
reflected.

11
• Reflection coefficient R and transmission
coefficient T for a potential step V0 high versus
energy E (in units V0).

12
• A particle of energy E0 traveling in a region in
which the potential energy is zero is incident on
a potential barrier of height V00.2E0. Find the
probability that the particle will be reflected.

13
Lets consider a rectangular potential barrier of
height V0 and with a given byU(x) 0, xlt0U(x)
V0, 0ltxltaU(x) 0, xgta
14
Barrier Potential
• We consider a particle of energy E , which is
slightly less than V0, that is incident on the
barrier from the left. Classically, the particle
would always be reflected. However, a wave
incident from the left does not decrease
immediately to zero at the barrier, but it

will instead decay exponentially in the
classically forbidden region 0ltxlta. Upon reaching
the far wall of the barrier (xa), the wave
function must join smoothly to a sinusoidal wave
function to the right of barrier.
15
Barrier Potential
• If we have a beam of particle incident from
left, all with the same energy EltV0, the general
solution of the wave equation are, following the
example for a potential step,
• where k1 v2mE/h and a v2m(V0-E)/h
• This implies that there is some probability of
the particle (which is represented by the wave
function) being found on the far side of the
barrier even though, classically, it should never
pass through the barrier.

16
Barrier Potential
• For the case in which the quantity
• aa v2ma2(V0 E)/h2
• is much greater than 1, the transmission
coefficient is proportional to e-2aa, with
• a v2m(V0 E)/h2
• The probability of penetration of the barrier
thus decreases exponentially with the barrier
thickness a and with the square root of the
relative barrier height (V0-E). This phenomenon
is called barrier penetration or tunneling. The
relative probability of its occurrence in any
given situation is given by the transmission
coefficient.

17
• A wave packet representing a particle incident
on two barriers of height just slightly greater
than the energy of the particle. At each
encounter, part of the packet is transmitted and
part reflected, resulting in part of the packet
being trapped between the barriers from same
time.

18
• A 30-eV electron is incident on a square
barrier of height 40 eV. What is the probability
that the electron will tunnel through the barrier
if its width is (a) 1.0 nm?
• (b) 0.1nm?

19
• The penetration of the barrier is not unique to
quantum mechanics. When light is totally
reflected from the glass-air interface, the light
wave can penetrate the air barrier if a second
peace of glass is brought within a few
wavelengths of the first, even when the angle of
incidence in the first prism is greater than the
critical angle. This effect can be demonstrated
with a laser beam and two 45 prisms.

20
a- Decay
The theory of barrier penetration was used by
George Gamov in 1928 to explain the enormous
variation of the half-lives for a decay of
radioactive nuclei. Potential well shown on the
diagram for an a particle in a radioactive
nucleus approximately describes a strong
attractive force when r is less than the nuclear
radius R. Outside the nucleus the strong nuclear
force is negligible, and the potential is given
by the Coulombs law, U(r) k(2e)(Ze)/r, where
Z is the nuclear charge and 2e is the charge of a
particle.
21
a- Decay
An a-particle inside the nucleus oscillates back
and forth, being reflected at the barrier at R.
Because of its wave properties, when the
a-particle hits the barrier there is a small
chance that it will penetrate and appear outside
the well at r r0. The wave function is similar
to that for a square barrier potential.
22
• The probability that an a-particle will tunnel
through the barrier is given by
• which is a very small number, i.e., the a
particle is usually reflected. The number of
times per second N that the a particle approaches
the barrier is given by

where v equals the particles speed inside the
nucleus.
The decay rate, or the probability per second
that the nucleus will emit an a particle, which
is also the reciprocal of the mean life time
, is given by
23
• The decay rate for emission of a particles from
radioactive nuclei of Po212. The solid curve is
the prediction of equation
• The points are the experimental results.

24
Applications of Tunneling
• Nanotechnology refers to the design and
application of devices having dimensions ranging
from 1 to 100 nm
• Nanotechnology uses the idea of trapping
particles in potential wells
• One area of nanotechnology of interest to
researchers is the quantum dot
• A quantum dot is a small region that is grown in
a silicon crystal that acts as a potential well
• Nuclear fusion
• Protons can tunnel through the barrier caused by
their mutual electrostatic repulsion

25
Resonant Tunneling Device
• Electrons travel in the gallium arsenide
semiconductor
• They strike the barrier of the quantum dot from
the left
• The electrons can tunnel through the barrier and
produce a current in the device

26
Scanning Tunneling Microscope
• An electrically conducting probe with a very
sharp edge is brought near the surface to be
studied
• The empty space between the tip and the surface
represents the barrier
• The tip and the surface are two walls of the
potential well

27
Scanning Tunneling Microscope
• The STM allows highly detailed images of surfaces
with resolutions comparable to the size of a
single atom
• At right is the surface of graphite viewed with
the STM

28
Scanning Tunneling Microscope
• The STM is very sensitive to the distance from
the tip to the surface
• This is the thickness of the barrier
• STM has one very serious limitation
• Its operation is dependent on the electrical
conductivity of the sample and the tip
• Most materials are not electrically conductive at
their surfaces
• The atomic force microscope (AFM) overcomes this
limitation by tracking the sample surface
maintaining a constant interatomic force between
the atoms on the scanner tip and the samples
surface atoms.

29
SUMMARY
• 1. Time-independent Schrödinger equation
• 2.In the simple harmonic oscillator
• the ground wave function is given
• where A0 is the normalization constant and
am?0/2h.
• 3. In a finite square well of height V0, there
are only a finite number of allowed energies.

30
SUMMARY
• 4.Reflection and barrier penetration
• When the potentials changes abruptly over a
small distance, a particle may be reflected even
though EgtU(x). A particle may penetrate a region
in which EltU(x). Reflection and penetration of
electron waves are similar for those for other
kinds of waves.

31
The Schrödinger Equation in Three Dimensions
• The one-dimensional time-independent Schrödinger
equation

• (1)
• is easily extended to three dimensions. In
rectangular coordinates, it is

• (2)
• where the wave function ? and the potential
energy U are generally functions of all three
coordinates , x, y, and z.

32
The Schrödinger Equation in Three Dimensions
• To illustrate some of the features of problems
in three dimensions, we consider a particle in
three-dimensional infinity square well given by
• U(x,y,z) 0 for 0ltxltL, 0ltyltL, and 0ltzltL.
• Outside this cubical region, U(x,y,z)8. For
this problem, the wave function must be zero at
the edges of the well.

33
The Schrödinger Equation in Three Dimensions
• The standard method for solving this partial
differential equation is guess the form of the
solution using the probability. For a
one-dimensional box along the x axis, we have
found the probability that the particle is in the
region dx at x to be
• A12sin2(k1x)dx (3)
• where A1 is the normalization constant, and
k1np/L is the wave number. Similarly, for a box
along y axis, the probability of a particle being
in a region dy at y is
• A22sin2(k2y)dy
(4)
• The probability of two independent events
occurring is the product of probabilities of each
event occurring.

34
The Schrödinger Equation in Three Dimensions
• So, the probability of a particle being in
region dx at x and in region dy at y is
• The probability of a particle being in the
region dx, dy, and dz is ?2(x,y,x)dxdydz, where
?(x,y,z) is the solution of equation
• (2)

35
The Schrödinger Equation in Three Dimensions
(2)
The solution is of the form
where the constant A is determined by
normalization. Inserting this solution in the
equation (2), we obtain for the energy
which is equivalent to
with pxhk1 and so on.
36
The Schrödinger Equation in Three Dimensions
• The wave function will be zero at xL if
k1n1p/L, where n1 is the integer. Similarly, the
wave function will be zero at yL if k2n2p/L,
and the wave function will be zero at zL if
k3n3p/L. It is also zero at x0, y0, and z0.
The energy is thus quantized to the values
• where n1, n2, and n3 are integers and E1 is the
ground-state energy of the one dimensional well.
• The lowest energy state (the ground state) for
the cubical well occurs when n12n22n321 and
has the value

37
The Schrödinger Equation in Three Dimensions
• The first excited energy level can be obtained
in three different ways
• 1) n12, n2n31 2) n22, n1n31 3) n32,
n1n21.
• Each way has a different wave function . For
example, the wave function for n12, n2n31 is
• There are thus three different quantum states
as described by three different wave functions
corresponding to the same energy level. The
energy level with more than one wave function are
associate is said to be degenerate. In this case,
there is threefold degeneracy.

38
The Schrödinger Equation in Three Dimensions
• Degeneracy is related to the spatial symmetry
of the system. If, for example, we consider a
noncubic well, where U0 for 0ltxltL1, 0ltyltL2, and
0ltzltL3, the boundary conditions at the edges
would lead to the quantum conditions k1L1n1p,
k2L2n2p, and k3L3n3p and the total energy would
be

39
• This energy level are not degenerate if L1, L2,
and L3 are all different.

Figure shows the energy levels for the ground
state and first two excited levels for an
infinity cubic well in which the excited states
are degenerated and for a noncubic infinity well
in which L1, L2, and L3 are all slightly
different so that the excited levels are slightly
split apart and the degeneracy is removed.
40
The Degenerate States
• The ground state is the state where the quantum
numbers n1, n2, and n3 are all equal to 1. Non
of the three quantum numbers can be zero. If any
one of n1, n2, and n3 were zero, the
corresponding wave number k would also equal to
zero and corresponding wave function would equal
to zero for all values of x,y, and z.

41
• Example 1.
• A particle is in three-dimensional box with
L3L22L1. Give the quantum numbers n1, n2, and
n3 that correspond to the thirteen quantum states
of this box that have the lowest energies.

42
• Example 2.
• Write the degenerate wave function for the
fourth and fifth excited states (level 5 and 6)
from the Example1.

43
The Schrödinger Equation for Two Identical
Particles
• Thus far our quantum mechanical consideration
was limited to situation in which a single
particle moves in some force field characterized
by a potential energy function U.
• The most important physical problem of this
type is the hydrogen atom, in which a single
electron moves in the Coulomb potential of the
proton nucleus.

44
The Schrödinger Equation for Two Identical
Particles
• This problem is actually a two-body problem,
since the proton also moves in the field of
electron. However, the motion of the much more
massive proton requires only a very small
correction to the energy of the atom that is
easily made in both classical and quantum
mechanics.
• When we consider more complicated problems,
such as the helium atom, we must apply the
quantum mechanics to two or more electrons moving
in an external field.

45
The Schrödinger Equation for Two Identical
Particles
The interaction of two electrons with each other
is electromagnetic and is essentially the same,
that the classical interaction of two charged
particles. The Schrödinger
equation for an atom with two or more electrons
cannot be solved exactly, so approximation method
must be used. This is not very different from
classical problem with three or more particles,
however, the complications arising from the
identity of electrons.

46
The Schrödinger Equation for Two Identical
Particles

There are due to the fact that it is impossible
to keep track of which electron is
which. Classically, identical particles can be
identified by their position, which can be
determined with unlimited accuracy. This is
impossible quantum mechanically because of the
uncertainty principle.
47
The Schrödinger Equation for Two Identical
Particles
• The undistinguishability of identical particles
has important consequence. For instance,
consider the very simple case of two identical,
noninteracting particles in one-dimensional
infinity square well.
• The time independent Schrödinger equation for
two particles, each mass m, is
• where x1 and x2 are the coordinates of the two
particles.

48
The Schrödinger Equation for Two Identical
Particles
• If the particles interact, the potential energy
U contains terms with both x1 and x2 that can not
be separated. For example, the electrostatic
repulsion of two electrons in one dimension is
represented by potential energy ke2/(x2-x1).
• However if the particles do not interact, as we
assuming here, we can write U U(x1) U(x2).
• For the infinity square well, we need only
solve the Shrödinger equation inside the well
where U0, and require that the wave function be
zero at the walls of the well.

49
The Schrödinger Equation for Two Identical
Particles
• With U0, equation
• looks just like the expression for a
two-dimensional
• well
• with no z and with y replaced by x2.

50
The Schrödinger Equation for Two Identical
Particles
• Solution of this equation can be written in the
form
• ?n,m ?n(x1)?m(x2)
• where ?n and ?m are the single particle wave
function for a particle in the infinity well and
n and m are the quantum numbers of particles 1
and 2. For example, for n1 and m2 the wave
function is

51
The Schrödinger Equation for Two Identical
Particles
• The probability of finding particle 1 in dx1
and particle 2 in dx2 is ?2n,m(x1,x2)dx1dx2,
which is just a product of separate probabilities
?2n(x1)dx1 and ?2m(x2)dx2. However, even though
we label the particles 1 and 2, we can not
distinguish which is in dx1 and which is in dx2
if they are identical. The mathematical
description of identical particles must be the
same if we interchange the labels. Therefore, the
probability density
• ?2(x2,x1) ?2(x1,x2)
• This equation is satisfied if ? is either
symmetric or antisymmetric
• ?2(x2,x1) ?2(x1,x2), symmetric
• or
• ?2(x2,x1) -?2(x1,x2), antisymmetric

52
The Schrödinger Equation for Two Identical
Particles
• For example, the symmetric and antisymmetric
wave function for the first exited state of two
identical particles in a infinity square well
• and

53
The Schrödinger Equation for Two Identical
Particles
• There is an important difference between
antisymmetric and symmetric wave functions. If
nm, the antisymmetric wave function is
identically zero for all values of x1 and x2 ,
whereas the symmetric function is not. Thus, the
quantum numbers n and m can not be the same for
antisymmetric function.
• Pauli exclusion principle
• No two electrons in an atom can have same
quantum numbers.

54
The Schrödinger Equation in Spherical Coordinates
• In quantum theory, the electron is described by
its wave function ?. The probability of finding
the electron in some volume dV of space is equals
the product of absolute square of the electron
wave function ?2 and dV.
• Boundary conditions on the wave function lead
to quantization of the wavelengths and
frequencies and thereby to the quantization of
the electron energy.

55
The Schrödinger Equation in Spherical Coordinates
• Consider a single electron of mass m moving in
three dimensions in a region in which the
potential energy is V. The time independent
Schrödinger Equation for such a particle
• For a single isolated atom, the potential
energy V depends only on the radial distance r
vx2 y2 z2 . The problem is then most
conveniently treated using the spherical
coordinates.

56
The Schrödinger Equation in Spherical Coordinates
We will use coordinates r, ?, and f, which
related to the rectangular coordinates x, y, and
z by z r cos?, x r sin?cosf, and y r
sin?sinf
57
The Schrödinger Equation in Spherical Coordinates
• (1)
• The transformation of the wave term in the
equation
• Substitution in equation (1) gives (2)

2
58
The Schrödinger Equation in Spherical Coordinates
• The first step in solving this partial
differential equation is to separate the
variables by writing the wave function ?(r,?,f)
as a product of functions of each single
variable
• ?(r,?,f) R(r) f(?) g(f),
• where R represent only the radial coordinate r
f depends only of ?, and g depends only of f.
When this form of ?(r,?,f) is substituted into
equation (2) the partial differential equation
can be transformed into three ordinary
differential equations, one for R(r), one for
f(?) and one for g(f).

59
• The potential energy U(r) appears only in
equation for R(r), which is called the radial
equation

60
(No Transcript)
61
The Schrödinger Equation in Spherical Coordinates
• In three dimensions, the requirement that the
wave function be continuous and normalizable
introduces three quantum numbers, one associated
with each spatial dimension. In spherical
coordinates the quantum number associated with r
is labeled n, that associated with ? is labeled
l, and that associated with f is labeled ml.
• For the rectangular coordinates x,y, and z the
corresponded quantum numbers n1, n2, and n3 for a
particle in a three-dimensional square well were
independent of one other, but the quantum numbers
associated with wave function for spherical
coordinates are interdependent.

62
Summary of the Quantum Numbers
• The possible values of this quantum numbers are
• n 1,2,3,..
• l 0,1,2,3,,(n-1)
• ml -l, (-l 1),,-2,-1,0,1,2,.,(l 1),l0,
1, 2,,l
• The number n is called the principal quantum
number. It is associated with the dependence of
the wave function on the distance r and therefore
with the probability of finding the electron at
various distances from the nucleus.
• The quantum numbers l and ml are associated with
the angular momentum of the electron and with the
angular dependence of the electron wave function.
• The quantum number l is called the orbital
quantum number. The magnitude L of the orbital
angular momentum is related to l by
• L v l (l 1) h

63
Summary of the Quantum Numbers
• The quantum number ml is called the magnetic
quantum number, it is relate to the z-component
of angular momentum. Since there is not preferred
direction for the z-axis for any central force,
all spatial directions are equivalent for an
isolated atom.
• However, if we will place the atom in an
external magnetic field the direction of the
field will be separated out from the other
directions. If z-direction is chosen for the
magnetic field direction, than z-component of the
angular momentum of the electron is given by the
quantum condition
• LZ mlh
• This quantum condition arises from the boundary
conditions on the azimuth coordinate f that the
probability of finding the electron at some angle
f1 must be the same as that of finding the
electron at angle f12p because these are the
same points in the space.

64
Summary of the Quantum Numbers
• If we measure the angular momentum of the
electron in units of h, we see that the
angular-momentum magnitude is quantized to the
value
• v l (l1) units and that its component along any
direction can have only the 2l 1 values ranging
from -l to l units. On the figure we can see the
possible orientation of angular momentum vector
for l2.

Vector-model diagram illustrating the possible
values of z-component of the angular momentum
vector for the case l2. The magnitude of Lhv6.
65
The direction of the angular momentum
• If the angular momentum is characterized by the
quantum number l 2, what are the possible
values of LZ, and what is the smallest possible
angle between L and the z axis?

66
Summary of the Quantum Numbers

That is, n can be any positive integer l can be
zero or any positive integer up to (n-1) and ml
can have (2l1) positive values, ranging from -l
to l in integral steps. In order to explain
the fine structure and to clear up some
difficulties with explanation the table of
elements Pauli suggested that in addition to the
quantum numbers n, l, and ml the electron should
have a fourth quantum number, which could take on
just two values. This fourth quantum number is
the z-component, mz, of an intrinsic angular
momentum of the electron, called spin. The spin
vector S relate to this fourth quantum number s
by S vs(s1) h and can take only two
values ½.
67
Quantum Theory of the Hydrogen Atom
• We can treat the simplest hydrogen atom as a
stationary nucleus, a proton, that has a single
moving particle, an electron, with kinetic energy
p2/2m. The potential energy U(r) due to the
electrostatic attraction between the electron and
the proton is
• In the lowest energy state, which is the ground
state, the principal quantum number n1, l0, and
ml0.
• The allowed energies

68
• Potential energy of an electron in a hydrogen
atom. If the total energy is greater than zero,
as E, the electron is not bound and the energy
is not quantized. If the total energy is less
than zero, as E, the electron is bound, than, as
in one-dimensional problems, only certain
discrete values of the total energy lead to
well-behaved wave function.

69
• Energy-level diagram for hydrogen. The diagonal
lines show transitions that involve emission or
absorption of radiation that obey the selection
rules ?l 1, ml0 or 1. States with the same
value of n but with different values of l have
the same energy E0/n2,
• where E013.6 eV.
• The wavelength of the light emitted by the atom
relate to the energy levels by
• hf hc Ei-Ef

70
• Energy-level diagram for the hydrogen atom,
showing transitions obeying the selection rule ?l
1. States with the same n value but different
l value have the same energy, -E1/n2, where
E113.6 eV, as in the Bohr theory. The wavelength
of the Lyman a(n 2 ? n 1) and Balmer a(n 3 ?
n 2) lines are shown in nm. Note that the
latter has two possible transactions due to the l
degeneracy.

71
Wave Function and the probability density
• The ground state In the lowest energy state,
the ground state of the hydrogen, n1, l0, and
ml0, E013.6eV, and the angular momentum is
zero. The wave function for the ground state is
• is the Bohr radius and C1,0,0 is a constant that
is determined by the normalization. In three
dimensions, the normalization condition is
• ??2dV 1
• where dV is a volume element and the integration
is performed over all space.

where
72
Volume Element in Spherical Coordinates
Volume element in spherical coordinates
73
• In spherical coordinates the volume element is
• dV (r sin?df)(rd?)dr r2 sin?d?dfdr
• We integrate over f, from f0 to f2p over ?,
from ?0 to ?p and over r from r0 to r8. The
normalization conditions is thus
• Since there is no ? or f dependence in ?1,0,0
the triple integral can be factored in a product
of three integrals

74
• This gives
• The remaining integral is of the form
• with n a positive integer and with agt0.

75
• This integral can be looked up in a table of
integrals
• so
• Than

so
76
Probability Densities
• The normalized ground-state wave function is
thus
• The probability of finding the electron in a
volume dV is
• ?2dV

77
Computer generated probability density ?2 for
the ground state of the hydrogen. The quantity -e
?2 can be though of as the electron charge
density in the atom. The density is spherically
symmetric (it depends only on r and independent
of ? or f), is greatest at the origin , and
decrease exponentially with r.
78
Probability Densities
• We are more often interested in the probability
of finding the electron at some radial distance r
between r and rdr. This radial probability
P(r)dr is ?2dV, where dV is the volume of the
spherical shell of thickness dr, which is
dV4pr2dr.
• The probability of finding the electron in the
range from r to
• rdr is thus
• and the radial probability density is

79
Probability Densities
• For the hydrogen atom in the ground state, the

80
Radial probability density P(r) versus r / a0 for
the ground state of the hydrogen atom. P(r) is
proportional to r2?2. The value of r for which
P(r) is maximum is the most probable distance
ra0, which is the first Bohr radius.
81
• Radial probability density P(r) vs. r/a0 for
the n2 states in hydrogen. P(r) for l1 has a
maximum at the Bohr value 22a0. For l 0 there
is a maximum near this value and a smaller
submaximum near the origin. The markers on the
r/a0 axis denote the values of (r/a0).

82
• P(r) vs. r/a0 for the n 3 state in hydrogen.

83
• Probability density ?? for the n2 states
in hydrogen. The probability is spherically
symmetric for l0. It is proportional to cos2?
for l1, m0, and to sin2 for l1, m1. The
probability densities have rotational symmetry
about the z axis. Thus, the three-dimensional
charge density for l1, m0 state is shaped
roughly like a dumbbell, while that for the l1,
m1 states resembles a doughnut, or toroid. The
shapes of these distributions are typical for all
atoms in S states (l0) and P states (l1) and
play an important role in molecular bounding.

84
• A particle moving in a circle has angular
momentum L. If the particle have a positive
charge, the magnetic moment due to the current is
parallel to L.

85
• Bar-magnet model of magnetic moment.
• (a) In an external magnetic field, the moment
experiences a torque which tends to align it with
the field. If the magnet is spinning (b), the
torque caused the system to precess around the
external field.

86
Example Probability that electron is in a thin
spherical shell
• Find the probability of finding the electron in
a thin spherical shell of radius r and thickness
?r0.06a0 at (a) ra0 and (b) r2a0 for the
ground state of the hydrogen atom.

87
The spin-orbit effect and fine structure
• The total angular momentum of an electron in an
atom is a combination of the orbital angular
momentum and spin angular momentum. It is
characterized by the quantum number j, which can
be either l - ½ or l ½. Because of
interaction of the orbital and spin magnetic
moments, the state j l - ½ has lower energy
than the state j l ½, for l 1. This small
splitting of the energy states gives rise to a
small splitting of the spectral lines called fine
structure.

88
The Table of Elements
• We can treat the simplest hydrogen atom as a
stationary nucleus, a proton, that has a single
moving particle, an electron, with kinetic energy
p2/2m. The potential energy U(r) due to the
electrostatic attraction between the electron and
the proton is
• In the lowest energy state, which is the ground
state, the principal quantum number n1, l0, and
ml0.
• The allowed energies

89
The Table of Elements
• For atoms with more than one electron, the
Schrödinger equation cannot be solved exactly.
However, the approximation methods allow to
determine the energy levels of the atoms and wave
functions of the electrons with high accuracy.
• As a first approximation, the Z electrons in an
atom are assumed to be noninteracting. The
Schrödinger equation can then be solved, and the
resulting wave function used to calculate the
interaction of the electrons.

90
The Table of Elements
• The state of each electron in an atom is
described by four quantum numbers n,l,m, and ms.
• Beginning with hydrogen, each larger neutral
atom adds one electron. The electrons go into
those states that will give the lowest energy
consistent with the Pauli exclusion principle
• No two electrons in an atom can have the same
set of values for the quantum numbers n, l, m,
and ms
• The energy of the electron is determined mainly
by the principal quantum number n, which is
relate to the radial dependence of the wave
function, and by the orbital angular-momentum
quantum number l.
• The dependence of the energy on l is due to the
interaction of the electrons in the atoms with
each other.

91
The Table of Elements
• The specification of n and l for each electron
in an atom is called the electron configuration.
• The l values are specified by a code
• s p d f g h
• l values 0 1 2 3 4 5
• The n values are referred as shells, which are
identified by another letter code
• shell K L M N ..
• n values 1 2 3 4 ..
• Using the exclusion principle and the
restriction of the quantum numbers (n is a
positive integer, l ranged from
• 0 to n-1, m changed from -l to l in integral
steps, and ms can be either ½ or -½), we can
understand much of the structure of the periodic
table.

92
The Periodic Table
• The energy required to remove the most loosely
electron from an atom in the ground state is
called the ionization energy. This energy is the
binding energy of the last electron placed in the
atom. The ionization energy can be found from
• Hydrogen (Z 1) n1, l 0, m 0, ms ½ -
1s
• Helium (Z 2) two electrons, in the ground
state both electrons are in the K shell,
n1, l0, m0, ms1½, ms2-½ - 1s2
• Lithium (Z3) K shell (n1) is completely full,
one electron on the L-shell 2p1

93
States of Hydrogen Atom
n 1 2 3
l 0 0 1 0 1 2
m 0 0 0,1 0 0 ,1 0, 1, 2
ms ½ ½ ½ ½ ½ ½
Sub shell 2 2 6 2 6 10
Total States 2 8 18
94
The Structure of Atom
• The electrons in atom that have same principal
quantum number n form an electron shell
• total
• K n1 2
• L n2 8
• M n3 18
• N n4 32
• O n5 50

2n2
95
Depending from orbital quantum number l the
electrons forms subshells
n Shell Total elect.
n Shell s (l0) p (l 1) d (l 2) f (l 3) g (l 4) number
1 K 2 - - - - 2
2 L 2 6 - - - 8
3 M 2 6 10 - - 18
4 N 2 6 10 14 - 32
5 O 2 6 10 14 18 50
Number of electrons in subshell
96
Distribution Electrons in atoms
Z Element K L M nlZ
1s 2s 2p 3s3p3d
1 H 1 - - 1s
2 He 2 - - 1s2
3 Li 2 1 - 1s2, 2s
4 Be 2 2 - 1s2, 2s2
5 B 2 2 1 1s2, 2s2, 2p
6 C 2 2 2 1s2, 2s2, 2p2
7 N 2 2 3 1s2, 2s2, 2p3
8 O 2 2 4 1s2, 2s2, 2p4
9 F 2 2 5 1s2, 2s2, 2p5
10 Ne 2 2 6 1s2, 2s2, 2p6
97
• For example the structure for oxygen, O, 1s2,
2s2, 2p4 it mean that 2 electrons are in the
state with n1 and l0 2 electrons
in the state with n2 and l0 and 4 electrons in
the state with n2 and l1.

98
1. Effective Nuclear Charge for an Outer Electron
• Suppose the electron cloud of the outer
electron in the lithium atom in the ground state
were completely outside the electron clouds of
the two inner electrons, the nuclear charge would
be shielded by the two inner electrons and the
effective nuclear charge would be Z'e1e. Then
the energy of the outer electron would be
(13.6eV)/22-3.4eV. However, the ionization
energy of lithium is 5.39eV, not 3.4eV. Use this
fact to calculate the effective nuclear charge
Zeff seen by the outer electron in lithium.

99
2. The Effective Charge of the Rb Ion
• The 5s electron in rubidium sees an effective
charge of 2.771e. Calculate the ionization energy
of this electron.

100
3. Determining Zeff experimentally
• The measured energy of a 3s state of sodium is
-5.138eV. Calculate the value of Zeff.

101
• Example
• The double charged ion N2 is formed by
removing two electrons from a nitrogen atom. (a)
What is the ground state electron configuration
for the N2 ion? (b) Estimate the energy of the
least strongly bond level in the L shell of N2.
• The double charged ion P2 is formed by
removing two electrons from a phosphorus atom.
(c) What is the ground-state electron
configuration for the P2 ion? (d) estimate the
energy of the least strongly bound level in the M
shell of P2.

102
Electron Interaction Energy in Helium
• The ionization energy for helium is 24.6 eV.
• (a) Use this value to calculate the energy of
interaction of the two electrons in the ground
state of the helium atom.
• (b) Use your result to estimate the average
separation of the two electrons.

103
Angular Momentum of the Exited Level Of Hydrogen
• Consider the n4 state of hydrogen. (a) What is
the maximum magnitude L of the orbital angular
momentum? (b) What is the maximum value of LZ?
(c) What is the minimum angle between and
of .

104
A Hydrogen Wave Function.
• The groundstate wave function for the hydrogen
(1s state) is
• (a) Verify that this function is normalized. (b)
What is the probability that the electron will be
found at a distance less than a from the nucleus?

105
Atomic Spectra
• Atomic spectra include optical spectra and
X-ray spectra. Optical spectra result from
transmissions between energy levels of a single
outer electron moving in the field of the
nucleus and core electrons of the atom.
• Characteristic X-ray spectra result from the
excitation of a inner core electron and the
subsequent filling of the vacancy by other
electrons in the atom.

106
Selection Rules
• Transition between energy states with the
emission of a photon are governed by the
following selection rules
• ?ml 0 or 1
• ?l 1