Download Chapter 28 Atomic Physics

Physics Including Human Applications
614 Chapter 28
Atomic Physics
GOALS
After you have mastered the contents of this chapter, you will be able to achieve the
following goals:
Definitions
Define each of the following terms and use it in an operational definition:
atomic number
ionization energy
Bohr radius
Beer's law
quantum number
Lambert's law
energy level
optical density
spectra absorbance
laser
Bohr Model
State the assumptions and predictions of the Bohr model of the hydrogen atom. Deduce
the concept of ionization potential.
Pauli Exclusion Principle
Use the Pauli exclusion principle and the four quantum numbers; n, l, ml, and ms to
account for the periodic chart of the elements.
Bohr Model Problems
Solve problems using the Bohr equations for hydrogen.
Energy Level Diagrams
Use energy level diagrams to explain the emission and absorption of radiation by
atomic systems.
Spectrometry and Lambert's Law
Sketch a simple spectrometer, and discuss how it can be used in measuring the
concentration of an element in a solution.
Laser
Explain the principle of laser action.
PREREQUISITES
Before you begin this chapter you should have achieved the goals in Chapter 4, Forces
and Newton's Laws, Chapter 5, Energy, Chapter 21, Electrical Properties of Matter, and
Chapter 27, Quantum and Relativistic Physics.
Chapter 28 Atomic Physics
Physics Including Human Applications
615 Chapter 28
Atomic Physics
28.1 Introduction
The realm of quantum physics is not an obvious part of your daily experience. Where
do you expect to find observable evidence of the quantum model? Although the
quantum model was originated to solve the problem of thermal radiation, it was the
atomic model of matter that gave rise to the impressive early successes of the quantum
model. Do you know the atomic basis for the operation of the common fluorescent
light?
Emission and absorption spectroscopy play a central role in the chemical analysis of
specimens in medical laboratories. The atomic and quantum models have been
combined to provide physical explanations for these and other atomic phenomena. In
this chapter you will be introduced to a quantum model of the atom and some of its
important applications.
28.2 Rutherford's Nuclear Atom Model
The early model of the atom proposed by J.J. Thomson in 1904 consisted of a positively
charged spherical volume with point particle electrons distributed throughout the
volume. (This was referred to as the "plum pudding" model of the atom, with the
electrons playing the role of raisins.) This model, with its various symmetrical
distributions of electrons attempted to explain the periodicity of chemical properties of
elements as well as the electromagnetic spectra of atoms. The empirical equations for
the various atomic spectral series could not be derived from the then existing models.
The model met with very little success, and in 1911 Lord Rutherford published a
paper which showed that the atom was much different than the Thomson model.
Rutherford's model was based on the scattering of alpha particles (discovered to be
helium nuclei of atomic mass 4) by gold foil(Figure 28.1).
Chapter 28 Atomic Physics
Physics Including Human Applications
616 Since the alpha particle is much more massive than the electron, (m /me ≈ 7360), you
should expect very little deflection of the alpha particle by the electrons in the gold.
Rutherford found, however, that some of the alpha particles were scattered at very
large angles, as though they had collided head on with a small, positively charged
particle. Rutherford concluded that the atom must be made up of a small, positively
charged nucleus surrounded by the electrons. Rutherford estimated the positive
nucleus radius to be about 1/3000 times the atomic radius. (He was about a factor of 3
too large in his estimate.) Rutherford imagined the electrons circling the nucleus, held
in circular orbits by the Coulomb force. The atom as a whole has a neutral charge; that
is, the positive charge on the nucleus is equal in magnitude to the sum of the electron
charges. In his early work Rutherford represented the nuclear charge as Ze, where Z is
the atomic number of the element and e is the electron charge. (Z is also equal to the
number of electrons in the atom.) Elements are arranged in the periodic table in
accordance with the individual atomic number.
α
28.3 Bohr's Model for the Hydrogen Atom
Niels Bohr, a young Danish theoretical physicist, had the insight and courage to
propose a model that applied quantum physics concepts to the hydrogen atom in 1913.
Planck's quantum hypothesis had accounted for blackbody radiation. Einstein had used
the quantum concept and the photon to explain the photoelectric effect. Bohr's atomic
model was even more remarkable. His model was designed to explain the stability of
the atom and to account for the characteristic line spectra of hydrogen. Bohr's model
incorporated Ernest Rutherford's nuclear model of the atom. Rutherford's alpha-particle
scattering experiments had shown that most of the mass was associated with the
positive charge of the atom in a small volume at the center of the atom. In Rutherford's
nuclear model, electrons revolved around the positive nucleus. However, there was a
serious flaw in this classical physics model. In classical electromagnetism, any
accelerating charge becomes a source of electromagnetic radiation. Rutherford's model
described the electron as being under constant centripetal acceleration, and classical
physics predicted the ultimate spiral of the electron into the nucleus. To solve this
dilemma, Bohr presented the following postulates:
Postulate 1: Classical dynamical equilibrium of the atom involves stationary states given by the
quantum condition on the angular momentum for stable orbits of the electron.
L = nh/2π
(28.1)
where n is the quantum number, n = 1,2,3,..., and L is angular momentum, mvr. These
stationary-state orbits are not required to satisfy the radiation laws of classical physics.
Postulate 2: Radiation is emitted or absorbed when the system makes transition from one
stationary state to another.
hƒ = Ef -Ei
(28.2)
where Ef is the final energy and Ei is the initial energy of the states involved when a
photon of energy hf is emitted or absorbed. Postulate 1 coupled with classical physics
Chapter 28 Atomic Physics
Physics Including Human Applications
617 can be used to derive equations for the energy value and the circular orbit radius
associated with Bohr's allowed stationary states for the hydrogen atom.
The energy of the nth state of the hydrogen atom is derived as follows(Figure 28.2).
From Newton's and Coulomb's laws
force of attraction = centripetal force
Fc = (1/4πεo)e2/r2 =mv2/r
From conservation of energy
kinetic energy + potential energy =Etotal
(1/2) mv2 - (1/4πεo)e2/r =Etotal
From the force equation we have
mv2 = (1/4πεo)e2/r
Thus
Etotal = (½) ((1/4πεo)e2/r) - ((1/4πεo)e2/r)
= -(1/2) ((1/4πεo)e2/r)
The negative sign for Etotal indicates that electron is bound to the nucleus. The quantum
condition of angular momentum is
L = mvr = nh/2π where n = 1,2,...
and from the force equation we have
mv = (1/4πεo) (me2/r)1/2 and (me2r/4πεo)1/2 = nh/2π
Thus, the radius of the nth electron orbit of hydrogen is given by
rn = n2 [(h/2π)2(4πεo/me2)] =n2ao
Chapter 28 Atomic Physics
(28.3)
Physics Including Human Applications
618 where ao = h2εo/πme2 = 0.529 x 10-10 m and is called the Bohr radius of the hydrogen atom.
The quantized energies for the hydrogen atom are given by
En = (1/n2)(me4/8εo 2h2) = -13.6 eV/n2
(28.4)
where n = 1,2, ... (Bohr's quantum number), m the electron mass, e, the electron charge,
h, Planck's constant, and εo, the permittivity of free space. (What is the significance of
the negative sign for this energy?)
EXAMPLE
The energy level of the ground state (n = 1) for the hydrogen atom is given by E1 = -13.6
eV/12. If the hydrogen atom receives this much energy from the external source, the
electron will be freed from the atom, and we say the atom is ionized. The ionization
energy of hydrogen in its lowest energy or ground state is 13.6 eV.
The second postulate of Bohr's model for the hydrogen atom states that if the orbit of
the electron remains the same, the atom will neither gain nor lose energy. In order for
the atom to either absorb or emit energy, the principle quantum number (n) must
change to some other positive integer, say m. The energy change is given by
Em - En = (- 13.6/m2) - (-13.6/n2) = -13.6(1/m2 - 1/n2) eV
(28.5)
On an energy level diagram an arrow pointing vertically upward indicates an increase
in the quantum number for the atom. This corresponds to the absorption of energy, say
a photon of energy hf. What energy must be absorbed to raise the hydrogen atom from
the n = 1 state to the n = 3 state?
On an energy level diagram an arrow pointing vertically downward indicates a
decrease in the quantum number of the atom. This corresponds to the emission of
energy, a photon of light.
How much energy is emitted when an n = 5 hydrogen atom goes into the n = 2 state?
Transitions are shown for the various energy levels from n = 1 n = 6 in Figure 28.3.
We calculate the absorption and emission energies given by the arrows 1 through 6.
Chapter 28 Atomic Physics
Physics Including Human Applications
619 Assume that the energy changes above correspond to the absorption or emission of
a photon. Remember that λ (nm) ≈ 1240(eV-nm)/(eV).
The measured wavelengths of the six photons emitted are:
λ1 = 93.80 nm
λ2 = 1097.35 nm
λ3 = 2638.30 nm
λ4 = 1278.35 nm
λ5 = 433.57 nm
λ6 = 97.25 nm
Which of these photons correspond to visible light?
Bohr's model was successful because it predicted experimentally verifiable atomic
spectra. Bohr's model could be used to derive the previously-known empirical
equations for spectral series. Unlike Planck's model, Bohr had no adjustable parameters
to use to fit the experimental data!
28.4 Atoms Other than Hydrogen
The Bohr model served well as a first approximation for a single electron atom. Bohr
himself supported his model as only the first step toward a comprehensive quantum
theory. In spite of refinements (such as elliptical orbits), the Bohr model could not
account for some of the fine structure, and line intensities of atomic spectra of systems
with more than one electron such as helium remained unexplained. On the
philosophical level, the Bohr model was a mixture of classical (orbits) and quantum
(photons) concepts. The quantum jumps as electrons go between orbits were very
mysterious indeed. In 1925 Schrodinger and Heisenberg developed their equivalent
versions of modern quantum theory of atoms. Schrodinger's formulation has become
known as wave mechanics, reflecting the mathematical form of its equations. In modern
quantum theory the electron in the atom is characterized by a probability amplitude
(known as a wave function). These wave functions, which are solutions to Schrodinger's
quantum wave equation, give stationary states that are characterized by four quantum
numbers (n, l, ml, and ms) instead of the single Bohr quantum number. Like the Bohr
model the stationary states have definite energies and are the only allowed states for the
electron in the atom. We can assign the following significance to the quantum numbers:
n, the principle quantum number, corresponds most closely to Bohr's quantum number. n
largely determines the energy of the stationary states; l, the orbital angular momentum
quantum number, determines the spatial symmetry of the electron probability
distribution in the atom; ml, the magnetic quantum number, determines the orientation of
the electron distribution in an external magnetic field; ms, the spin quantum number,
determines the intrinsic magnetic property of the electron. The word "spin" is derived
from a model of the electron as an electric charge spinning on its axis. These quantum
numbers are found to be governed by the following rules:
1. The principle quantum number n must be an integer, n = 1, 2, ...
2. The orbital angular momentum quantum number l must be an integer; it may take all
values up to n - 1, l = 0, 1, 2, ..., n - 1
3. The magnetic quantum number ml must be an integer, taking all values from - l to + l,
ml - l, - l + 1, ... 0, 1, ... + l
Chapter 28 Atomic Physics
Physics Including Human Applications
620 4. The spin quantum number ms is either +1/2 or -1/2, corresponding to spin up and
spin down.
Wolfgang Pauli formulated the exclusion principle in 1922. The exclusion principle
states that no two electrons in a given atom can be in the same quantum state. This
means that no two electrons can have identical sets of quantum numbers. We can think
of the set of four quantum numbers as a set of unique coordinates for each electron in
an atom (n, l, ml, ms).
Consider the following example. The lowest state in an atom corresponds to n = 1, l
= 0, ml = 0 with ms being either +1/2 or -1/2. We see that the n = 1 states are (1, 0, 0,
+1/2) and (1, 0, 0, -1/2). Thus two electrons exhaust the available n = 1 states consistent
with exclusion principle. The two electron atom is helium, an inert gas.
Pauli used the exclusion principle to explain the periodic table of elements. The
periodicity in the appearance of chemical properties of an atom is primarily associated
with electrons in the highest quantum states. These electrons are said to be in outer
orbits and more loosely bound than "inner" electrons. When the electrons of an atom fill
the allowed states for a given atom for a given n value, the atom is relatively inert (as is
helium). Atoms with one electron outside an inert core of electrons are said to be
hydrogenlike. Sodium is an example of such an atom. List three other hydrogenlike
atoms.
The order in which atomic systems fill the allowed quantum states is determined by
the rule that lowest energy states are filled first. The situation is complicated by the fact
that lower n values have lower energies, but smaller l values have lower energy values
than large l values. For example, the n = 4, l = 0 states have low energy than the n = 3, l
= 2 states.
EXAMPLE
Find the element in which atoms have all n = 2 states filled. From our previous example,
we know that there are two n = 1 states. Then = 2 states are: (2, 1, 0, +1/2), (2, 1, 0, -1/2),
(2, 1, 1, +1/2), (2, 1, 1, -1/2), (2, 1, -1, +1/2), (2, 1, -1, -1/2), (2, 0, 0, +1/2), (2, 0, 0, -1/2).
The total number of states for an n = 2 system is thus 10. The atom with 10 electrons is
neon, another inert gas.
The charge distribution for electron clouds associated with a given l value are called
orbitals. Each value of ml for a given l value provides a different orientation for the
orbital. The directional and shape properties of atomic orbitals provides a basis for a
chemical bonding theory as developed by Heitler, London, Slater, and Pauling. Figure
28.4 shows some examples of electron orbitals.
Chapter 28 Atomic Physics
Physics Including Human Applications
621 28.5 Spectra: The Fingerprints of Nature
The basic model is the same for all the elements. Every atom has a series of energy
levels that may be occupied by electrons. In its lowest energy state the atom will have
all of its electrons in the lowest possible electron states. If an atom is to emit or absorb
energy, the electrons must change from one electron state to another, and so the atom is
only capable of absorbing or emitting energy of certain values. It is not possible for
hydrogen to emit or absorb energy of 12 eV. The other elements can emit or absorb
energy only at certain discrete energies. The energy level diagram for sodium is shown
in Figure 28.5.
Chapter 28 Atomic Physics
Physics Including Human Applications
622 The energy level diagram for atoms of each element is absolutely unique. It is not
duplicated by the atoms of any other element. So the emission and absorption
characteristics of a material are indications of the kinds of atoms that are present in that
material. The chemical composition of a material can be deduced from an examination
of the wavelengths of optical photons (spectra), which are emitted or absorbed by a
material. In general, to reduce the interaction effects that can occur between
neighboring atoms, these spectral properties of materials are best examined with the
material in the gaseous state, so that the atoms of the material are as far apart as
feasible.
28.6 Emission Photometry
To observe the emission spectra of an atom, energy must be absorbed by the atom so
that at least one of its electrons is in an excited energy state. Then as these electrons
make a transition to a lower energy level, characteristic photons will be emitted, and
these photons can be observed (Figure 28.6).
Chapter 28 Atomic Physics
Physics Including Human Applications
623 In the usual laboratory setting the excitation energy is supplied by a heat source
such as a flame. A schematic diagram of a typical flame photometry system is shown in
Figure 28.7.
The physics of such a system is now within your grasp. The atomizer is used to
spray a fine mist of a solution containing the material to be observed into the flame. The
thermal energy of the flame excites some of the electrons in the atoms in the solution.
Some of the excited atoms will make transitions to lower energy states with the
emission of characteristic photons. The observation of the numbers and energies of the
emitted photons gives an indication of the kinds of atoms (elements) that are present in
the material as well as the concentration of each element.
In order to perform quantitative analyses using a flame photometer, an internal
standard element is introduced into the solution in known amount Cs. The unknown
concentration of the element Cx is calculated from the ratio of the intensity of the
emission of the unknown concentration to the intensity of the light emitted by the
standard:
Cx/Cs =Ix/Is
(28.6)
where Ix is the intensity of emission from the unknown concentration at wavelength λs
and Is is the intensity of emission from the known concentration of the internal standard
at wavelength λs.
Chapter 28 Atomic Physics
Physics Including Human Applications
624 28.7 Absorption Photometry
Most of the atoms in a flame remain in their lowest energy states; so the sample as a
whole is more likely to absorb radiation than to emit it. Therefore, the measurement of
the amount of radiation absorbed by a substance in the flame will be an order of
magnitude more sensitive than the measurement of flame emission. Once again on the
basis of your study of the hydrogen atom, the physics of absorption photometry your
study of the hydrogen atom, the physics of absorption photometry can be understood. Just
as all elements emit characteristics wavelengths of light when returning from an excited
state to a lower energy state, elements will only absorb characteristic wavelengths of
light as they are excited from their lowest energy states. The quantitative use of
absorption photometry relies upon the use of Beer's law:
The change in light intensity is proportional to the product of the concentration of the absorbing
atoms and the sample thickness.
Atomic absorption photometry may be used for the chemical analysis of blood and
urine samples. This technique not only makes use of the inherent sensitivity of the
absorption technique, but uses the spectral characteristics of a cold-cathode excitation
lamp to predetermine the element whose concentration will be measured. For example,
when it is desired to know the amount of sodium in a sample, a sodium lamp is used to
irradiate the vapor, and only the absorption of a characteristic sodium line is measured.
The presence of other substances in the host materials will not influence the
measurement of the sodium concentration. Atomic absorption measurements usually
proceed in three steps. First, the sample to be examined is vaporized, usually by a
flame, so that most of its constituents are in atomic form. Second, the atomic vapor is
irradiated by light characteristic of the element being sought. Finally, the absorption of
the characteristic light is related to the concentration of the element. An example of an
atomic absorption system used to measure sodium concentration is sketched in Figure
28.8
According to Beer's law the absorbance or optical density
is proportional to the concentration of material in the solution
(Figure 28.9). The absorbance is given by,
A = log10Io/I
(28.7)
where I0 is the intensity of the light with none of the unknown
element present, and I is the intensity with some of the
unknown material present. Hence, I is a smaller number than
I0, and the ratio of I0 to I is a positive number larger than one.
Chapter 28 Atomic Physics
Physics Including Human Applications
625 Therefore, A is always a number greater than zero. In practice, an atomic absorption
system is calibrated with a series of known solutions (Cs). The concentrations of
unknown (Cx) is determined directly from a graph or ratio equation:
Cx = AxCs /As
(28.8)
where As is the absorbance of the known solution and Ax is the absorbance of the
unknown.
28.8 The Colorimeter and the Spectrophotometer
As we have noted, the absorption of light takes place when the atoms in the sample
absorb photons. In this process the photon energy is used to excite atomic electrons to
higher energy levels. Each atomic system has its own unique absorption spectrum
which can be used in the chemical analysis of complex systems.
For a given wavelength of incident light, the intensity of light that is transmitted
through a sample of thickness x is given by Lambert's law which is written as follows:
I =Ioexp(-µx)
(28.9)
where I0 is the incident intensity, µ, the absorption coefficient, x, the thickness of
sample.
Beer's law states that for low concentrations the absorption coefficient is
proportional to the concentration of the absorbing atomic system in the sample. This
can be expressed in equation form as:
µ = αC
where a is a constant depending on the material and the wavelength of light used and C
is the concentration of the absorbing system. From Lambert's law we find the following
equation for x:
x = (1/αC) ln(Io/I)
or
αC = (1/0.434x) log10 (Io/I)
(28.10)
where log10(I0/I) is called the optical density (OD) of the sample, or the absorbance.
The colorimeter is an instrument designed to measure the absorption of light at
certain selected wavelengths. The wavelength used is determined by the particular
atomic system for which you are looking in the sample. In many colorimeters the
output is the optical density of the sample for the wavelength selected. In a simple
colorimeter the output may simply be a linear intensity scale. In this case the solvent is
first placed in the sample cell and the output reading is the value to be used as I0. The
sample is then placed in the instrument and the value of I is obtained. Consider the
following example.
Chapter 28 Atomic Physics
Physics Including Human Applications
626 EXAMPLE
The sample cell and pure solvent give a reading of 95 (linear scale), while the sample
cell with absorbing atomic system NaCl added gives a reading of 38.
a. Find the optical density of the absorbing sample.
OD = log10 (Io/I) = log10 (95/38) = .39
b. If solvent is added to the sample until the reading is 45, find the ratio of the original
concentration of NaCl to the final concentration.
αC1(0.434) = log10 (Io/I1) = log10 (95/38) = 0.39
αC2(0.434) = log10 (Io/I2) = log10 (95/45) = 0.32
C1 / C2 = .39/.32 = 1.23
The spectrophotometer is an instrument that is designed to measure and display the
absorption spectra, that is, the transmission intensity versus wavelength, for different
samples. Figure 28.10 shows both a single beam and a double beam spectrophotometer.
In the single beam instrument a blank sample cell is used to give an I0 reading for each
wavelength, and the sample gives a corresponding value for I (transmitted intensity).
The more sophisticated (and also much more expensive) double beam instrument uses a
chopped beam and electronics that determine the ratio of I/I0 for each wavelength as
the instrument scans through a wavelength range.
28.9 Quantum Efficiency in Photobiology
Interactions between light and matter seem to involve single photon absorption. This
model predicts a direct relationship between the number of incident photons and the
photoproducts resulting from the interaction. The concept ofquantum yield or quantum
efficiency is defined as
Chapter 28 Atomic Physics
Physics Including Human Applications
quantum efficiency ≡ photo-reaction rate / photon incidence rate
627 (28.11)
The photon incidence rate can be determined from the intensity of the incident
radiation as follows:
intensity = (energy / photon) x (photons / (unit area x sec))
Then we find
photons/sec = intensity/(energy/photon) x area of sample
or
ΔN / Δt (photons/sec) =I(watts/m2)A(m2)/hƒ(joules/photon)
(28.12)
A determination of quantum efficiency is very helpful in
studying the relative importance of light and living matter
reactions. An example is given in Figure 28.11 which shows the
relative effectiveness of different photons in producing
photosynthesis reactions in a sample of Euglena cells.
28.10 A Model for Laser Operation
As we pointed out in the chapter on the wave properties of light, the laser is a most
significant light source. We can now describe an energy level model that serves as the
basis for many laser systems.
Atoms of lasting material are excited by an external energy source (usually by a
flash tube or radio-frequency excitation). This input energy raises electrons to an excited
state. Some of these electrons end up in an excited state with an unusually long lifetime
(called a metastable state). This produces a population inversion with more electrons in
the metastable state than normal. When an electron does drop back to the ground state,
the emitted photon stimulates other excited atoms to decay. The ends of the laser cavity
have reflecting coatings so that these photons make several passes down the tube
producing an avalanche of photons. This represents a great amplification of the initial
photon by emitted radiation. The unique feature of this radiation is that the atoms emit
radiation in phase with each other. The high intensity is due to the collective behavior
of the atoms, and the monochromatic nature of the radiation is due to the sharpness of
the energy level transition involved in the emission process. An energy level diagram of
a typical laser system is shown in Figure 28.12
Chapter 28 Atomic Physics
Physics Including Human Applications
628 ENRICHMENT 28.11 Mathematical Derivation of Lambert's Law
We consider a beam of light passing through a light absorbing solution. Imagine a layer
of this solution dx thick (perpendicular to the incident beam). If the intensity of the
incident beam is I, and I' is the intensity of the transmitted beam, then the fraction of
light absorbed is given as
(I - I')/I
This should be proportional to the number of absorbing systems in the beam in the
solution. If C is the concentration, that is the amount of solute per unit volume, then the
quantity of absorbers per unit area of solution is
amount of solute (absorbers)/unit area = Cdx
Thus
(I - I')/I ∝ C dx
or, since I - I' = dI and I' is less than I, we have
- dl / I = αC dx
where α is a proportionality constant. If we let αC =µ, the absorption coefficient,
-dl/I = µdx or I = -µx + constant
If I0 is the intensity of incident beam (x = 0), then the constant equals ln I0, and we have
ln I = ln (Io –µx) or I =Ioexp(-µx)
If the solution contains several different absorbing solutes with different concentrations
we get the following equation:
-dl/I = (α1C1 +α2C2 + ... )dx
I = Io exp [-(α1C1 +α2C2 + ... )x]
The optical density OD or absorbance is defined as follows:
OD = log10 Io/Ix = (α/2.3) Cx
since ln y = 2.3 log10y. For multiple absorbing solutes this becomes
OD = ∑i=1N(x/2.3)α1C1
where x is the thickness of the absorbing solution.
Chapter 28 Atomic Physics
Physics Including Human Applications
629 SUMMARY
Use these questions to evaluate how well you have achieved the goals of this chapter.
The answers to these questions are given at the end of this summary with the number of
the section where you can find related content material.
Bohr Model
1. Bohr's quantum postulate stated that one of the following quantities came in discrete
units of h/2π:
a. linear momentum
b. energy
c. position
d. angular momentum
e. mass
2. Bohr's model for the hydrogen atom predicts that the absorption spectra involves
a. accelerating electrons
b. same wavelengths as emission spectra
c. electrons going to higher energy levels
d. electrons dropping to lower energy levels
e. deaccelerating electrons
3. Which of the following characteristics of hydrogen atoms was most difficult for
classical physics to interpret:
a. mass
b. charge
c. line spectra
d. ionization
e. size
Pauli Exclusion Principle
4. Give the possible sets of quantum numbers for n = 2.
Bohr Model Problems
5. The series of hydrogen spectral lines corresponding to energy transitions ending on
the n = 2 level is called the Balmer series. The shortest wavelength of the Balmer
series is
a. 122 nm
b. 91.2 nm
c. 656 nm
d. 3.40 nm
e. 365 nm
Chapter 28 Atomic Physics
Physics Including Human Applications
630 6. The longest wavelength of the Balmer series will be
a. 122 nm
b. 91.2 nm
c. 656 nm
d. 13.4 nm
e. 365 nm
7. Since room temperature thermal energy corresponds to about 0.03 eV, you would
expect hydrogen atoms at room temperature to be
a. ionized
b. in n = 2 state
c. in n = 20 state
d. in n = 1 state
e. in n = 10 state
Energy Level Diagrams
8. Which of the following transitions would give the shortest wavelength emission line:
a. n = 1 to n = 2
b. n = 1 to n = 3
c. n = 3 to n = 1
d. n = 4 to n = 2
e. n = 5 to n = 4
9. Which of the following transitions would give the longest wavelength absorption
line:
a. n = 1 ton = 2
b. n = 2 to n = 3
c. n = 1 to n = 3
d. n = 4 to n = 2
e. n = 5 to n = 4
Spectrometry and Lambert's Law
10. For low concentrations Beer's law predicts that the absorption coefficient of a
solution will be linear with
a. wavelength
b. frequency
c. concentration
d. energy
e. intensity
Chapter 28 Atomic Physics
Physics Including Human Applications
631 11. If the optical density of a sample is 1.0, then the ratio of transmitted to incident
radiation is
a. 0.1
b. 10
c. 100
d. 0.01
e. 100
12. The spectrophotometer is an instrument designed to measure and display
a. frequency vs. wavelength
b. transmission intensity vs. wavelength
c. intensity vs. thickness
d. intensity vs. concentration
e. absorption coefficient vs. concentration
Answers
1. d (Section 28.3)
2. b, c (Section 28.3)
3. c (Section 28.3)
4. l = 1, ml = -1, 0, + 1, mS = ±1/2; l = 0, ml = 0,
mS = ±1/2; total of 8 (Section 28.4)
5. e (Section 28.3)
6. c (Section 28.3)
7. d (Section 28.3)
8. c (Section 28.3)
9. e (Section 28.3)
10. c (Section 28.8)
11. a (Section 28.8)
12. b (Section 28.8)
ALGORITHMIC PROBLEMS
Listed below are the important equations from this chapter. The problems following the
equations will help you learn to translate words into equations and to solve singleconcept problems.
Equations
L = nh/2π
(28.1)
hƒ = Ef -Ei
(28.2)
rn = n2 [(h/2π)2(4πεo/me2)] =n2ao
where ao = 0.529 x 10-10 m
(28.3)
En = (1/n2)(me4/8εo 2h2) = -13.6 eV/n2
(28.4)
Em - En = (- 13.6/m2) - (-13.6/n2) = -13.6(1/m2 - 1/n2) eV
(28.5)
A = OD = log10Io/I
(28.7)
Cx = AxCs /As
(28.8)
I =Ioexp(-µx)
(28.9)
Chapter 28 Atomic Physics
Physics Including Human Applications
632 Problems
1. What would be the velocity of an electron in an orbit for n = 1 according to the Bohr
model? The radius r1 = .053 nm. How does it compare with the velocity in an orbit
for n = 2?
2. If radiation is emitted by an atom in going from an energy level of - 1.90 eV (3.04 x 1019
J) to an energy level of -3.40 eV (5.44 x 10-19 J), what is
a. the frequency of the radiation
b. its wavelength
3. What is the absorbance or optical density of a sample in which the ratio of I0/I is 3?
4. What is the ratio of I/I0 for a specimen 0.5 cm thick of a material which has an
absorption coefficient of 2 per centimeter?
Answers
1. 2.19 x 106 m/sec; v1/v2 = 2
2. a. 3.62 x 1014 Hz; b. 829 nm
3. 0.477
4. 0.37
EXERCISES
These exercises are designed to help you apply the ideas of a section to physical
situations. When appropriate the numerical answer is given in brackets at the end of
each exercise.
Section 28.3
1. Suppose you could observe the hydrogen atom through a magnifying system such
that the hydrogen nucleus appeared to be the size of a baseball (r ≈ 4 cm). What
would the apparent radius of the n = 1 electron orbit? Assume the radius of a proton
= 1.5 x 10-15 m. [1.42 x 104 cm]
2. Assume you want to find an electron orbit that is as far away from the n = 1 electron
orbit as is the nucleus. What is the n value for such a fictitious orbit? [SQR RT(2)]
3. It has been suggested that the volume of an atom is mostly empty space. What
percentage of the volume of an n = 1 hydrogen atom is occupied by matter? Radius
of proton = 1.5 x 10-15 m [Volume of atom about 4 x 1013 times as large as volume of
proton]
4. What is the velocity of the electron in a hydrogen atom for n = 2? The average time in
the excited state is usually about 10-8 sec. How many revolutions does this electron
make during this time? [v = 0.6 x 106 m/sec; 4.73 x 106 revolutions]
Chapter 28 Atomic Physics
Physics Including Human Applications
633 5. Compute the energy associated with a hydrogen atom for n = 1, 2, 3, and 4. Draw an
energy level diagram. [n = 1, -13.6 eV; n = 2, -3.4 eV; n = 3, -1.51 eV; n = 4, -0.86 eV]
6. A large number of hydrogen atoms are excited to state n = 4. What transitions and
spectra lines are then possible? [4 → 3, 4 → 2, 4 → 1, 3 → 2, 3 → 1, 2 → 1]
Section 28.4
7. Find the n and l numbers for the final electron ground state of krypton (Z = 36). [n =
4, l = 1.]
Section 28.7
8. An atomic absorption instrument is being used to measure the concentration of
sodium in an aqueous solution. The following calibration measurements are taken:
Solution
Readout Current (mA)
Distilled water
272.
10 ppm Na in H2 O
100.
20 ppm Na in water
36.8
40 ppm Na in water
4.97
60 ppm Na in water
0.67
80 ppm Na in water
0.09
What are the concentrations of sodium in unknown solutions whose readout
currents are given by: unknown solution 1, 200 mA; unknown solution 2, 40.0 mA;
unknown solution 3, 8.00 mA.
Na in 1 = __________ppm
Na in 2 = __________ppm
Na in 3 = __________ppm
[4, 19, 35]
9. The absorbance of a 10-ml sample is found to be 3. The sample is accidently diluted.
The lab technician (who had completed this physics course) ran another absorbance
spectrum and found the new absorbance to be 2.7.
a. Find the ratio of sample concentrations and volume of the diluted sample.
b. Find the ratio of transmitted intensities for these two measurements. [a. C2/C1 =
0.9,V2 = 11.1 ml; b. I2/I1 = 2]
10. A blood sugar study shows the absorbance of a morning sample to be 2 and that of
an evening sample to be 1.8. Find the ratio of sugar concentration in the two
samples. [CE = 0.9CM]
Chapter 28 Atomic Physics
Physics Including Human Applications
634 PROBLEMS
Each of the following problems may involve more than one physical concept. The
answer is given in parentheses at the end of the problem.
11. Find the atom that completes the third period in the periodic table (all states up to n
= 3, l = 1 states filled). [Argon]
12. A helium He+ ion is similar to a hydrogen atom. Compare the radii of orbits for n =
1 for He+ ion and the hydrogen atom. (Use reduced the mass correction from
problem 14.) [rHe = 1/2 rH ]
13. Assume the sodium atom is hydrogen type (a nucleus with +11e charge surrounded
by the 10 inner electrons) and that it has an ionization potential of 5.12 eV. What is
the radius of the eleventh electron orbit? Compare this with the size of the Na atom.
[r11 = 14 nm]
14. One of the first corrections that was made in the simple Bohr model was the
correction for the relative motion of the proton and the electron. This correction
involves the relative motion of the proton and electron about the center of mass of
the system. The results show that this correction calls for the use of the reduced
mass in place of the electron mass. The reduced mass m is given by µ = meMp/me + Mp
where me = electron mass, Mp = proton mass = 1840 me. Use this correction to
compare the Balmer series limit wavelength (n = ∞ → n = 2) of deuterium with the
corresponding hydrogen line. [Differs by about 30 parts in 109,707]
15. Imagine a new kind of atom in which the electron is attached to the nucleus of the
hydrogen atom by a quantum mechanical spring. Hence the force of attraction is
given by F = -kr where k has a value of 32.4π2 N/m. (This atom is not the usual Bohr
atom where the force of attraction is given by Coulomb's law!)
a. Compute the value of r in terms of m, k, and v for an electron traveling in a
circular orbit of radius r with a velocity v in this new kind of atom.
b. Use Bohr's postulate to quantize the angular momentum of the electron, and
express the radius of the electron orbit in this new atom as a function of n, the
principle quantum number where n = 1,2,3, ....
c. Since you know that for a force of the form kr, the potential energy is given by
(½)kr2, calculate the total energy of the electron in terms of n.
d. Sketch the energy level diagram for the new atom. How does it differ from the
energy level diagram for the Bohr model of the hydrogen atom?
[a. r = (SQR RT(m/k))v; b. r = (n(h/2π)SQR RT(1/mk))1/2 = [n(h/2π)√1/mk]1/2; c. E = Eo +n(h/2π) SQR RT(k/m) = -Eo + (n(h/2π)/2π)SQR RT(k/m)]
Chapter 28 Atomic Physics
Physics Including Human Applications
635 16. Light from an ultraviolet light source consists of all wavelengths between 100 nm
and 200 nm. If this light is incident upon a cell containing hydrogen gas, discuss
quantitatively the physical phenomena you expect to result from this interaction.
17. Beer's law may be stated in the following form:
Itransmitted = Iincidentexp(-kx)
Where x is length of path in absorbing medium and k is the absorption coefficient.
What are the units of k? If Itransmitted is 75 percent of Iincident for a given thickness x, what
thickness is required to reduce Itransmitted to 50 percent? [x1 = x + (1/k) ln(3/2)]
Chapter 28 Atomic Physics