More Reasons the Hydrogen Atom
is so Important
• The hydrogen atom is an ideal system for
performing precise comparisons of theory with
experiment
Chapter 28
– Also for improving our understanding of atomic
structure
• Much of what we know about the hydrogen atom
can be extended to other single-electron ions
Atomic Physics
– For example, He+ and Li2+
Early Models of the Atom
•
•
•
Scattering Experiments
J.J. Thomson’s model of the atom
– A volume of positive charge
– Electrons embedded throughout the
volume
A change from Newton’s model of the
atom as a tiny, hard, indestructible sphere
Rutherford, 1911
– Planetary model
– Based on results of thin foil
experiments
– Positive charge is
concentrated in the center of
the atom, called the nucleus
– Electrons orbit the nucleus like
planets orbit the sun
• The source was a naturally radioactive material that
produced alpha particles
• Most of the alpha particles passed though the foil
• A few deflected from their original paths
– Some even reversed their direction of travel
In a Rutherford scattering experiment, an α-particle (charge = +2e)
heads directly toward a gold nucleus (charge = +79e).
The α-particle had a kinetic energy of 5.0 MeV when very far (r → ∞) from the nucleus.
Assuming a head-on collision, the a-particle comes to rest momentarily
at the point of closest approach. From conservation of energy,
With ri → ∞
rf =
0+
ke ( 2e)( 79e)
rf
= KEi +
ke ( 2e)( 79e)
The wavelengths of hydrogen’s
spectral lines can be found from
Equation found by Balmer
RH is the Rydberg constant
RH = 1.097 373 2 x 107 m-1
n is an integer, n = 3, 4, 5 …
The spectral lines correspond to different values of n
•
ri
•
, this gives the distance of closest approach as
9
2
2
−19
158ke e2 158( 8.99× 10 N ⋅ m C )( 1.60× 10 C )
=
KEi
5.0 M eV ( 1.60 × 10-13 J M eV )
= 4.5× 10−14 m = 45 fm
•
1
1 ⎞
⎛ 1
= RH⎜ 2 − 2 ⎟
n ⎠
λ
⎝2
Assuming the gold nucleus to be fixed in space,
determine the distance of closest approach.
KE f + PE f = KEi + PEi , or
Emission Spectrum
2
The Balmer Series has lines whose
wavelengths are given by the preceding
equation
Examples of spectral lines
– n = 3, λ = 656.3 nm
– n = 4, λ = 486.1 nm
Absorption Spectrum
1
Bohr’s Assumptions for Hydrogen
•
Mathematics of Bohr’s Assumptions and Results
The electron moves in circular orbits around the
proton under the influence of the Coulomb force
of attraction
– The Coulomb force produces the centripetal
acceleration
Only certain electron orbits are stable
– These are the orbits in which the atom does not emit energy in the form of
electromagnetic radiation
– Therefore, the energy of the atom remains constant and classical mechanics
can be used to describe the electron’s motion
• Radiation is emitted by the atom when the electron “jumps” from a more energetic
initial state to a lower state
– The “jump” cannot be treated classically
• The electron’s “jump,” continued
– The frequency emitted in the “jump” is related to the change in the
atom’s energy
– It is generally not the same as the frequency of the electron’s orbital
motion
– The frequency is given by Ei – Ef = h ƒ
• The size of the allowed electron orbits is determined by a condition imposed
on the electron’s orbital angular momentum
•
•
•
•
The ionization energy is the energy needed to completely remove the electron
from the atom
– The ionization energy for hydrogen is 13.6 eV
The uppermost level corresponds to E = 0 and n → ∞
Following are four possible transitions for a hydrogen atom
I. ni = 2; nf = 5
II. ni = 5; nf = 3
III. ni = 7; nf = 4
IV. ni = 4; nf = 7
(a) Which transition will emit the shortest-wavelength photon?
(b) For which transition will the atom gain the most energy?
(c) For which transition(s) does the atom lose energy?
The change in the energy of the electron is
⎛ 1
1⎞
ΔE = E f − Ei = 13.6 eV ⎜ 2 − 2 ⎟
⎜ ni n f ⎟
⎝
⎠
1
⎝4
1⎞
⎟ = 2.86 eV (absorption)
25 ⎠
Transition II: ΔE = 13.6 eV ⎛⎜ 1 − 1 ⎞⎟ = − 0.967 eV(emission)
25 9
⎝
Transition
⎠
1
1
III:ΔE = 13.6 eV ⎛⎜ − ⎞⎟ = − 0.572 eV
⎝ 49 16 ⎠
Transition IV:
1⎞
⎛ 1
ΔE = 13.6 eV ⎜ − ⎟ = 0.572 eV
⎝ 16 49 ⎠
hc
hc
=
(a) Since , λ =
Eγ − ΔE
The total energy of the atom
•
•
E = KE + PE =
1
e2
m v 2 − ke
2 e
r
The energy of the atom can also be
expressed as
–
E =−
ke
e2
v2
= me
r2
r
kee2
2r
The radii of the Bohr orbits are quantized
rn =
2 2
nh
me kee2
v 2 = ke
e2
n2 h2
= 2 2
me r 2
me r
n = 1, 2, 3, K
– This is based on the assumption that the electron can only exist in certain
allowed orbits determined by the integer n
• When n = 1, the orbit has the smallest radius, called the Bohr radius, ao
• ao = 0.052 9 nm
•
A general expression for the radius of any
orbit in a hydrogen atom is
– rn = n2 ao
The energy of any orbit is
– En = - 13.6 eV/ n2
The lowest energy state is called the ground state
– This corresponds to n = 1
– Energy is –13.6 eV
The next energy level has an energy of –3.40 eV
– The energies can be compiled in an energy level
diagram
⎛
Transition I: ΔE = 13.6 eV ⎜ −
•
p =h/λ
Energy Level Diagram
Radii and Energy of Orbits
•
Electron’s orbital angular momentum
– me v r = n ħ where n = 1, 2, 3, …
–
•
•
•
(emission)
(absorption)
transition II emits the shortest wavelength photon.
(b) The atom gains the most energy in
transition I
(c) The atom loses energy intransitions II and III
•
•
•
•
The value of RH from Bohr’s analysis is in excellent
agreement with the experimental value
A more generalized equation can be used to find the
wavelengths of any spectral lines
⎛ 1
1
1⎞
= RH ⎜⎜ 2 − 2 ⎟⎟
For the Balmer series, nf = 2
λ
⎝ n f ni ⎠
– For the Lyman series, nf = 1
Whenever an transition occurs between a state,
ni to another state, nf (where ni > nf), a photon is emitted
– The photon has a frequency f = (Ei – Ef)/h and wavelength λ
Modifications of the Bohr Theory – Elliptical Orbits
•
Sommerfeld extended the results to include elliptical orbits
– Retained the principle quantum number, n
• Determines the energy of the allowed states
– Added the orbital quantum number, ℓ
• ℓ ranges from 0 to n-1 in integer steps
– All states with the same principle quantum number are said to form a shell
– The states with given values of n and ℓ are said to form a subshell
(a) If an electron makes a transition from the n = 4 Bohr orbit to the n = 2 orbit,
determine the wavelength of the photon created in the process.
(b) Assuming that the atom was initially at rest, determine the recoil speed of
the hydrogen atom when this photon is emitted.
(a) The energy emitted by the atom is
⎛ 1 1⎞
ΔE = E4 − E2 = − 13.6 eV ⎜ 2 − 2 ⎟ = 2.55 eV
⎝4 2 ⎠
The wavelength of the photon produced is then
λ=
−34
8
hc hc ( 6.63× 10 J⋅ s)( 3.00× 10 m s)
=
=
Eγ ΔE
( 2.55 eV ) (1.60× 10−19 J eV )
= 4.88× 10−7 m = 488 nm
(b) Since momentum must be conserved, the photon and
the atom go in opposite directions with equal magnitude momenta.
Thus,
or
p= m v= h λ
atom
v=
h
6.63× 10−34 J⋅ s
=
= 0.814 m s
m atom λ ( 1.67 × 10−27 kg )( 4.88 × 10−7 m )
2
Modifications of the Bohr Theory
Spin Magnetic Quantum Number
Zeeman Effect
•
Another modification was needed to account for the Zeeman effect
– The Zeeman effect is the splitting of spectral lines in a
strong magnetic field
– This indicates that the energy of an electron is slightly modified when
the atom is immersed in a magnetic field
– A new quantum number, m ℓ, called the orbital magnetic quantum
number, had to be introduced
• m ℓ can vary from - ℓ to + ℓ in integer steps
• It is convenient to think of the
electron as spinning on its
axis
• There are two directions for
the spin
– Spin up, ms = ½
– Spin down, ms = -½
Fine Structure
•
High resolution spectrometers show that spectral lines are, in fact,
two very closely spaced lines, even in the absence of a magnetic field
– This splitting is called fine structure
– Another quantum number, ms, called the spin magnetic quantum
number, was introduced to explain the fine structure
• There is a slight energy
difference between the two
spins and this accounts for the
doublet in some lines
de Broglie Waves
in the Hydrogen Atom
•
•
•
•
One of Bohr’s postulates was the angular
momentum of the electron is quantized, but
there was no explanation why the restriction
occurred
de Broglie assumed that the electron orbit
would be stable only if it contained an
integral number of electron wavelengths
By applying wave theory to the electrons in an
atom, de Broglie was able to explain the
appearance of integers in Bohr’s equations as
a natural consequence of standing wave
patterns
In general, the circumference must equal
some integer number of wavelengths
2πr=nλ
n = 1, 2, …
Quantum Number Summary
•
•
•
Principle Quantum Number, n: The values
of n can range from 1 to ∞ in integer steps
Orbital Quantum Number, ℓ: The values of ℓ
can range from 0 to n-1 in integer steps
Orbital Magnetic Quantum Number, m ℓ :
The values of m ℓ can range from -ℓ to ℓ in
integer steps
• Spin Magnetic Quantum Number, ms:
The values of m s can range from - ½ to ½
me v r = n ħ
Schrödinger’s wave equation was subsequently applied to atomic systems
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
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
The Pauli Exclusion Principle
• No two electrons in an atom can ever have the same set of values of
the quantum numbers n, ℓ, m ℓ, and ms
• This explains the electronic structure of complex atoms as a
succession of filled energy levels with different quantum numbers
Filling Shells
• As a general rule, the order that electrons fill an atom’s subshell is:
- Once one subshell is filled, the next electron goes into the vacant
subshell that is lowest in energy
• Otherwise, the electron would radiate energy until it reached the
subshell with the lowest energy
- A subshell is filled when it holds 2(2ℓ+1) electrons
3
Zirconium (Z = 40) has two electrons in an incomplete d subshell.
(a) What are the values of n and ℓ for each electron?
(b) What are all possible values of mℓ and ms?
(c) What is the electron configuration in the ground state of zirconium?
(a) Zirconium, with 40 electrons, has 4 electrons outside a closed Krypton core.
The Krypton core, with 36 electrons, has all states up through the subshell filled.
Normally, one would expect the next 4 electrons to go into the subshell.
However, an exception to the rule occurs at this point, and the subshell fills
(with 2 electrons) before the subshell starts filling.
The two remaining electrons in Zirconium are in an incomplete subshell.
Thus, n = 4, and l = 2
for each of these electrons.
(b)For electrons in the 4d
ml
are
subshell, with
m l = 0,± 1,± 2
l=2
and those for
, the possible values of
ms
are
m s = ±1 2
(c) We have 40 electrons, so the electron configuration is:
1s2 2s2 2p6 3s2 3p6 3d10 4s2 4p6 4d2 5s2 = [K r]4d2 5s2
Table 28-3, p.918
The Periodic 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’s
Exclusion Principle explains the configuration
Table 28-4, p.919
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
Atomic Transitions – Stimulated
Absorption
• The blue dots represent
electrons
• When a photon with
energy ∆E is absorbed,
one electron jumps to a
higher energy level
– These higher levels are
called excited states
– ∆E = hƒ = E2 – E1
– In general, ∆E can be the
difference between any two
energy levels
4
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
Atomic Transitions – Stimulated
Emission
• 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
Population Inversion
• 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 this situation can be inverted, you can cause
more atoms to be in excited states than in the
ground state, a net emission of photons can
result
– This situation is called a population inversion
Lasers
• To achieve laser action, three conditions must
be met
– The system must be in a state of population inversion
• More atoms in an excited state than the ground state
– 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
• This is achieved by using reflecting mirrors
Laser Beam – He Ne Example
Production of a Laser Beam
• The energy level diagram for
Ne in a He-Ne laser
• 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
from E*3 in Ne, a 632.8 nm
photon is emitted
5
Holography
Holography, cont
• Holography is the
production of threedimensional 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
• 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 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 three-dimensional
perspective
Energy Bands
Energy Bands in Solids
•
•
Sodium example
Blue represents energy bands
occupied by the sodium electrons
when the atoms are in their ground
states
Gold represents energy bands that
are empty
White represents energy gaps
Electrons can have any energy within
the allowed bands
Electrons cannot have energies in the
gaps
•
•
•
•
•
•
In solids, the discrete energy levels of isolated atoms broaden into
allowed energy bands separated by forbidden gaps
The separation and the electron population of the highest bands
determine whether the solid is a conductor, an insulator, or a
semiconductor
•
•
•
The valence band is the highest filled band
The conduction band is the next higher empty band
The energy gap has an energy, Eg, equal to the difference in energy between
the top of the valence band and the bottom of the conduction band
Conductors
•
•
•
When a voltage is applied to a
conductor, the electrons accelerate
and gain energy
In quantum terms, electron energies
increase if there are a high number
of unoccupied energy levels for the
electron to jump to
For example, it takes very little
energy for electrons to jump from the
partially filled to one of the nearby
empty states
Semiconductors
•
•
•
Insulators
•
•
•
Fig. 28-25, p.926
The valence band is completely full of
electrons
A large band gap separates the valence
and conduction bands
A large amount of energy is needed for
an electron to be able to jump from the
valence to the conduction band
– The minimum required energy is Eg
•
A semiconductor has a small energy
gap
Thermally excited electrons have
enough energy to cross the band
gap
The resistivity of semiconductors
decreases with increases in
temperature
The white area in the valence band
represents holes
6
Semiconductors, cont
• Holes are empty states in the valence band
created by electrons that have jumped to the
conduction band
• Some electrons in the valence band move to fill
the holes and therefore also carry current
• The valence electrons that fill the holes leave
behind other holes
– It is common to view the conduction process in the
valence band as a flow of positive holes toward the
negative electrode applied to the semiconductor
Doping in Semiconductors
• Doping is the adding of impurities to a
semiconductor
– Generally about 1 impurity atom per 107
semiconductor atoms
• Doping results in both the band structure
and the resistivity being changed
p-type Semiconductors
• Acceptor atoms are doping materials that
contain one less electron than the
semiconductor material
– A hole is left where the missing electron would be
• The energy level of the hole lies in the energy
gap, just above the valence band
• An electron from the valence band has enough
thermal energy to fill this impurity level, leaving
behind a hole in the valence band
Movement of Charges in
Semiconductors
• An external voltage is
supplied
• Electrons move toward
the positive electrode
• Holes move toward the
negative electrode
• There is a symmetrical
current process in a
semiconductor
n-type Semiconductors
• Donor atoms are doping materials that contain
one more electron than the semiconductor
material
– This creates an essentially free electron with an
energy level in the energy gap, just below the
conduction band
• Only a small amount of thermal energy is
needed to cause this electron to move into the
conduction band
A p-n Junction
• A p-n junction is
formed when a p-type
semiconductor is
joined to an n-type
• Three distinct regions
exist
– A p region
– An n region
– A depletion region
7
The Depletion Region
• Mobile donor electrons from the n side nearest
the junction diffuse to the p side, leaving behind
immobile positive ions
• At the same time, holes from the p side nearest
the junction diffuse to the n side and leave
behind a region of fixed negative ions
• The resulting depletion region is depleted of
mobile charge carriers
– There is also an electric field in this region that sweeps
out mobile charge carriers to keep the region truly
depleted
Diode Action
•
•
•
The p-n junction has the ability to pass current in only one
direction
When the p-side is connected to a positive terminal, the
device is forward biased and current flows
When the n-side is connected to the positive terminal, the
device is reverse biased and a very small reverse current
results
Applications of Semiconductor
Diodes
• Rectifiers
– Change AC voltage to DC voltage
– A half-wave rectifier allows current to flow during half the
AC cycle
– A full-wave rectifier rectifies both halves of the AC cycle
• Transistors
– May be used to amplify small signals
• Integrated circuit
– A collection of interconnected transistors, diodes, resistors
and capacitors fabricated on a single piece of silicon
8