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Atomic Physics

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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?
  • The classical answer is
  • 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
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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
    radial probability density is

80
Radial Probability Density
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
    Z-axis? Give your answers to (a) and (b) in terms
    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
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