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General Physics (PHY 2140)
Lightning Review
Last lecture:
Lecture 17
1. Quantum physics
9 Atomic Descriptions
9 Atomic Spectra
9 Bohr’
Bohr’s Atomic Theory
9 Quantum Mechanics
9 Quantum Numbers
¾ Modern Physics
9Atomic Physics
9Electron Clouds
9The Pauli Exclusion Principle
9Characteristic X-Rays
9Atomic Transitions
9Lasers and Holography
http://www.physics.wayne.edu/~alan/2140Website/Main.htm
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λ
⎛ 1 1 ⎞
= RH ⎜ 2 − 2 ⎟
⎜ n f ni ⎟
⎝
⎠
If hydrogen obeyed classical physics, we would have no quantized
electron orbits. Therefore the transitions between orbits (energy
levels) could be arbitrarily large or small. This leads to a continuous
spectrum of emitted light.
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Electron Clouds
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Electron Clouds
The graph shows the solution
to the wave equation for
hydrogen in the ground state
„ The curve peaks at the
Bohr radius
„ The electron is not
confined to a particular
orbital distance from the
nucleus
The probability of finding the
electron at the Bohr radius is a
maximum
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me vr = n , n = 1, 2,3,...
Review Problem: Suppose that the electron in the hydrogen atom obeyed
classical rather then quantum mechanics. Why should such an atom emit a
continuous rather then discrete spectrum?
Chapter 28
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2π r = nλ , n = 1, 2,3,...
Ei − E f = hf
The wave function for
hydrogen in the ground state is
symmetric
„ The electron can be found
in a spherical region
surrounding the nucleus
The result is interpreted by
viewing the electron as a cloud
surrounding the nucleus
„ The densest regions of the
cloud represent the highest
probability for finding the
electron
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1
radial probability distribution (r 2 ψ2) = probability of finding
electron at a distance r from the center of the nucleus
90% Probability contours showing relative size of orbitals
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Quantum Number Summary
The values of n can increase from 1 in integer steps
The values of ℓ can range from 0 to nn-1 in integer steps
The values of m ℓ can range from -ℓ to ℓ in integer steps
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28.9 The Pauli Exclusion Principle
Examples
Recall Bohr’
Bohr’s model of an atom. Why don’
don’t all the
electrons stay on the lowest possible orbit?
Pauli’
Pauli’s exclusion principle: no two electrons in an atom
can ever be in the same quantum state
„
In other words, no two electrons in the same atom can have
exactly the same values for n, ℓ, m ℓ, and ms
1. Hydrogen (one electron), 1s1
n = 1, = 0, m = 0, ms = ±1/ 2
2. Helium (two electrons), 1s2
n = 1, = 0, m = 0, ms = + 1 2
n = 1, = 0, m = 0, ms = −1 2
This explains the electronic structure of complex atoms
as a succession of filled energy levels with different
quantum numbers
3. Lithium (three electrons), 1s22s1 n = 1, = 0, m = 0, ms = + 1 2
n = 1, = 0, m = 0, ms = −1 2
n = 2, = 0, m = 0, ms = ± 1 2
See Table 28.4 for other the configurations of other elements.
elements.
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The Periodic Table
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Bit of history: Mendeleev’s original table
The outermost electrons are
primarily responsible for the
chemical properties of the
atom
Mendeleev arranged the
elements according to their
atomic masses and chemical
similarities
The electronic configuration of
the elements explained by
quantum numbers and Pauli’
Pauli’s
Exclusion Principle explains
the configuration:
1s,2s,2p,3s,3p,4s,3d,4p,5s,4d
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3
(a) Write out the electronic configuration of the ground state for
for oxygen (Z
(Z = 8). (b)
Write out values for the set of quantum numbers n, l, ml, and ms for each of the
electrons in oxygen.
Problem: electron configuration of O
Given:
Z=8
(a) Write out the electronic configuration of the ground state
for oxygen (Z
(Z = 8). (b) Write out values for the set of
quantum numbers n, l, ml, and ms for each of the
electrons in oxygen.
Recall that the number of electrons is the same as the
charge of the nucleus. Thus, we have 8 electrons.
n = 1, = 0, m = 0, ms = ± 1 2
n = 2, = 0, m = 0, ms = ± 1 2
n = 2, = 1, m = (0, 1), ms = ± 1 2
Find:
Thus, the electron configuration is:
1s 2 2s 2 2 p 4
structure
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QUICK QUIZ
Krypton (atomic number 36) has how many electrons in its next
to outer shell (n
(n = 3)?
(a) 2
(c) 8
= 2n2
(b) 4
(d) 18
+
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(d). Krypton has a closed configuration consisting of filled n=1, n=2,
and n=3 shells as well as filled 4s
4s and 4p
4p subshells.
subshells. The filled n=3
shell (the next to outer shell in Krypton) has a total of 18 electrons, 2
in the 3s
3s subshell,
subshell, 6 in the 3p
3p subshell and 10 in the 3d
3d subshell.
subshell.
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Recall: Kr needs 36 electrons, the remainder are in the N shell.
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4
Characteristic X-Rays
Explanation of Characteristic X-Rays
When a metal target is
bombarded by highhigh-energy
electrons, xx-rays are emitted
The xx-ray spectrum typically
consists of a broad continuous
spectrum and a series of sharp
lines
„ The lines are dependent on
the metal
„ The lines are called
characteristic xx-rays
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The details of atomic structure can be used to explain
characteristic xx-rays
A bombarding electron collides with an electron in the target
metal that is in an inner shell
If there is sufficient energy, the electron is removed from the
target atom
The vacancy created by the lost electron is filled by an electron
electron
falling to the vacancy from a higher energy level
The transition is accompanied by the emission of a photon
whose energy is equal to the difference between the two levels
„
„
„
„
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Modifications to Bohr’s Theory
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Modifications to Bohr’s Theory, cont.
For atoms with a larger nuclear charge, but with a single
electron (He+, Li2+, Be3+), we must modify the energy to
be:
Since the energy of an emitted photon is the difference
between energy levels, we can write the wavelength of
such a photon as:
me ke2 Z 2 e 4
Z 2 (13.6)
En = −
=−
(eV) n = 1, 2, 3, …
2 2
n2
2 n
1
λ
(Note the difference is the inclusion of Z, the number of protons
protons in
the nucleus)
nucleus)
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Or as:
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=
me ke2 Z 2 e 4 ⎛ 1
1 ⎞
⎜⎜ 2 − 2 ⎟⎟ n = 1, 2, 3, …
3
4π c
⎝ n f ni ⎠
1
λ
=Z
me ke2 e 4 ⎛ 1 1 ⎞
⎜ − ⎟
4π c 3 ⎜⎝ n 2f ni2 ⎟⎠
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Moseley Plot
Problem: X-Rays from Nickel
λ is the wavelength of the Kα
line
„ Kα is the line that is
produced by an electron
falling from the L shell
(n=2) to the K shell (n=1)
From this plot, Moseley was
able to determine the Z values
of other elements and produce
a periodic chart in excellent
agreement with the known
chemical properties of the
elements
The Kα x-ray is emitted when an electron undergoes a
transition form the L shell (n=2) to the K shell (n=1) in a
metal. Calculate the wavelength of the Kα x-ray from a
nickel target, Z=28.
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The Kα x-ray is emitted when an electron undergoes a transition form the L shell (n=2)
to the K shell (n=1) in a metal. Calculate the wavelength of the
the Kα x-ray from a nickel
target, Z=28.
The atomic number for nickel is Z = 28. Using eq. 28.18 and
28.20 we have:
Given:
Z = 28
En = −
Z eff2 (13.6)
n2
eV
EK = −(28 − 1) 2 (13.6 eV) = −9.91×103 eV
(13.6 eV)
EL = −(28 − 3) 2
= −2.13 ×103 eV
2
(2)
hc
Eγ = EL − EK = 7.78 keV =
λ
Find:
λ
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Atomic Transitions – Energy Levels
An atom may have many
possible energy levels
At ordinary temperatures, most
of the atoms in a sample are in
the ground state
Only photons with energies
corresponding to differences
between energy levels can be
absorbed
Thus, the wavelength is:
λ=
hc (6.63 ×10−34 Jis)(3.00 ×108 m/s)
=
Eγ
7.78 keV(1.60 ×10−16 J/keV)
= 1.60 ×10−10 m = 0.160 nm
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Atomic Transitions – Stimulated Absorption
Atomic Transitions – Spontaneous Emission
Once an atom is in an excited
state, there is a constant
probability that it will jump back
to a lower state by emitting a
photon
This process is called
spontaneous emission
The blue dots represent
electrons
When a photon with energy
ΔE is absorbed, one electron
jumps to a higher energy
level
„
„
„
These higher levels are
called excited states
ΔE = hƒ
hƒ = E2 – E1
In general, ΔE can be the
difference between any two
energy levels
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Atomic Transitions – Stimulated Emission
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Population Inversion
An atom is in an excited stated
and a photon is incident on it
The incoming photon
increases the probability that
the excited atom will return to
the ground state
There are two emitted
photons, the incident one and
the emitted one
„ The emitted photon is in
exactly in phase with the
incident photon
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When light is incident on a system of atoms, both stimulated
absorption and stimulated emission are equally probable
Generally, a net absorption occurs since most atoms are in the
ground state
If you can cause more atoms to be in excited states, a net emission
emission
of photons can result
„
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This situation is called a population inversion
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Lasers
Production of a Laser Beam
To achieve laser action, three conditions must be met
„
„
„
The system must be in a state of population inversion
The excited state of the system must be a metastable state
Its lifetime must be long compared to the normal lifetime of
an excited state
The emitted photons must be confined in the system long
enough to allow them to stimulate further emission from other
excited atoms
¾ This is achieved by using reflecting mirrors
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A More Detailed Diagram of a HeHe-Ne Laser Operation
Laser Beam – He Ne Example
The energy level diagram for Ne
The mixture of helium and neon is
confined to a glass tube sealed at
the ends by mirrors
A high voltage applied causes
electrons to sweep through the
tube, producing excited states
When the electron falls to E2 in
Ne, a 632.8 nm photon is emitted
(3s2 → 2p4)
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Holography
Holography, cont
Holography is the production of
threethree-dimensional images of an
object
Light from a laser is split at B
One beam reflects off the object
and onto a photographic plate
The other beam is diverged by
Lens 2 and reflected by the
mirrors before striking the film
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The two beams form a complex interference pattern on the
photographic film
„
„
It can be produced only if the phase relationship of the two waves
waves
remains constant
This is accomplished by using a laser
The hologram records the intensity of the light and the phase
difference between the reference beam and the scattered beam
The image formed has a threethree-dimensional perspective
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