Chapter 41
Atomic Structure
Sir Joseph John Thomson
“J. J.” Thomson
1856 - 1940
Discovered the
electron
Did extensive work
with cathode ray
deflections
1906 Nobel Prize for
discovery of electron
Early Models of the Atom
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
Scattering Experiments
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
Early Models of the Atom, 2
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
Difficulties with the
Rutherford Model
Atoms emit certain discrete characteristic
frequencies of electromagnetic radiation
The Rutherford model is unable to explain this
phenomena
Rutherford’s electrons are undergoing a
centripetal acceleration and so should
radiate electromagnetic waves of the same
frequency
The radius should steadily decrease as this
radiation is given off
The electron should eventually spiral into the
nucleus, but it doesn’t
Emission Spectra
A gas at low pressure has a voltage
applied to it
A gas emits light characteristic of the gas
When the emitted light is analyzed with a
spectrometer, a series of discrete bright
lines is observed
Each line has a different wavelength and color
This series of lines is called an emission
spectrum
Examples of Emission Spectra
Emission Spectrum of
Hydrogen – Equation
The wavelengths of hydrogen’s spectral
lines can be found from
1
1
 1
= RH  2 − 2 
λ
n 
2
RH is the Rydberg constant
RH = 1.097 373 2 x 107 m-1
n is an integer, n = 1, 2, 3, …
The spectral lines correspond to different
values of n
Spectral Lines of Hydrogen
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
Other Series
Lyman series
Far ultraviolet
Ends at energy level 1
Paschen series
Infrared (longer than Balmer)
Ends at energy level 3
General Rydberg Equation
The Rydberg equation can apply to
any series
1
1 
 1
= RH  2 − 2 
λ
n 
m
m and n are positive integers
n>m
Absorption Spectra
An element can also absorb light at specific
wavelengths
An absorption spectrum can be obtained by
passing a continuous radiation spectrum
through a vapor of the gas
The absorption spectrum consists of a
series of dark lines superimposed on the
otherwise continuous spectrum
The dark lines of the absorption spectrum
coincide with the bright lines of the emission
spectrum
Absorption Spectrum of
Hydrogen
Applications of Absorption
Spectrum
The continuous spectrum emitted
by the Sun passes through the
cooler gases of the Sun’s
atmosphere
The various absorption lines can be
used to identify elements in the solar
atmosphere
Led to the discovery of helium
Niels Bohr
1885 – 1962
Participated in the early
development of
quantum mechanics
Headed Institute in
Copenhagen
1922 Nobel Prize for
structure of atoms and
radiation from atoms
The Bohr Theory of Hydrogen
In 1913 Bohr provided an
explanation of atomic spectra that
includes some features of the
currently accepted theory
His model includes both classical
and non-classical ideas
His model included an attempt to
explain why the atom was stable
Bohr’s Assumptions for
Hydrogen
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
Bohr’s Assumptions, cont
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
Bohr’s Assumptions, final
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
Mathematics of Bohr’s
Assumptions and Results
Electron’s orbital angular momentum
The total energy of the atom
me v r = n ħ where n = 1, 2, 3, …
2
1
e
E = KE + PE = mev 2 − ke
r
2
The energy of the atom can also be
expressed as
kee2
E =−
2r
Bohr Radius
The radii of the Bohr orbits are
quantized
n2 h2
rn =
n = 1, 2, 3, K
2
me kee
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
Radii and Energy of Orbits
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
Specific Energy Levels
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
Specific Energy Levels, cont
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 → ∞
Energy Level Diagram
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
Generalized Equation
 1
1
1
= RH  2 − 2 
λ
 nf ni 
For the Balmer series, nf = 2
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
Bohr’s Correspondence
Principle
Bohr’s Correspondence Principle
states that quantum mechanics is
in agreement with classical physics
when the energy differences
between quantized levels are very
small
Similar to having Newtonian
Mechanics be a special case of
relativistic mechanics when v << c
Successes of the Bohr Theory
Explained several features of the
hydrogen spectrum
Can be extended to “hydrogen-like”
atoms
Those with one electron
Ze2 needs to be substituted for e2 in
equations
Z is the atomic number of the element
Quantum Mechanics and
the Hydrogen Atom
One of the first great achievements of
quantum mechanics was the solution of
the wave equation for the hydrogen
atom (U = -ke2/r)
The energies of the allowed states are
in exact agreement with the values
obtained by Bohr when the allowed
energy levels depend only on the
principle quantum numbers
Quantum Numbers
A set of number the “state” of the atom
n – principle quantum number
Two other quantum numbers emerge
from the solution of Schrödinger
equation (spherical coordinates)
– orbital quantum number
m – orbital magnetic quantum number
separation of variables
Ψ(r,θ,φ) = R(r) Θ(θ) Φ(φ)
Quantum Number Summary
The values of n can range from 1
to ∞ in integer steps
The values of can range from 0
to n-1 in integer steps
The values of m can range from to in integer steps
Angular momentum in
spherical polar coordinates
Spherical polar coordinates are the natural coordinate system in
which to describe angular momentum. In these coordinates,
∇ = eˆ r
∂ 1
∂
1
∂
+ eˆ θ
+
eˆ φ
∂r r ∂θ r sin θ ∂φ
z
θ
y r = reˆ r
r
So the full (vector) angular momentum
operator can be written
φ
x
L = −ihr × ∇ = −ihreˆ r × ∇ =
To find z-component, note that unit
vector k in z-direction satisfies
∂
Lz = −ih
∂φ
L2 in spherical polar coordinates
L2 = L ⋅ L = −h 2 (r × ∇) ⋅ (r × ∇)
= − h 2  r 2∇ 2 − (r ⋅∇)(∇ ⋅ r ) 
 1 ∂ 
∂ 
1
∂2 
L = −h 
 sin θ
+
2
2 
sin
θ
∂
θ
∂
θ
sin
θ
∂
φ




2
2
Depends only on angular behaviour of wavefunction.
Closely related to angular part of Laplacian
1 ∂  2 ∂  L2
∇ = 2 r
− 2 2
r ∂r  ∂r  h r
2
Eigenvalues and eigenfunctions
Look for simultaneous eigenfunctions of L2 and one
component of L (conventional to choose Lz)
Eigenvalues and
eigenfunctions of Lz:
Physical boundary condition: wave-function
must be single-valued
Φ (φ + 2π ) = Φ (φ )
Quantization of angular
momentum about z-axis
(compare Bohr model)
Eigenvalues and eigenfunctions
Now look for eigenfunctions of L2, in the form
f (θ , φ ) = Θ (θ )Φ (φ ) = exp(imφ )Θ (θ )
(ensures solutions remain eigenfunctions of
Lz, as we want)
Eigenvalue condition becomes
Lˆ2 [exp(imφ )Θ (θ )] = β h 2 exp(imφ )Θ (θ )
Let eigenvalue = β h 2
The Legendre equation
Make the substitution
µ = cos θ
∂
dµ ∂
∂
⇒
=
= − sin θ
∂θ d θ ∂µ
∂µ
This is exactly the Legendre equation, can be
solved using the Frobenius method.
Legendre polynomials and
associated Legendre functions
In order for solutions to exist that remain finite at µ=±1 (i.e. at
θ=0 and θ=π) we require that the eigenvalue satisfies
β = l (l + 1), where l = 0,1, 2,K
(like SHO, where we found restrictions on
energy eigenvalue in order to produce
normalizable solutions)
The finite solutions are then the associated Legendre functions, which
can be written in terms of the Legendre polynomials:
m
m /2  d 
Pl m ( µ ) = (1 − µ 2 )

 Pl ( µ )
d
µ


where m is an integer constrained to lie between –l and +l.
Spherical harmonics
The full eigenfunctions can also be written as spherical harmonics:
Because they are eigenfunctions of Hermitian operators with different eigenvalues,
they are automatically orthogonal when integrated over all angles (i.e. over the surface
of the unit sphere). The constants C are conventionally defined so the spherical
harmonics obey the following important normalization condition:
First few examples
Remember
x = r sin θ cos φ
y = r sin θ sin φ
z = r cos θ
Y00 (θ , φ ) =
Y11 (θ , φ ) = −
1
4π
3
3 ( x + iy )
sin θ exp(iφ ) = −
8π
8π
r
Y10 (θ , φ ) =
3
3 z
cos θ =
4π
4π r
Y1−1 (θ , φ ) =
3
3 ( x − iy )
sin θ exp(−iφ ) =
8π
8π
r
Shapes of the spherical harmonics
z
Y11
Y00
Y10
y
l =1
x
l=0
m=0
Yl m =
1
4π
Re[Y11 ]
m=0
l =1
Yl m =
m =1
Yl m = −
3
cos θ Imaginary
4π
3
sin θ exp(iφ )
8π
To read plots: distance from origin corresponds to magnitude (modulus) of
plotted quantity; colour corresponds to phase (argument).
Real
Shapes of spherical harmonics
z
y
2
2
Y
Y20
1
2
Y
x
2
2
Re[Y ]
Re[Y21 ]
l=2
l=2
m=2
m =1
Yl m =
15
sin 2 θ exp(2iφ )
32π
Yl m = −
l=2
m=0
Yl m =
5
(3cos 2 θ − 1)
16π
Imaginary
15
sin θ cos θ exp(iφ )
8π
To read plots: distance from origin corresponds to magnitude (modulus) of
plotted quantity; colour corresponds to phase (argument).
Real
Vector model for angular momentum
To summarize:
Eigenvalues of Lˆ2 are l (l + 1) h 2 , with l = 0,1, 2,K
l is known as the principal angular momentum quantum number:
determines the magnitude of the angular momentum
Eigenvalues of Lˆ z are mh, with m = −l ,K , −1, 0,1,K + l
m is known as the magnetic quantum number: determines the
component of angular momentum along a chosen axis (the z-axis)
These states do not correspond to well-defined values of Lx and Ly, since these
operators do not commute with Lz.
Semiclassical picture: each solution corresponds to a cone of angular
momentum vectors, all with the same magnitude and the same z-component.
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
Electron Clouds, cont
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
Imposing a magnetic field
When = 2, there are five degenerate states that we
can label with a new quantum number ml. These
states will “spilt” when an external field is imposed.
L=rxp
L2Ψ = ℓ(ℓ+1)ћ2Ψ
L = √ℓ(ℓ+1) ћ
Lz = mℓ ћ
mℓ = −ℓ to ℓ
z-axis :
external field
Shells and Subshells
All states with the same principle
quantum number, n, are said to form a
shell
Shells are identified as K, L, M, …
Correspond to n = 1, 2, 3, …
The states with given values of m and are said to form a subshell
s, p, d, f, g states referred to
= 0, 1, 2, 3, 4
Example – H atom
Radial function & probabilities
Result of probability plots for all components
•
Once we have a wave function for each component, we can square it
and look at a graph of that function. This variety of shapes and sizes forms
atomic orbitals and forms the basis for reactivity and bonding models.
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
This is seen in the quantum number m
U = - µ·B = - mµBB
µB is a constant = eћ/2me = 57.88 µeV/T
called the Bohr magneton
Magnetic moments,
the Zeeman Effect, and splitting
•
•
Orbitals that look
equivalent in the
absence of an
external field
(degenerate) will
separate as they
interact with that
field (split or lift
the degeneracy).
e.g. at 1 T
∆E = 0.058 eV
Selection rules
∆ =+/-1, ∆m =0,+/-1
•
Some transitions are allowed and some forbidden. The selection
rules are different depending on the type of transition, but they will
always deal with the energies and angular momenta involved in the
change.
Electron spin
•
Anomalous Zeeman effects led spectroscopists to wonder if all the
variables (quantum numbers or quantum states) had been
accounted for.
Spin Magnetic Quantum Number
Some spectral lines were
found to actually be two
very closely spaced lines
This splitting is called
fine structure
A fourth quantum
number, spin magnetic
quantum number, was
introduced to explain
fine structure
S=½
S2Ψ = S(S+1)ћ2Ψ
S = √S(S+1) ћ
Sz = msћ
ms = ± ½
U = - µzB ≈ msµBB
z-axis :
external field
Spin Magnetic Quantum Number
It is convenient to think of
the electron as spinning
on its axis
There are two directions
for the spin
The electron is not
physically spinning
Spin up, ms = ½
Spin down, ms = -½
There is a slight energy
difference between the
two spins and this
accounts for the doublet in
some lines
Spin Notes
A classical description of electron
spin is incorrect
Since the electron cannot be located
precisely in space, it cannot be
considered to be a spinning solid
object
P. A. M. Dirac developed a
relativistic quantum theory in
which spin appears naturally
Spin Angular Momentum
The magnitude of the spin angular
3
momentum is
S =
s ( s + 1) h =
2
h
The spin angular momentum can have two
orientations relative to a z axis, specified by
the spin quantum number ms = ± ½
ms = + ½ corresponds to the spin up case
ms = - ½ corresponds to the spin down case
Spin Angular Momentum,
final
The z component
of spin angular
momentum is
Sz = msh = ± ½ h
Spin angular
moment is
quantized
Spin Magnetic Moment
The spin magnetic moment µspin is
related to the spin angular momentum
r
r
e
by µ = − S
spin
me
The z component of the spin magnetic
moment can have values
µ spin , z
eh
=±
2me
Wolfgang Pauli
1900 – 1958
Contributions include
Major review of
relativity
Exclusion Principle
Connect between
electron spin and
statistics
Theories of relativistic
quantum
electrodynamics
Neutrino hypothesis
Nuclear spin
hypothesis
The Pauli Exclusion Principle
No two electrons in an atom can
ever have the same set of values
of the quantum numbers n, , m ,
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
Orbitals
An orbital is defined as the atomic state
characterized by the quantum numbers n, ℓ
and mℓ
From the exclusion principle, it can be seen
that only two electrons can be present in any
orbital
One electron will have spin up and one spin down
Each orbital is limited to two electrons, the
number of electrons that can occupy the
various shells is also limited
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
Periodic Table, element
The periodic table groups elements with the
same structure in their outermost shells
together in columns. The description of
each element is included.
Quantum states for the first four shells
Ground state electron
configurations to Zn
Allowed Quantum States,
Example with n = 3
In general, each shell can accommodate
up to 2n2 electrons
Hund’s Rule
Hund’s Rule states that when an
atom has orbitals of equal energy,
the order in which they are filled
by electrons is such that a
maximum number of electrons
have unpaired spins
Some exceptions to the rule occur in
elements having subshells that are
close to being filled or half-filled
Configuration of Some
Electron States
Multielectron Atoms
For multielectron atoms, the positive nuclear
charge Ze is largely shielded by the negative
charge of the inner shell electrons
The outer electrons interact with a net charge
that is smaller than the nuclear charge
Allowed energies are
E
n
1 3 .6 Z
= −
n 2
2
e ff
eV
Zeff depends on n and ℓ
Screening
Characteristic X-Rays
When a metal target is
bombarded by highenergy electrons, x-rays
are emitted
The x-ray spectrum
typically consists of a
broad continuous
spectrum and a series of
sharp lines
The lines are dependent on
the metal of the target
The lines are called
characteristic x-rays
X-ray spectroscopy
•
X-rays have enough
energy to probe any
electron within an
atom. Moseley’s
results with x-ray
studies are
summarized in the
next two slides.
Explanation of
Characteristic X-Rays
The details of atomic structure can be
used to explain characteristic x-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 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
Moseley Plot
λ is the wavelength of the
Kα line
Kα is the line that is
produced by an electron
falling from the L shell to
the K shell
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
Moseley’s Law
f ~ (Z-1)2
(Z-1) due to screening
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
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 you can cause more atoms to be in
excited states, a net emission of
photons can result
This situation is called a population
inversion
Question
The only valid electron state and shell designation among the
following is:
1s, L
2s, K
3f, M
1p, K
2p, L
Question
The only valid electron state and shell designation among the
following is:
1s, L
2s, K
3f, M (no f)
1p, K
2p, L
Question
The correct ground state electron configuration of boron
(Z = 5) is:
1s22p3
1s21p22s
1s22s22p
1s22p23s
1s22s22p3
Question
The correct ground state electron configuration of boron
(Z = 5) is:
1s22p3
1s21p22s
1s22s22p
1s22p23s
1s22s22p3
Question
Perhaps the most famous observation in spectroscopy was
the recognition that the yellow-orange line in the spectrum
of sodium is in fact a narrowly separated doublet. The
explanation of this splitting is most closely related to
the existence of spin angular momentum of the electron.
the existence of quantized orbital angular momentum of
the electron.
the exclusion principle.
solution of the Schrodinger equation.
the de Broglie wavelength of the electron.
Question
Perhaps the most famous observation in spectroscopy was
the recognition that the yellow-orange line in the spectrum
of sodium is in fact a narrowly separated doublet. The
explanation of this splitting is most closely related to
the existence of spin angular momentum of the electron.
the existence of quantized orbital angular momentum of the
electron.
the exclusion principle.
solution of the Schrodinger equation.
the de Broglie wavelength of the electron.
Question
Consider an atom with the electron configuration
1s22s22p63s23p6. Which of the following is an accurate
statement concerning this atom?
This atom is in an excited state.
This atom is most likely to give rise to an ion with charge +2e.
The atomic number of this atom is Z = 11.
This atom has a nonzero angular momentum.
This atom would probably be very inert chemically.
Question
Consider an atom with the electron configuration
1s22s22p63s23p6. Which of the following is an accurate
statement concerning this atom?
This atom is in an excited state.
This atom is most likely to give rise to an ion with charge +2e.
The atomic number of this atom is Z = 11.
This atom has a nonzero angular momentum.
This atom would probably be very inert chemically.
Question
This illustration shows the possible
orientations of the angular
momentum vector in a hydrogen
atom state with l = 2. For a given
value of Lz,
A. the angular momentum vector can point in any direction
tangent to the cone for that value of Lz.
B. the electron orbits along the corresponding red circle, so the
orbit may or may not have the nucleus at its center.
C. Both A. and B. are true.
D. Neither A. nor B. is true.
Question
This illustration shows the possible
orientations of the angular
momentum vector in a hydrogen
atom state with l = 2. For a given
value of Lz,
A. the angular momentum vector can point in any direction
tangent to the cone for that value of Lz.
B. the electron orbits along the corresponding red circle, so the
orbit may or may not have the nucleus at its center.
C. Both A. and B. are true.
D. Neither A. nor B. is true.
Question
This illustration shows radial
probability distribution functions
for three hydrogen-atom wave
functions, plotted versus r/a (r =
distance from the center of the
atom and a = 0.0529 nm). It
follows that
A. an electron in a 4p state is always farther from the center of
the atom than is an electron in a 2p state.
B. an electron in a 2p state can be found at the atom’s center.
C. a 3p state has 3 units of orbital angular momentum.
D. none of the above is true.
Question
This illustration shows radial
probability distribution functions
for three hydrogen-atom wave
functions, plotted versus r/a (r =
distance from the center of the
atom and a = 0.0529 nm). It
follows that
A. an electron in a 4p state is always farther from the center of
the atom than is an electron in a 2p state.
B. an electron in a 2p state can be found at the atom’s center.
C. a 3p state has 3 units of orbital angular momentum.
D. none of the above is true.
Question
If a sample of gas atoms is placed in a strong, uniform magnetic
field, the spectrum of the atoms changes. Why is this?
A. Electrons have magnetic moments due to their spin and their
orbital motion.
B. The nucleus and the electrons are pushed in opposite
directions by a magnetic field.
C. Electrons are drawn into regions of strong magnetic field.
D. Electrons are repelled from regions of strong magnetic field.
E. none of the above
Question
If a sample of gas atoms is placed in a strong, uniform magnetic
field, the spectrum of the atoms changes. Why is this?
A. Electrons have magnetic moments due to their spin and their
orbital motion.
B. The nucleus and the electrons are pushed in opposite
directions by a magnetic field.
C. Electrons are drawn into regions of strong magnetic field.
D. Electrons are repelled from regions of strong magnetic field.
E. none of the above
Question
Which statement about electron spin is correct?
A. The spin angular momentum has two possible magnitudes
and two possible values of its z-component.
B. The spin angular momentum has only one possible
magnitude but two possible values of its z-component.
C. The spin angular momentum two possible magnitudes but
only one possible value of its z-component.
D. The spin angular momentum has only one possible
magnitude and only one possible value of its z-component.
E. none of the above
Question
Which statement about electron spin is correct?
A. The spin angular momentum has two possible magnitudes
and two possible values of its z-component.
B. The spin angular momentum has only one possible
magnitude but two possible values of its z-component.
C. The spin angular momentum two possible magnitudes but
only one possible value of its z-component.
D. The spin angular momentum has only one possible
magnitude and only one possible value of its z-component.
E. none of the above
Question
Potassium has 19 electrons. It is relatively easy to remove one
electron, but substantially more difficult to then remove a
second electron. Why is this?
A. The second electron feels a stronger attraction to the
other electrons than did the first electron that was removed.
B. When the first electron is removed, the other electrons
readjust their orbits so that they are closer to the nucleus.
C. The first electron to be removed was screened from more
of the charge on the nucleus than is the second electron.
D. all of the above
E. none of the above
Question
Potassium has 19 electrons. It is relatively easy to remove one
electron, but substantially more difficult to then remove a
second electron. Why is this?
A. The second electron feels a stronger attraction to the
other electrons than did the first electron that was removed.
B. When the first electron is removed, the other electrons
readjust their orbits so that they are closer to the nucleus.
C. The first electron to be removed was screened from more
of the charge on the nucleus than is the second electron.
D. all of the above
E. none of the above
Question
Ordinary hydrogen has one electron and one proton. It requires
10.2 eV of energy to take an electron from the innermost (K)
shell in hydrogen and move it into the next (L) shell.
Uranium has 92 electrons and 92 protons. The energy required to
move an electron from the K shell to the L shell of uranium is
A. (91)(10.2 eV).
B. (92)(10.2 eV).
C. (91)2(10.2 eV).
D. (92)2(10.2 eV).
E. none of the above
Question
Ordinary hydrogen has one electron and one proton. It requires
10.2 eV of energy to take an electron from the innermost (K)
shell in hydrogen and move it into the next (L) shell.
Uranium has 92 electrons and 92 protons. The energy required to
move an electron from the K shell to the L shell of uranium is
A. (91)(10.2 eV).
B. (92)(10.2 eV).
C. (91)2(10.2 eV).
D. (92)2(10.2 eV).
E. none of the above