Grade 12 Unit 9
Grade 12 Unit 9
SCIENCE 1209 ATOMIC AND NUCLEAR PHYSICS
CONTENTS I. QUANTUM THEORY . . . . . . . . . . . . . . . . . . . .
2
ELECTROMAGNETIC RADIATION . . . . . . . . . . . . . . . . . . . . MATTER WAVES . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . ATOMIC MODELS . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . ATOMIC SPECTRA . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . BOHR MODEL . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
2 8 11 12 14
II. NUCLEAR THEORY . . . . . . . . . . . . . . . . . . . .
22
BUILDING BLOCKS OF THE NUCLEUS . . . . . . . . . . . . . . . PROPERTIES OF THE NUCLEUS . . . . . . . . . . . . . . . . . . . .
22 23
III. NUCLEAR REACTIONS . . . . . . . . . . . . . . . . .
30
NUCLEAR FISSION . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . FUSION . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . NUCLEAR REACTION APPLICATIONS . . . . . . . . . . . . . . . .
30 32 33
GLOSSARY . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . FORMULAS . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
39 41
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ATOMIC AND NUCLEAR PHYSICS You are about to begin the study of modern physics, the physics of very small objects. You will learn more about how atoms and their subparts interact. The laws of physics that you have already learned will be used, but you will notice that at times some new concepts will be considered. OBJECTIVES Read these objectives. The objectives tell you what you will be able to do when you have successfully completed this LIFEPAC®. When 1. 2. 3. 4. 5. 6. 7. 8. 9. 10. 11. 12. 13. 14.
you have finished this LIFEPAC, you should be able to: Explain when wave theory is applicable and when quantum theory is applicable. Calculate the energy of photoelectrons and X-rays. Explain de Broglie waves. State and apply the uncertainty principle. Define emission spectra and absorption spectra. Define line spectra and continuous spectra. Describe the Bohr model of the hydrogen atom. Define nucleon and isotopes. Explain mass defect in terms of mass-energy equivalent. Explain how the binding energy curve shows that fission and fusion can release energy. Explain the concept of half-life. Explain how alpha and beta decay can stabilize a nucleus. Describe fission reaction and fusion reaction. Explain a chain reaction and nuclear energy.
Survey the LIFEPAC. Ask yourself some questions about this study. Write your questions here.
The last page of this LIFEPAC contains formulas and constants that may be used for problem-solving, both on activities and on tests.
1
I. QUANTUM THEORY For many centuries, the primary concern of physics was to describe the behavior of large objects. Newtonian mechanics describes the behavior of matter influenced by gravitational fields or by electromagnetic interactions. Newtonian physics had been successful; so successful, in fact, that most physicists believed they were approaching the ultimate description of nature. They thought that in just a matter of time they would be able to work out the details of a perfect, ordered world. Many believed that the future work of physics would be essentially just measuring the last decimal place. Problems with this neat and supposedly complete theoretical description began to arise about the end of the nineteenth century. More precise experiments opened a whole new world to
the physicist who was beginning to investigate the properties of very small particles. The behavior of these microscopic systems could not be explained in terms of the classical physics and Newtonian mechanics, which had been the physicists’ stock and trade up to this point. In classical physics the physicist had explained the behavior of objects in terms of direct and compelling deductions from a well defined set of experiments. Attempts to do the same from new observations were not successful. The new physics, or quantum physics, was too abstract to make deductive explanations possible. The main reasons were twofold: The basic constructs of this quantum theory were removed from everyday experience, and the microscopic particles could not be observed directly.
SECTION OBJECTIVES Review these objectives. When you have completed this section, you should be able to: 1. Explain when wave theory is applicable and when quantum theory is applicable. 2. Calculate the energy of photoelectrons and X-rays. 3. Explain de Broglie waves. 4. State and apply the uncertainty principle. 5. Define emission spectra and absorption spectra. 6. Define line spectra and continuous spectra. 7. Describe the Bohr model of the hydrogen atom. VOCABULARY Study these words to enhance your learning success in this section. absorption spectrum
energy levels
proton
alpha particle
excited atom
quanta
continuous spectrum
ground state
quantum number
de Broglie wave
line spectrum
radiation
electron volt
photoelectric effect
uncertainty principle
emission spectrum
photon
X-ray
Note: All vocabulary words in this LIFEPAC appear in boldface print the first time they are used. If you are unsure of the meaning when you are reading study the definitions given.
ELECTROMAGNETIC RADIATION The wave theory of light was well recognized by the end of the nineteenth century. Experimental work was being conducted in which both light rays and high speed electrons were allowed to strike a metal surface. The results of these experiments
could not be explained by existing theory. As a result a new theory was developed, called the quantum theory. At the time, physicists did not fully realize what was happening; but modern physics had been born.
2
Photoelectric effect. During the late 1800s several experiments disclosed that electrons were emitted when light fell on a metal plate. This phenomenon is the photoelectric effect. At first no one was surprised: Light waves were known to carry energy; some of this energy, therefore, could somehow be absorbed by electrons to give them enough kinetic energy to escape from the metal surface.
The symbol v (the Greek nu) represents the frequency of incident light. The value of h was always the same value, but v0 changed for each metal. An explanation of this phenomenon was given in 1905 by Albert Einstein. He used a result of earlier work by the German physicist Max Planck, who described the radiation emitted by a glowing object. Planck could get his theory to work if he assumed that the light was emitted in little “lumps” of energy. He called these little bursts of energy quanta and showed that they had energy of
LIGHT
E = hv, where v is the frequency and h is a constant known as Planck’s constant. The value of Planck’s constant h is given as h = 6.63
•
10-34 joule•sec.
AMMETER
Few physicists really believed in Planck’s quanta, but Einstein pointed out that the photoelectric effect was experimental verification of Planck’s equation. He suggested that the photoelectric equation be rewritten:
Soon some problems developed: First, the energy of the photoelectrons was not related to the intensity of the light. When the light was made brighter, more electrons were emitted; but they all had the same average energy as those emitted by dim light. Second, the energy of the photoelectrons was related to the frequency of the light. If the frequency was below a certain value, no electrons were emitted even if the light was intense. Conversely, if the frequency was above the value, electrons were emitted even in very dim light. Photoelectrons emitted by any given frequency were observed to have energy that ranged from zero to some maximum value. If the frequency of the light was increased, the maximum electron energy increased. In other words, a faint blue light produced electrons with higher energy than did a bright red light even though the bright red light produced more electrons. The relationship between KEmax, the maximum photoelectron energy, and the frequency of the incident light can be stated:
hv = KEmax + hv0, and be stated: quantum (incident photon) energy = maximum electron energy + energy required to eject an electron from the metal surface. A large portion of the photoelectron population will have less than maximum energy for the following reasons: 1. The photon (the quantum of incident light) may have shared its energy with more than one electron. 2. The electron may have lost part of its energy before it got out of the metal. The energy of photons is so small that the joule is cumbersome. A unit of energy suited to the atomic scale is the electron volt. One electron volt is the energy change realized by an electron moving through a potential change of one volt. Numerically,
maximum photoelectron energy= h (frequency - frequency0) KEmax = h (v - v0).
1 ev = 1.6
3
•
10-19 joules
LlFEPACs 1204 and 1205 dealt with diffraction and interference. The observed phenomena could be explained in terms of the wave theory of light. Now you have seen an argument that light behaves like a series of packets of energy, called photons, or quanta, each photon small enough to interact with a single electron. The wave theory cannot explain the photoelectric effect and the quantum theory cannot explain interference and diffraction. Which is the correct theory? Many times before in physics, one theory has been replaced by a new theory; but this apparent conflict was the first time that two
✍
different theories were needed to describe a phenomenon. Developing a model for light is a situation in which science must acknowledge that nature is not an “open book” to man. These two theories are complementary: One is correct in some experiments, but the other theory must be used in others. Interestingly, light can exhibit either a wave or a particle nature, but never exhibits both in any single situation. Since we cannot reconcile the two descriptive models, we have no choice but to accept them both.
Answer these questions.
1.1
What is the photoelectric effect?
1.2
Why do photoelectrons have a maximum energy?
1.3
Why does faint light not appear as a series of flashes?
1.4
Why can the photoelectric effect not be explained without quantum theory?
✍
Complete these sentences.
1.5
The phenomenon that describes the emission of electrons from a metal surface when light shines on it is the .
1.6
The maximum energy of photoelectrons a. with frequency and b. with increasing intensity of the light. (increases, decreases, remains constant)
1.7
Photon energy is proportional to
1.8
Photons are also known as
✍ 1.9
. .
Solve these problems. A yellow lamp emits light with a wavelength of 6
•
10-7 m.
a. Write the common value for the speed of light, in meters per second.
4
b. Calculate the frequency of the yellow light from the equation given in Science LIFEPAC 1204.
c. Use Planck’s equation to calculate the energy of a single photon with the given wavelength.
d. How many such photons are required to produce 10 joules?
1.10
A 10,000-watt radio station transmits at 880 kHz. a. Determine the number of joules transmitted per second.
b. Calculate the energy of a single photon at the transmitted frequency.
c. Calculate the number of photons that are emitted per second.
1.11
Light with a wavelength of 5 • 10-7 m strikes a surface that requires 2 ev to eject an electron. a. Calculate the frequency of the given wave.
b. Calculate the energy, in joules, of one incident photon at this frequency.
c. Calculate the energy, in electron volts, of one incident photon.
5
d. Calculate the maximum KE, in electron volts, of the emitted photoelectron.
1.12
Photoelectrons with a maximum speed of 7 • 105 m/sec are ejected from a surface in the presence of light with a frequency of 8 • 1014 Hz.
a. If the mass of an electron is 9.1 • 10-31 kg, calculate the maximum kinetic energy of a single electron, in joules.
b. Calculate, in joules, the energy of incident photons.
c. Calculate the energy required to eject photoelectrons from the surface.
X-rays. When photons of light were found to give up their energy to electrons, the question was asked: Can the energy of moving electrons be converted into photons? Further investigation disclosed that not only could electrons produce photons, but that several years previously the German physicist Roentgen had in fact produced the effect. Roentgen had conducted experiments in which he allowed high speed electrons to strike the surface of metal. He had observed that the metal target emitted some very penetrating radiation. These Roentgen rays, or X-rays, as they are now called, exposed photographic plates. Roentgen was certain that they were not particles; they were affected by neither electric nor magnetic fields. He also found that as the energy of the electrons increased, the X-rays became more penetrating. The density of emitted X-rays increased when the number of electrons was greater. In addition Roentgen established that X-rays could exhibit both interference and polarization effects. He believed these phenomena were evidence that the emissions were electromagnetic radiations.
TARGET ANODE
FILAMENT
ee-
V
X-RAYS
When a charged particle accelerates, it emits electromagnetic radiation. X-rays are identified as electromagnetic radiation emitted when a very fast electron is slowed suddenly as it strikes the metal. Most of the incident electrons probably lose their energy too slowly for X-rays to be emitted. Some of the electrons, however, lose much or all of their energy in a single collision with an atom in the target metal. This radiation is the inverse of the photoelectric effect. In fact, the highest frequency found in X-rays corresponds to the photon energy,
6
and all frequencies below the maximum X-ray frequency also exist. When the electron energy is above a certain value, the emission curves possess strange, sharp spikes. The spikes indicate that some frequencies are emitted preferentially. These emissions originate from rearrangement of the electron structure of the target atom.
30,000 v
X-RAY INTENSITY
15,000 v
5,000 v 1
2 3 4 5 FREQUENCY x 1018 cycles/sec
6
AN X-RAY SPECTRUM
✍
Complete these sentences.
1.13
When the number of electrons striking the anode of an X-ray tube increases, the of the emitted X-rays increases.
1.14
When the speed of electrons striking the anode of an X-ray tube increases, the of the emitted X-rays increases.
✍ 1.15
Complete these activities. An X-ray tube emits X-rays with a wavelength of 10-11 m. a. Calculate the photon energy, in joules, that the emitted X-rays possess.
b. Calculate the energy, in electron volts, that the X-rays possess.
c. Determine the energy, in electron volts, possessed by the incident electrons.
d. Calculate the potential that must be applied across the X-ray tube to give each incident electron its energy.
7
1.16
Calculate the highest frequency X-rays produced by 8
1.17
A television tube can accelerate electrons to 2 X-rays with the highest energy.
1.18
Calculate the energy, in electron volts, of X-rays that have a frequency of 1019 Hz.
•
•
104 ev electrons.
104 ev. Calculate the wavelength of emitted
MATTER WAVES and the relationship wavelength, and velocity,
X-rays are used to study the configuration of atoms in crystals. Atoms in a crystal interact with X-rays, causing the scattered waves to interfere constructively in some regions and destructively in others. The interference pattern thus produced is characteristic of the spatial arrangement of the atoms in the crystal. The exact pattern is determined by the wavelength of the X-rays and the distance between the atoms. If, instead of X-rays, electrons are used, a pattern is produced almost exactly the same as the X-ray pattern. This experiment is known as the Davisson-Germer experiment. The implication of this experiment is that electrons behave as waves. X-rays and their diffraction pattern are used to measure the distance between atoms in crystals.
frequency,
λv = c Substituting the second equation into the first gives momentum =
/c = h/λ
hv
Solving this relationship for the wavelength gives λ=
/momentum
h
for a photon. The new idea that de Broglie suggested was that this formula is much more general and that it applies to particles as well as to waves. The expression for the momentum of a particle is
De Broglie waves. By using crystals with known atomic spacing, the experiment can be used to determine the wavelength of the electron. These wavelengths are called de Broglie wavelengths; and the waves themselves are de Broglie waves, or matter waves. The wavelength of the de Broglie wave is inversely proportional to the momentum of the particle (generally an electron) that is being observed. De Broglie took the formula for the momentum of a photon, momentum =
between
momentum = mv. Substituting this equation into the former relationship gives the expression for the wavelength of a particle: λ=
/mv
h
Again the relationship between the wavelength and the momentum involves the quantity h, Planck’s constant.
/c
hv
8
The value of h is extremely small, which explains why waves are not observed for large objects. The momentum must be small compared to h for the wavelength to be appreciable. Particles at times have the properties of a wave; at other times they exhibit the properties of a particle. As in the case of electromagnetic radiation, the particle is never observed to exhibit both particle characteristics and wave characteristics in the same interaction; and we cannot determine which description is the correct one. This phenomenon is sometimes called the wave particle duality, or just duality. The reason we cannot observe this property of duality is because the wavelengths are so small. For instance, the wavelength of an automobile traveling 60 miles an hour is on the order of 10-38 feet. This wavelength is much much smaller than the automobile. On the other hand, the wavelength of an electron traveling about 10 million meters per second is about 10-8 meters. This very small wavelength is still the magnitude of atomic dimensions. Since the wave length is comparable to the size of the electron, the wavelength is observable.
ELECTRON e-
N IO CT E R DI VE A W
PHOTON INITIAL MOMENTUM
ELECTRON OBSERVER
e-
PHOTON
FINAL MOMENTUM
OBSERVER
Even the position is not known with absolute accuracy since, even with the best of instruments, the position will be off by at least part of a wavelength. Thus,
Uncertainty principle. One of the tenets of quantum physics is an inherent limitation on our capability for making observations. To measure something, some way is needed to carry information from the object to the observer. You must be able to touch it, shine a light on it, or perform some other manipulation for information about the object’s size or its position to be carried back to the observer. Observing a large object is fairly easy. For instance, you can measure a table with a ruler. When the object being observed is an electron, the problem is different. About the only way that information about the electron’s size or its position can be gathered is to shine light on the electron and to reflect it to the observer. The difficulty is that the light, to reflect from the electron, must strike the electron. In striking the electron, some of the light energy is transferred to the electron. The reflected light enables the position of the electron to be determined precisely: Project back along the path of light and determine where the electron was when struck by the light. However, when the electron was struck by the photon, the energy given up by the light changed the direction of the electron. The very act of observing the electron changes its momentum. Although you now know where the electron was at that instant, you do not know in exactly what direction it is now going.
∆x ⬇ λ is a reasonable estimate of the uncertainty about the position of the particle. The photon had a momentum: momentum = h/λ . Although you cannot know exactly how much the momentum of the electron will change, it is reasonably the same magnitude as the photon momentum; thus, ∆mv ⬇ h/λ . Combining the two equations of uncertainty gives the result: ∆x
•
∆mv ⬇ h.
The uncertainty in position times the uncertainty in momentum is approximately equal to Planck’s constant, which means that we can never hope to measure both position and momentum 9
simultaneously with perfect accuracy. In 1927 the German physicist Heisenberg first proposed this result, which is usually referred to as the Heisenberg uncertainty principle. Uncertainty does not result from deficiencies in measuring instruments; it is a constraint of nature. Remember again that Planck’s constant is very small; matter waves have significance for atomic particles, but not for large, “Newtonian” objects.
the relationships can be verified with absolute accuracy. Are these relationships, then, not true? Unfortunately, we cannot make as positive a statement as we might wish. Individual particles may not behave according to the Newtonian laws of classical physics; but for statistically large groups of particles, these expressions become statements of probability. Many many small particles obey classical laws of physics on the average, but individual particles may deviate from this average behavior. Fortunately, everything that we physically encounter is made up of such large numbers of particles that their average behavior is all that is observed. Causality is reasonably reliable for macroscopic events, but we cannot expect it to hold true in situations that we cannot experience directly.
Causality. Previously we assumed that the laws of physics were cause-and-effect relationships: that a given cause always produces a specific effect. The equations studied in mechanics give relationships; generally, that force is proportional to something else. The uncertainty principle, however, shows that all the factors that go into an equation cannot be measured with total precision. None of
✍
Complete these sentences. divided
1.19
The wavelength of a moving particle is equal to a. . by b.
1.20
The uncertainty principle states that the product of uncertainty in the momentum and in the . position of a particle is approximately equal to
✍
Complete these activities
1.21
Calculate the de Broglie wavelength of a 5,000 kg truck traveling at 80 kph. (Hint: convert everything to fundamental units.)
1.22
Calculate the de Broglie wavelength of an electron traveling at 107 m/sec (me = 9.1
1.23
Calculate the approximate momentum change in a particle of mass 1.7 initially at rest, whose position is located to within 10-4 m.
1.24
Calculate the uncertainty of the velocity of a particle confined to a space of 10-9 m if the particle is: a. an electron
b. a proton
10
•
•
10-31 kg).
10-27 kg (a proton),
ATOMIC MODELS By the year 1900 the field of chemistry had accumulated a large amount of evidence that material consisted of atoms. Very little evidence existed, however, about the nature of the atoms themselves. J.J. Thomson had recently discovered the electron; this discovery suggested that electrical forces were of significant strength on the atomic scale.
of Thomson’s model. This experiment was conducted by two men named Geiger and Marsden. They placed in a lead box material that emits alpha particles. (Alpha particles will be discussed in detail later. All we have to know about them now is that they are small, positively charged radiations emitted by some natural substances such as radium.) A small hole drilled in the box allowed a stream of alpha particles to escape. This beam of alpha particles was directed at a very thin metal foil. Geiger and Marsden found that, just as the Thomson model predicted, most of the alpha particles passed through the foil or were deflected through relatively small angles. The particles were observed as they struck a zinc sulphide screen surrounding the foil. When alpha particles struck the screen a small amount of light was emitted. Geiger and Marsden made one interesting observation: Most of the alpha particles passed through the screen, but a few were scattered through rather large angles. In fact, some were scattered back nearly 180°. Alpha particles are much larger than electrons, nearly 8,000 times greater in mass; only very strong forces could change the direction of the alpha particles so drastically.
Thomson model. Thomson proposed what is referred to as the Thomson model, sometimes called the plum pudding model, of the atom. He suggested that atoms were spheres of positively charged matter and the electrons were embedded in this matter much as plums are embedded in plum pudding. This model was quite reasonable using the evidence available at the beginning of this century. It provided an explanation for most of the atomic behavior that had been observed. If this model were the nature of atoms, then one would expect that charged particles as they moved through an atom would not be appreciably affected. Electric fields in atoms would be quite weak, because the positive charge in an atom would be diffused by negative “plums,” and vice versa. Shortly afterwards, however, another experiment was done that caused some concern as to the validity
ZINC-SULFIDE SCREEN
ALPHA EMITTER
LIGHT FLASHES
LEAD BOX THIN METAL FOIL
THE GEIGER-MARSDEN EXPERIMENTS
Rutherford model. To explain this large angle scattering, the British physicist Ernest Rutherford proposed a new model of the atom. His theory was that the positive charge was not distributed throughout the entire atom, but was concentrated in a very very small portion near the center. The atom was mostly empty space. The negative electrons moved around in this empty
space; but they were very small, very light, and could not have an appreciable effect upon the heavy alpha particle. The center of the atom, the nucleus, was very small and would effect only those alpha particles that came very near; most of the alpha particles traveled through the empty space. Alpha particles that approached a nucleus, however, would experience strong electric forces 11
and would be scattered through a large angle. If an alpha particle were to strike an electron, it would knock the electron out of the way; the nucleus, however, being much heavier than the alpha particle, caused the alpha particle to change its
✍ 1.25
direction. Using this idea and the rules of electrostatics, Rutherford was able to derive a formula that explained the backward scattering of alpha particles. As a result, Rutherford is credited with the discovery of the dense atomic nucleus.
Complete this activity. Using the description of the Thomson and Rutherford models of the atom, draw a labeled diagram of each model that illustrates the differences between the two.
THOMSON MODEL
✍ 1.26
RUTHERFORD MODEL
Prepare a report. Some of the men involved in the development of early atomic physics worked in the same laboratory. Look up the following men: Ernest Rutherford, J. J. Thomson, and Neils Bohr. Write a five-page, double-spaced, typewritten report on the men and their contributions to atomic theory. Submit your report for evaluation. Score Adult check
______________________ Initial
Date
ATOMIC SPECTRA A spectrum is an array of the different wavelengths of radiant energy emitted by a source. Different sources produce different wavelengths and exhibit one of the four different kinds of spectra: 1. 2. 3. 4.
Emission spectra. Heating a block of metal produces no noticeable change until its temperature reaches about 1000°K, at which point the block begins to glow a dull red. As heating continues, the glow becomes bright red and then orange. When the temperature of the block is about 1500°K, the spectrum contains red, orange, and yellow. At 2,000°K the block appears yellow, and the spectrum now includes green. When the temperature reaches 3,000°K, the block glows white hot; and all the colors from red through violet are present in the spectrum along with invisible infrared and ultraviolet radiations.
Continuous emission spectra, Line emission spectra, Continuous absorption spectra, or Line absorption spectra.
12
INTENSITY
VISIBLE
ULTRAVIOLET INFRARED
WAVELENGTH
RADIATION EMITTED
BY A
HEATED SOLID
When hot gas is the source of the radiation, quite a different spectrum results. Instead of a continuous spectrum, a number of bright lines constitute the spectrum. SLIT HOT GAS
PRISM
LENS
ARRANGEMENT
LENS
TO
DEMONSTRATE
The most intense line spectra are obtained from metallic arc lamps, such as a carbon arc, when one of the carbon rods has been impregnated with a metal like mercury or sodium. Continuous spectra come from high temperature solids but line spectra always come from an incandescent gas.
A
LINE EMISSION SPECTRUM
SCREEN
wavelengths will be missing. When light from an incandescent bulb (hot solid filament) passes through colored glass, the glass absorbs all wavelengths except the color of the glass. When a continuous-emission spectrum is passed through gas below its temperature of incandescence, certain discrete wavelengths are removed. The resulting absorption spectrum has dark lines in the positions of the missing wavelengths.
Absorption spectra. When a continuous emission spectrum passes through a solid and the radiation is then analyzed, a wide band of SLIT
SODIUM GAS HOT GAS LENS
LENS
LENS
PRISM
SCREEN
SODIUM
GAS BURNER
ARRANGEMENT
TO
DEMONSTRATE LINE ABSORPTION SPECTRUM
13
When the sodium metal is heated, the chamber is filled with relatively cool sodium vapor; the spectrum on the screen displays two dark lines in the yellow region. Alkali metals are about the only substances that produce lines in the visible region of the spectrum, but practically all substances produce lines in the ultraviolet.
✍
1.27
Define these terms. spectrum
1.28
emission spectrum
1.29
absorption spectrum
1.30
continuous spectrum
1.31
line spectrum
Again we see that solids produce a continuous spectrum and that gases produce a line spectrum. The reason for the linear display is the slit through which the spectrum is focused. If the light is focused instead through a small hole, small disks would replace the line.
BOHR MODEL A new era in physics began in 1913 when Niels Bohr proposed a new structure of the hydrogen atom. This theory gave a satisfactory explanation for the spectrum emitted by hydrogen and was later successfully extended to other elements.
v
e+ e-
Hydrogen model. Hydrogen is the simplest of all atoms; its nucleus is a single proton. Bohr proposed a model with the following properties: Postulate One:
Fe
a. The hydrogen atom contains one electron in a circular orbit. b. The hydrogen nucleus is assumed to be at rest. c. The centripetal force on the electron is equal to the electrostatic force between a unit positive charge and a unit negative charge.
BOHR’S HYDROGEN MODEL From Coulomb’s law, Fe = k
/
ee 2 r
and from the centripetal force, Fc =
14
mv2
/r
Here n is the principal quantum number and is allowed to have only whole number values, 1, 2, 3, 4, and so on; the orbit sizes and atomic energy levels are thereby fixed. Bohr calculated the value of k to be h/2π. Yes, Planck’s constant again. Now two equations describe the allowed orbits:
Since the centripetal force is an electrostatic force, mv2
/r = k ee/r2
Up to this point the physics has been classical physics, but here Bohr introduced his quantum hypothesis.
mv2
2
/r = k e /r2 and mvr = nh/2π
Postulate Two: a. The electron is forbidden all but a few orbits around the nucleus. b. The orbits that are allowed are definite and discrete. c. For an orbit to be valid, the preceding equation must be satisfied, and a stringent constraint must be placed on the angular momentum of the electron: its angular momentum must equal an integer times a small constant:
m v r k e n h π
angular momentum mvr = nk.
✍ 1.32
is the electron mass (9.1 • 10-31 kg), is the orbital velocity, is the orbital radius, is the electrostatic constant from Coulomb’s 2 law (9 • 109 N•m /coul2), is the size of the charge on an electron or a proton (1.6 • 10-19 coul), is the principal quantum number, is Planck’s constant (6.63 • 10-34 joule•sec.), and is taken to be 3.14.
Complete this activity. Solve these two equations for v, equate the results, and solve for r, the radius of a discrete orbit.
mv2
2
/r = ke /r2
Score Adult check
mvr =
/2π
nh
______________________ Initial
Date
velocity of light. The high velocity and small orbit results in a very large number of revolutions per second, about 1015. This frequency is the same order of magnitude as the frequency of visible light. A consequence of Bohr’s work up to this point is that the single electron in the hydrogen atom has several orbits that it can occupy. If the electron moves from one orbit it must go directly to another of the allowed orbits. Each orbit must be an integral multiple of the electron’s wavelength in circumference.
Solving this equation for the first orbit (n=1) gives a radius of 0.000,000,000,053 m. A more convenient unit to use for measuring atomic distances is the angstrom: 1 meter = 1010 angstroms 1 A = 10-10 m Thus, we have a radius for the innermost orbit of 0.53 A and a diameter of 1.06 A. If this value is put into the angular momentum equation, the velocity of the electron is calculated to be about 1/137 the
15
If a gas is excited with an electrical discharge, many electrons will go to higher orbits. These excited atoms do not stay excited for long. The electron is attracted to the nucleus and will soon jump to a lower orbit. It may go all the way to the lowest orbit, the ground state, in one jump; or it may make several jumps on its way down. The photons emitted in this process cause the gas to glow in discrete wavelengths that can be discerned with the use of a prism.
n=5
n=4
n=3 2
IONIZATION n n n n
= = = =
∞ 5 4 3
n=2
ev 0.0 -0.54 -0.85 -1.5
-3.4
The final assumption that Bohr made in his model concerned these orbit changes. Postulate Three: a. No radiation is emitted by an electron while it is in one of its allowed orbits. b. Electromagnetic radiation is emitted only when an electron jumps from one orbit to another. c. The frequency of this radiation is determined by the energy difference between the two orbits.
GROUND STATE
n=1
ENERGY LEVELS
FOR
-13.6
HYDROGEN
The success of Bohr’s theory is not that it gives us the “right answer,” for we now know that the theory is much too simplified. The model gave rise to an equation that agrees exactly with experimental results, and this equation is the significant contribution.
hv = E1 - E2 E1 is the energy of the electron in the initial orbit and E2 is its energy in the final orbit. Of course, hv is the energy of the radiated photons.
Quantum numbers. A man named Stoner worked with Bohr to extend the orbital model of the hydrogen atom to other elements. The helium atom has an atomic number Z = 2: Its nucleus contains two protons, which are electrically balanced by two orbiting electrons. Lithium, Z = 3, has three protons and three electrons. The allowed orbits are called shells and are the orbits for the quantum number n = 1, 2, 3 and so on. To explain all the experimental data, Bohr and Stoner had to limit the number of electrons in each shell. The maximum number of electrons per shell is 2n2.
Atomic excitation. Obviously, radiation is emitted only when an electron goes from the higher orbit to the lower energy orbit. As with a gravitational system the electron loses potential energy as it falls into inner orbits; hence the photons are emitted as the electrons move to orbits represented by smaller quantum numbers. In a sample of hydrogen gas, most of the electrons are in the n = 1 orbit. Some of the electrons, however, are knocked into higher orbits by collisions with other atoms or with other electrons. As the temperature of the hydrogen increases, this excitation is more likely to happen.
16
Quantum Number n
Maximum Number of Electrons
1
2
•
12 = 2
2
2
•
22 = 8
3
2
•
32 = 18
4
2
•
42 = 32
5
2
•
52 = 50
6
2
•
62 = 72
The shells are generally filled in order as Z increases, but exceptions occur for some of the heavier elements. For some reason the number of electrons that fill a shell is a “preferred number.” For example, mercury with Z = 80 has the following configuration: n = 1 shell
2 electrons
n = 2 shell
8 electrons
n = 3 shell
18 electrons
n = 4 shell
32 electrons
n = 5 shell
18 electrons
n = 6 shell
2 electrons
electrons enter the next shell before the incomplete shell (n = 5 in this case) finishes filling. The equation for orbit size is valid only for hydrogen. As Z gets larger, the electrons are attracted more strongly; consequently, the orbits get smaller. Therefore, all other atoms are smaller than the hydrogen model would predict. For instance, the helium (Z = 2) atom is only about 70 percent the size of the hydrogen atom. Lithium (Z = 3) is slightly larger than hydrogen; but since only two electrons can be in the n = 1 shell, one must be in the n = 2 shell. Beryllium (Z = 4) is again smaller than hydrogen. In fact, lithium (Z = 3), sodium (Z = 11) and mercury (Z = 80) are all about the same size. As a result the heaviest elements are not much larger in diameter than the lightest elements.
The n = 5 shell stops filling at eighteen, the number for a filled n = 3 shell, and two electrons go to the n = 6 shell even though the n = 5 shell has room for thirty-two more electrons. Not many
✍
Complete these activities.
1.33
Calculate the orbital radii of the hydrogen atom for the first six principal quantum numbers.
1.34
Assume that the radius of the hydrogen nucleus is 1.4 • 10-15 meters. How much larger than the nucleus is the entire hydrogen atom? (Calculate the atomic radius for n = 1.)
1.35
Assume that the radius of the sun is 865,000 miles. According to the atomic scale of distances for hydrogen, considering the earth to be an orbiting electron, calculate the earth’s orbital radius.
17
1.36
An atom that has six principal quantum numbers can yield several emission lines. An excited electron that jumps from the n = 1 to the n = 2 can only drop back to the n = 1. But, an excited electron that jumps from the n = 1 to the n = 6 can jump back 1, 2, 3, 4, or 5 orbits to drop back to the n = 1. a.
Write the number of possible electron shell transitions for which energy is radiated. (Hint: if an electron has jumped from n = 1 to n = 6, it can have these transitions: n = 6 to n = 5, n = 6 to n = 4, n = 6 to n = 3, n = 6 to n = 2, n = 6 to n = 1. That's 5 of them – can you find the rest? Remember, the electron didn't have to jump all the way to the n=6 level to start out with.)
b.
Use the Energy Levels for Hydrogen Chart to calculate the wavelength corresponding to each electron transition. Note that energy is given in electron volts.
TRANSITION
ENERGY
18
IN EV’S
EMITTED WAVELENGTHS
✍
Answer these questions.
1.37
What is a quantum number?
1.38
What are the Bohr postulates? a. b. c.
1.39
Why does a neon sign light up?
1.40
What forces must be balanced in the Bohr model?
✍
Solve these problems.
1.41
Calculate the speed of an electron in the innermost orbit of a hydrogen atom.
1.42
If an electron spends 10-8 seconds in the n = 2 orbit, calculate how many revolutions it makes around the nucleus.
Review the material in this section in preparation for the Self Test. This Self Test will check your mastery of this particular section. The items missed on this Self Test will indicate specific areas where restudy is needed for mastery.
19
SELF TEST 1 Match these items (each answer, 2 points). 1.01
small unit of energy
a. electron volt
1.02
lowest energy level
b. de Broglie wave
1.03
characterizes an electron orbit
c. ground state
1.04
smallest amount of light
d. uncertainty principle
1.05
wave associated with moving matter
e. quantum number
1.06
statement about limit of position measurement
f. continuous spectrum
1.07
light from a heated metal
g. photon
1.08
light from a heated gas
h. line spectrum
1.09
radiation from decelerated electrons
i. wave theory
1.010
6.63
j. X-ray
•
10
-34
joule
•
sec.
k. Planck’s constant Complete these sentences (each answer, 3 points). 1.011
When a photon collides with an electron and is deflected, the photon’s
1.012
According to Einstein’s interpretation, the maximum photoelectron energy is equal to the energy of the incident photon minus the energy required to .
1.013
When the number of electrons striking the anode of an X-ray tube is increased, the of the emitted X-rays increases.
1.014
When the speed of the electrons striking the anode is increased, the the emitted X-rays increases.
1.015
The lowest energy level in an atom is called its
1.016
The energy of a photon is equal to a.
1.017
The wavelength of a moving particle is equal to a. by b. .
1.018
De Broglie’s hypothesis was verified when electrons scattered by certain crystals were observed to experience the wave phenomenon called .
1.019
The orbit of an electron in a hydrogen atom is always an integral number of in circumference.
decreases.
of state.
times b.
. divided
Choose the correct answer (each answer, 2 points). 1.020
In the Rutherford model of the atom, the positive charge in an atom is a.
concentrated at its center
b.
spread uniformly throughout its volume
c.
in the form of positive electrons at some distance from its center
d. readily deflected by an incident alpha particle 1.021
Modern physics theories indicate that
.
a. all particles exhibit wave behavior b. only moving particles exhibit wave behavior c. only charged particles exhibit wave behavior d. only uncharged Particles exhibit wave behavior
20
.
1.022
1.023
1.024
1.025
Photoelectrons are emitted by a metal surface only when the light shining on the surface exceeds a certain minimum . a. wavelength
c. velocity
b. frequency
d. charge .
The transition in a hydrogen atom that absorbs the photon of highest frequency is a. n = 1 to n = 2
c. n = 2 to n = 6
b. n = 2 to n = 1
d. n = 6 to n = 2
When light shines on a metal surface, the maximum energies of emitted electrons a. vary with the intensity of the light
c. vary with the speed of the light
b. vary with the frequency of the light
d. are random
.
An electron can rotate around an atomic nucleus indefinitely without radiating energy if its . orbit a. is a perfect circle b. is sufficiently far from the nucleus c. is less than a de Broglie wavelength in circumference d. contains an integral number of de Broglie wavelengths
1.026
1.027
A neon sign does not produce
.
a. a line spectrum
c. an absorption spectrum
b. an emission spectrum
d. photons
The de Broglie wavelength of a particle is
.
a. proportional to its momentum
c. inversely proportional to its momentum
b. proportional to its energy
d. inversely proportional to its energy
Complete these items (each answer, 5 points). 1.028
If the electron of a hydrogen atom is in the ninth circular orbit of diameter 85.9 A, the frequency of revolution is .
1.029
Make a diagram of an arsenic atom (Z = 33) according to the Bohr-Stoner scheme of atomic structure. Show orbits with numbers indicating electrons in each.
Score Adult check
63 79
______________________ Initial
21
Date
SCIENCE 1209 ATOMIC AND NUCLEAR PHYSICS
CONTENTS I. QUANTUM THEORY . . . . . . . . . . . . . . . . . . . .
2
ELECTROMAGNETIC RADIATION . . . . . . . . . . . . . . . . . . . . MATTER WAVES . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . ATOMIC MODELS . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . ATOMIC SPECTRA . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . BOHR MODEL . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
2 8 11 12 14
II. NUCLEAR THEORY . . . . . . . . . . . . . . . . . . . .
22
BUILDING BLOCKS OF THE NUCLEUS . . . . . . . . . . . . . . . PROPERTIES OF THE NUCLEUS . . . . . . . . . . . . . . . . . . . .
22 23
III. NUCLEAR REACTIONS . . . . . . . . . . . . . . . . .
30
NUCLEAR FISSION . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . FUSION . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . NUCLEAR REACTION APPLICATIONS . . . . . . . . . . . . . . . .
30 32 33
GLOSSARY . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . FORMULAS . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
39 41
Author: Editor: Illustrations:
Robert Hughson, M.S. Alan Christopherson, M.S. Jim Lahowetz Alpha Omega Graphics
804 N. 2nd Ave. E., Rock Rapids, IA 51246-1759 © MM by Alpha Omega Publications, Inc. All rights reserved. LIFEPAC is a registered trademark of Alpha Omega Publications, Inc.
All trademarks and/or service marks referenced in this material are the property of their respective owners. Alpha Omega Publications, Inc. makes no claim of ownership to any trademarks and/or service marks other than their own and their affiliates’, and makes no claim of affiliation to any companies whose trademarks may be listed in this material, other than their own.
ATOMIC AND NUCLEAR PHYSICS You are about to begin the study of modern physics, the physics of very small objects. You will learn more about how atoms and their subparts interact. The laws of physics that you have already learned will be used, but you will notice that at times some new concepts will be considered. OBJECTIVES Read these objectives. The objectives tell you what you will be able to do when you have successfully completed this LIFEPAC®. When 1. 2. 3. 4. 5. 6. 7. 8. 9. 10. 11. 12. 13. 14.
you have finished this LIFEPAC, you should be able to: Explain when wave theory is applicable and when quantum theory is applicable. Calculate the energy of photoelectrons and X-rays. Explain de Broglie waves. State and apply the uncertainty principle. Define emission spectra and absorption spectra. Define line spectra and continuous spectra. Describe the Bohr model of the hydrogen atom. Define nucleon and isotopes. Explain mass defect in terms of mass-energy equivalent. Explain how the binding energy curve shows that fission and fusion can release energy. Explain the concept of half-life. Explain how alpha and beta decay can stabilize a nucleus. Describe fission reaction and fusion reaction. Explain a chain reaction and nuclear energy.
Survey the LIFEPAC. Ask yourself some questions about this study. Write your questions here.
The last page of this LIFEPAC contains formulas and constants that may be used for problem-solving, both on activities and on tests.
1
I. QUANTUM THEORY For many centuries, the primary concern of physics was to describe the behavior of large objects. Newtonian mechanics describes the behavior of matter influenced by gravitational fields or by electromagnetic interactions. Newtonian physics had been successful; so successful, in fact, that most physicists believed they were approaching the ultimate description of nature. They thought that in just a matter of time they would be able to work out the details of a perfect, ordered world. Many believed that the future work of physics would be essentially just measuring the last decimal place. Problems with this neat and supposedly complete theoretical description began to arise about the end of the nineteenth century. More precise experiments opened a whole new world to
the physicist who was beginning to investigate the properties of very small particles. The behavior of these microscopic systems could not be explained in terms of the classical physics and Newtonian mechanics, which had been the physicists’ stock and trade up to this point. In classical physics the physicist had explained the behavior of objects in terms of direct and compelling deductions from a well defined set of experiments. Attempts to do the same from new observations were not successful. The new physics, or quantum physics, was too abstract to make deductive explanations possible. The main reasons were twofold: The basic constructs of this quantum theory were removed from everyday experience, and the microscopic particles could not be observed directly.
SECTION OBJECTIVES Review these objectives. When you have completed this section, you should be able to: 1. Explain when wave theory is applicable and when quantum theory is applicable. 2. Calculate the energy of photoelectrons and X-rays. 3. Explain de Broglie waves. 4. State and apply the uncertainty principle. 5. Define emission spectra and absorption spectra. 6. Define line spectra and continuous spectra. 7. Describe the Bohr model of the hydrogen atom. VOCABULARY Study these words to enhance your learning success in this section. absorption spectrum
energy levels
proton
alpha particle
excited atom
quanta
continuous spectrum
ground state
quantum number
de Broglie wave
line spectrum
radiation
electron volt
photoelectric effect
uncertainty principle
emission spectrum
photon
X-ray
Note: All vocabulary words in this LIFEPAC appear in boldface print the first time they are used. If you are unsure of the meaning when you are reading study the definitions given.
ELECTROMAGNETIC RADIATION The wave theory of light was well recognized by the end of the nineteenth century. Experimental work was being conducted in which both light rays and high speed electrons were allowed to strike a metal surface. The results of these experiments
could not be explained by existing theory. As a result a new theory was developed, called the quantum theory. At the time, physicists did not fully realize what was happening; but modern physics had been born.
2
Photoelectric effect. During the late 1800s several experiments disclosed that electrons were emitted when light fell on a metal plate. This phenomenon is the photoelectric effect. At first no one was surprised: Light waves were known to carry energy; some of this energy, therefore, could somehow be absorbed by electrons to give them enough kinetic energy to escape from the metal surface.
The symbol v (the Greek nu) represents the frequency of incident light. The value of h was always the same value, but v0 changed for each metal. An explanation of this phenomenon was given in 1905 by Albert Einstein. He used a result of earlier work by the German physicist Max Planck, who described the radiation emitted by a glowing object. Planck could get his theory to work if he assumed that the light was emitted in little “lumps” of energy. He called these little bursts of energy quanta and showed that they had energy of
LIGHT
E = hv, where v is the frequency and h is a constant known as Planck’s constant. The value of Planck’s constant h is given as h = 6.63
•
10-34 joule•sec.
AMMETER
Few physicists really believed in Planck’s quanta, but Einstein pointed out that the photoelectric effect was experimental verification of Planck’s equation. He suggested that the photoelectric equation be rewritten:
Soon some problems developed: First, the energy of the photoelectrons was not related to the intensity of the light. When the light was made brighter, more electrons were emitted; but they all had the same average energy as those emitted by dim light. Second, the energy of the photoelectrons was related to the frequency of the light. If the frequency was below a certain value, no electrons were emitted even if the light was intense. Conversely, if the frequency was above the value, electrons were emitted even in very dim light. Photoelectrons emitted by any given frequency were observed to have energy that ranged from zero to some maximum value. If the frequency of the light was increased, the maximum electron energy increased. In other words, a faint blue light produced electrons with higher energy than did a bright red light even though the bright red light produced more electrons. The relationship between KEmax, the maximum photoelectron energy, and the frequency of the incident light can be stated:
hv = KEmax + hv0, and be stated: quantum (incident photon) energy = maximum electron energy + energy required to eject an electron from the metal surface. A large portion of the photoelectron population will have less than maximum energy for the following reasons: 1. The photon (the quantum of incident light) may have shared its energy with more than one electron. 2. The electron may have lost part of its energy before it got out of the metal. The energy of photons is so small that the joule is cumbersome. A unit of energy suited to the atomic scale is the electron volt. One electron volt is the energy change realized by an electron moving through a potential change of one volt. Numerically,
maximum photoelectron energy= h (frequency - frequency0) KEmax = h (v - v0).
1 ev = 1.6
3
•
10-19 joules
LlFEPACs 1204 and 1205 dealt with diffraction and interference. The observed phenomena could be explained in terms of the wave theory of light. Now you have seen an argument that light behaves like a series of packets of energy, called photons, or quanta, each photon small enough to interact with a single electron. The wave theory cannot explain the photoelectric effect and the quantum theory cannot explain interference and diffraction. Which is the correct theory? Many times before in physics, one theory has been replaced by a new theory; but this apparent conflict was the first time that two
✍
different theories were needed to describe a phenomenon. Developing a model for light is a situation in which science must acknowledge that nature is not an “open book” to man. These two theories are complementary: One is correct in some experiments, but the other theory must be used in others. Interestingly, light can exhibit either a wave or a particle nature, but never exhibits both in any single situation. Since we cannot reconcile the two descriptive models, we have no choice but to accept them both.
Answer these questions.
1.1
What is the photoelectric effect?
1.2
Why do photoelectrons have a maximum energy?
1.3
Why does faint light not appear as a series of flashes?
1.4
Why can the photoelectric effect not be explained without quantum theory?
✍
Complete these sentences.
1.5
The phenomenon that describes the emission of electrons from a metal surface when light shines on it is the .
1.6
The maximum energy of photoelectrons a. with frequency and b. with increasing intensity of the light. (increases, decreases, remains constant)
1.7
Photon energy is proportional to
1.8
Photons are also known as
✍ 1.9
. .
Solve these problems. A yellow lamp emits light with a wavelength of 6
•
10-7 m.
a. Write the common value for the speed of light, in meters per second.
4
b. Calculate the frequency of the yellow light from the equation given in Science LIFEPAC 1204.
c. Use Planck’s equation to calculate the energy of a single photon with the given wavelength.
d. How many such photons are required to produce 10 joules?
1.10
A 10,000-watt radio station transmits at 880 kHz. a. Determine the number of joules transmitted per second.
b. Calculate the energy of a single photon at the transmitted frequency.
c. Calculate the number of photons that are emitted per second.
1.11
Light with a wavelength of 5 • 10-7 m strikes a surface that requires 2 ev to eject an electron. a. Calculate the frequency of the given wave.
b. Calculate the energy, in joules, of one incident photon at this frequency.
c. Calculate the energy, in electron volts, of one incident photon.
5
d. Calculate the maximum KE, in electron volts, of the emitted photoelectron.
1.12
Photoelectrons with a maximum speed of 7 • 105 m/sec are ejected from a surface in the presence of light with a frequency of 8 • 1014 Hz.
a. If the mass of an electron is 9.1 • 10-31 kg, calculate the maximum kinetic energy of a single electron, in joules.
b. Calculate, in joules, the energy of incident photons.
c. Calculate the energy required to eject photoelectrons from the surface.
X-rays. When photons of light were found to give up their energy to electrons, the question was asked: Can the energy of moving electrons be converted into photons? Further investigation disclosed that not only could electrons produce photons, but that several years previously the German physicist Roentgen had in fact produced the effect. Roentgen had conducted experiments in which he allowed high speed electrons to strike the surface of metal. He had observed that the metal target emitted some very penetrating radiation. These Roentgen rays, or X-rays, as they are now called, exposed photographic plates. Roentgen was certain that they were not particles; they were affected by neither electric nor magnetic fields. He also found that as the energy of the electrons increased, the X-rays became more penetrating. The density of emitted X-rays increased when the number of electrons was greater. In addition Roentgen established that X-rays could exhibit both interference and polarization effects. He believed these phenomena were evidence that the emissions were electromagnetic radiations.
TARGET ANODE
FILAMENT
ee-
V
X-RAYS
When a charged particle accelerates, it emits electromagnetic radiation. X-rays are identified as electromagnetic radiation emitted when a very fast electron is slowed suddenly as it strikes the metal. Most of the incident electrons probably lose their energy too slowly for X-rays to be emitted. Some of the electrons, however, lose much or all of their energy in a single collision with an atom in the target metal. This radiation is the inverse of the photoelectric effect. In fact, the highest frequency found in X-rays corresponds to the photon energy,
6
and all frequencies below the maximum X-ray frequency also exist. When the electron energy is above a certain value, the emission curves possess strange, sharp spikes. The spikes indicate that some frequencies are emitted preferentially. These emissions originate from rearrangement of the electron structure of the target atom.
30,000 v
X-RAY INTENSITY
15,000 v
5,000 v 1
2 3 4 5 FREQUENCY x 1018 cycles/sec
6
AN X-RAY SPECTRUM
✍
Complete these sentences.
1.13
When the number of electrons striking the anode of an X-ray tube increases, the of the emitted X-rays increases.
1.14
When the speed of electrons striking the anode of an X-ray tube increases, the of the emitted X-rays increases.
✍ 1.15
Complete these activities. An X-ray tube emits X-rays with a wavelength of 10-11 m. a. Calculate the photon energy, in joules, that the emitted X-rays possess.
b. Calculate the energy, in electron volts, that the X-rays possess.
c. Determine the energy, in electron volts, possessed by the incident electrons.
d. Calculate the potential that must be applied across the X-ray tube to give each incident electron its energy.
7
1.16
Calculate the highest frequency X-rays produced by 8
1.17
A television tube can accelerate electrons to 2 X-rays with the highest energy.
1.18
Calculate the energy, in electron volts, of X-rays that have a frequency of 1019 Hz.
•
•
104 ev electrons.
104 ev. Calculate the wavelength of emitted
MATTER WAVES and the relationship wavelength, and velocity,
X-rays are used to study the configuration of atoms in crystals. Atoms in a crystal interact with X-rays, causing the scattered waves to interfere constructively in some regions and destructively in others. The interference pattern thus produced is characteristic of the spatial arrangement of the atoms in the crystal. The exact pattern is determined by the wavelength of the X-rays and the distance between the atoms. If, instead of X-rays, electrons are used, a pattern is produced almost exactly the same as the X-ray pattern. This experiment is known as the Davisson-Germer experiment. The implication of this experiment is that electrons behave as waves. X-rays and their diffraction pattern are used to measure the distance between atoms in crystals.
frequency,
λv = c Substituting the second equation into the first gives momentum =
/c = h/λ
hv
Solving this relationship for the wavelength gives λ=
/momentum
h
for a photon. The new idea that de Broglie suggested was that this formula is much more general and that it applies to particles as well as to waves. The expression for the momentum of a particle is
De Broglie waves. By using crystals with known atomic spacing, the experiment can be used to determine the wavelength of the electron. These wavelengths are called de Broglie wavelengths; and the waves themselves are de Broglie waves, or matter waves. The wavelength of the de Broglie wave is inversely proportional to the momentum of the particle (generally an electron) that is being observed. De Broglie took the formula for the momentum of a photon, momentum =
between
momentum = mv. Substituting this equation into the former relationship gives the expression for the wavelength of a particle: λ=
/mv
h
Again the relationship between the wavelength and the momentum involves the quantity h, Planck’s constant.
/c
hv
8
The value of h is extremely small, which explains why waves are not observed for large objects. The momentum must be small compared to h for the wavelength to be appreciable. Particles at times have the properties of a wave; at other times they exhibit the properties of a particle. As in the case of electromagnetic radiation, the particle is never observed to exhibit both particle characteristics and wave characteristics in the same interaction; and we cannot determine which description is the correct one. This phenomenon is sometimes called the wave particle duality, or just duality. The reason we cannot observe this property of duality is because the wavelengths are so small. For instance, the wavelength of an automobile traveling 60 miles an hour is on the order of 10-38 feet. This wavelength is much much smaller than the automobile. On the other hand, the wavelength of an electron traveling about 10 million meters per second is about 10-8 meters. This very small wavelength is still the magnitude of atomic dimensions. Since the wave length is comparable to the size of the electron, the wavelength is observable.
ELECTRON e-
N IO CT E R DI VE A W
PHOTON INITIAL MOMENTUM
ELECTRON OBSERVER
e-
PHOTON
FINAL MOMENTUM
OBSERVER
Even the position is not known with absolute accuracy since, even with the best of instruments, the position will be off by at least part of a wavelength. Thus,
Uncertainty principle. One of the tenets of quantum physics is an inherent limitation on our capability for making observations. To measure something, some way is needed to carry information from the object to the observer. You must be able to touch it, shine a light on it, or perform some other manipulation for information about the object’s size or its position to be carried back to the observer. Observing a large object is fairly easy. For instance, you can measure a table with a ruler. When the object being observed is an electron, the problem is different. About the only way that information about the electron’s size or its position can be gathered is to shine light on the electron and to reflect it to the observer. The difficulty is that the light, to reflect from the electron, must strike the electron. In striking the electron, some of the light energy is transferred to the electron. The reflected light enables the position of the electron to be determined precisely: Project back along the path of light and determine where the electron was when struck by the light. However, when the electron was struck by the photon, the energy given up by the light changed the direction of the electron. The very act of observing the electron changes its momentum. Although you now know where the electron was at that instant, you do not know in exactly what direction it is now going.
∆x ⬇ λ is a reasonable estimate of the uncertainty about the position of the particle. The photon had a momentum: momentum = h/λ . Although you cannot know exactly how much the momentum of the electron will change, it is reasonably the same magnitude as the photon momentum; thus, ∆mv ⬇ h/λ . Combining the two equations of uncertainty gives the result: ∆x
•
∆mv ⬇ h.
The uncertainty in position times the uncertainty in momentum is approximately equal to Planck’s constant, which means that we can never hope to measure both position and momentum 9
simultaneously with perfect accuracy. In 1927 the German physicist Heisenberg first proposed this result, which is usually referred to as the Heisenberg uncertainty principle. Uncertainty does not result from deficiencies in measuring instruments; it is a constraint of nature. Remember again that Planck’s constant is very small; matter waves have significance for atomic particles, but not for large, “Newtonian” objects.
the relationships can be verified with absolute accuracy. Are these relationships, then, not true? Unfortunately, we cannot make as positive a statement as we might wish. Individual particles may not behave according to the Newtonian laws of classical physics; but for statistically large groups of particles, these expressions become statements of probability. Many many small particles obey classical laws of physics on the average, but individual particles may deviate from this average behavior. Fortunately, everything that we physically encounter is made up of such large numbers of particles that their average behavior is all that is observed. Causality is reasonably reliable for macroscopic events, but we cannot expect it to hold true in situations that we cannot experience directly.
Causality. Previously we assumed that the laws of physics were cause-and-effect relationships: that a given cause always produces a specific effect. The equations studied in mechanics give relationships; generally, that force is proportional to something else. The uncertainty principle, however, shows that all the factors that go into an equation cannot be measured with total precision. None of
✍
Complete these sentences. divided
1.19
The wavelength of a moving particle is equal to a. . by b.
1.20
The uncertainty principle states that the product of uncertainty in the momentum and in the . position of a particle is approximately equal to
✍
Complete these activities
1.21
Calculate the de Broglie wavelength of a 5,000 kg truck traveling at 80 kph. (Hint: convert everything to fundamental units.)
1.22
Calculate the de Broglie wavelength of an electron traveling at 107 m/sec (me = 9.1
1.23
Calculate the approximate momentum change in a particle of mass 1.7 initially at rest, whose position is located to within 10-4 m.
1.24
Calculate the uncertainty of the velocity of a particle confined to a space of 10-9 m if the particle is: a. an electron
b. a proton
10
•
•
10-31 kg).
10-27 kg (a proton),
ATOMIC MODELS By the year 1900 the field of chemistry had accumulated a large amount of evidence that material consisted of atoms. Very little evidence existed, however, about the nature of the atoms themselves. J.J. Thomson had recently discovered the electron; this discovery suggested that electrical forces were of significant strength on the atomic scale.
of Thomson’s model. This experiment was conducted by two men named Geiger and Marsden. They placed in a lead box material that emits alpha particles. (Alpha particles will be discussed in detail later. All we have to know about them now is that they are small, positively charged radiations emitted by some natural substances such as radium.) A small hole drilled in the box allowed a stream of alpha particles to escape. This beam of alpha particles was directed at a very thin metal foil. Geiger and Marsden found that, just as the Thomson model predicted, most of the alpha particles passed through the foil or were deflected through relatively small angles. The particles were observed as they struck a zinc sulphide screen surrounding the foil. When alpha particles struck the screen a small amount of light was emitted. Geiger and Marsden made one interesting observation: Most of the alpha particles passed through the screen, but a few were scattered through rather large angles. In fact, some were scattered back nearly 180°. Alpha particles are much larger than electrons, nearly 8,000 times greater in mass; only very strong forces could change the direction of the alpha particles so drastically.
Thomson model. Thomson proposed what is referred to as the Thomson model, sometimes called the plum pudding model, of the atom. He suggested that atoms were spheres of positively charged matter and the electrons were embedded in this matter much as plums are embedded in plum pudding. This model was quite reasonable using the evidence available at the beginning of this century. It provided an explanation for most of the atomic behavior that had been observed. If this model were the nature of atoms, then one would expect that charged particles as they moved through an atom would not be appreciably affected. Electric fields in atoms would be quite weak, because the positive charge in an atom would be diffused by negative “plums,” and vice versa. Shortly afterwards, however, another experiment was done that caused some concern as to the validity
ZINC-SULFIDE SCREEN
ALPHA EMITTER
LIGHT FLASHES
LEAD BOX THIN METAL FOIL
THE GEIGER-MARSDEN EXPERIMENTS
Rutherford model. To explain this large angle scattering, the British physicist Ernest Rutherford proposed a new model of the atom. His theory was that the positive charge was not distributed throughout the entire atom, but was concentrated in a very very small portion near the center. The atom was mostly empty space. The negative electrons moved around in this empty
space; but they were very small, very light, and could not have an appreciable effect upon the heavy alpha particle. The center of the atom, the nucleus, was very small and would effect only those alpha particles that came very near; most of the alpha particles traveled through the empty space. Alpha particles that approached a nucleus, however, would experience strong electric forces 11
and would be scattered through a large angle. If an alpha particle were to strike an electron, it would knock the electron out of the way; the nucleus, however, being much heavier than the alpha particle, caused the alpha particle to change its
✍ 1.25
direction. Using this idea and the rules of electrostatics, Rutherford was able to derive a formula that explained the backward scattering of alpha particles. As a result, Rutherford is credited with the discovery of the dense atomic nucleus.
Complete this activity. Using the description of the Thomson and Rutherford models of the atom, draw a labeled diagram of each model that illustrates the differences between the two.
THOMSON MODEL
✍ 1.26
RUTHERFORD MODEL
Prepare a report. Some of the men involved in the development of early atomic physics worked in the same laboratory. Look up the following men: Ernest Rutherford, J. J. Thomson, and Neils Bohr. Write a five-page, double-spaced, typewritten report on the men and their contributions to atomic theory. Submit your report for evaluation. Score Adult check
______________________ Initial
Date
ATOMIC SPECTRA A spectrum is an array of the different wavelengths of radiant energy emitted by a source. Different sources produce different wavelengths and exhibit one of the four different kinds of spectra: 1. 2. 3. 4.
Emission spectra. Heating a block of metal produces no noticeable change until its temperature reaches about 1000°K, at which point the block begins to glow a dull red. As heating continues, the glow becomes bright red and then orange. When the temperature of the block is about 1500°K, the spectrum contains red, orange, and yellow. At 2,000°K the block appears yellow, and the spectrum now includes green. When the temperature reaches 3,000°K, the block glows white hot; and all the colors from red through violet are present in the spectrum along with invisible infrared and ultraviolet radiations.
Continuous emission spectra, Line emission spectra, Continuous absorption spectra, or Line absorption spectra.
12
INTENSITY
VISIBLE
ULTRAVIOLET INFRARED
WAVELENGTH
RADIATION EMITTED
BY A
HEATED SOLID
When hot gas is the source of the radiation, quite a different spectrum results. Instead of a continuous spectrum, a number of bright lines constitute the spectrum. SLIT HOT GAS
PRISM
LENS
ARRANGEMENT
LENS
TO
DEMONSTRATE
The most intense line spectra are obtained from metallic arc lamps, such as a carbon arc, when one of the carbon rods has been impregnated with a metal like mercury or sodium. Continuous spectra come from high temperature solids but line spectra always come from an incandescent gas.
A
LINE EMISSION SPECTRUM
SCREEN
wavelengths will be missing. When light from an incandescent bulb (hot solid filament) passes through colored glass, the glass absorbs all wavelengths except the color of the glass. When a continuous-emission spectrum is passed through gas below its temperature of incandescence, certain discrete wavelengths are removed. The resulting absorption spectrum has dark lines in the positions of the missing wavelengths.
Absorption spectra. When a continuous emission spectrum passes through a solid and the radiation is then analyzed, a wide band of SLIT
SODIUM GAS HOT GAS LENS
LENS
LENS
PRISM
SCREEN
SODIUM
GAS BURNER
ARRANGEMENT
TO
DEMONSTRATE LINE ABSORPTION SPECTRUM
13
When the sodium metal is heated, the chamber is filled with relatively cool sodium vapor; the spectrum on the screen displays two dark lines in the yellow region. Alkali metals are about the only substances that produce lines in the visible region of the spectrum, but practically all substances produce lines in the ultraviolet.
✍
1.27
Define these terms. spectrum
1.28
emission spectrum
1.29
absorption spectrum
1.30
continuous spectrum
1.31
line spectrum
Again we see that solids produce a continuous spectrum and that gases produce a line spectrum. The reason for the linear display is the slit through which the spectrum is focused. If the light is focused instead through a small hole, small disks would replace the line.
BOHR MODEL A new era in physics began in 1913 when Niels Bohr proposed a new structure of the hydrogen atom. This theory gave a satisfactory explanation for the spectrum emitted by hydrogen and was later successfully extended to other elements.
v
e+ e-
Hydrogen model. Hydrogen is the simplest of all atoms; its nucleus is a single proton. Bohr proposed a model with the following properties: Postulate One:
Fe
a. The hydrogen atom contains one electron in a circular orbit. b. The hydrogen nucleus is assumed to be at rest. c. The centripetal force on the electron is equal to the electrostatic force between a unit positive charge and a unit negative charge.
BOHR’S HYDROGEN MODEL From Coulomb’s law, Fe = k
/
ee 2 r
and from the centripetal force, Fc =
14
mv2
/r
Here n is the principal quantum number and is allowed to have only whole number values, 1, 2, 3, 4, and so on; the orbit sizes and atomic energy levels are thereby fixed. Bohr calculated the value of k to be h/2π. Yes, Planck’s constant again. Now two equations describe the allowed orbits:
Since the centripetal force is an electrostatic force, mv2
/r = k ee/r2
Up to this point the physics has been classical physics, but here Bohr introduced his quantum hypothesis.
mv2
2
/r = k e /r2 and mvr = nh/2π
Postulate Two: a. The electron is forbidden all but a few orbits around the nucleus. b. The orbits that are allowed are definite and discrete. c. For an orbit to be valid, the preceding equation must be satisfied, and a stringent constraint must be placed on the angular momentum of the electron: its angular momentum must equal an integer times a small constant:
m v r k e n h π
angular momentum mvr = nk.
✍ 1.32
is the electron mass (9.1 • 10-31 kg), is the orbital velocity, is the orbital radius, is the electrostatic constant from Coulomb’s 2 law (9 • 109 N•m /coul2), is the size of the charge on an electron or a proton (1.6 • 10-19 coul), is the principal quantum number, is Planck’s constant (6.63 • 10-34 joule•sec.), and is taken to be 3.14.
Complete this activity. Solve these two equations for v, equate the results, and solve for r, the radius of a discrete orbit.
mv2
2
/r = ke /r2
Score Adult check
mvr =
/2π
nh
______________________ Initial
Date
velocity of light. The high velocity and small orbit results in a very large number of revolutions per second, about 1015. This frequency is the same order of magnitude as the frequency of visible light. A consequence of Bohr’s work up to this point is that the single electron in the hydrogen atom has several orbits that it can occupy. If the electron moves from one orbit it must go directly to another of the allowed orbits. Each orbit must be an integral multiple of the electron’s wavelength in circumference.
Solving this equation for the first orbit (n=1) gives a radius of 0.000,000,000,053 m. A more convenient unit to use for measuring atomic distances is the angstrom: 1 meter = 1010 angstroms 1 A = 10-10 m Thus, we have a radius for the innermost orbit of 0.53 A and a diameter of 1.06 A. If this value is put into the angular momentum equation, the velocity of the electron is calculated to be about 1/137 the
15
If a gas is excited with an electrical discharge, many electrons will go to higher orbits. These excited atoms do not stay excited for long. The electron is attracted to the nucleus and will soon jump to a lower orbit. It may go all the way to the lowest orbit, the ground state, in one jump; or it may make several jumps on its way down. The photons emitted in this process cause the gas to glow in discrete wavelengths that can be discerned with the use of a prism.
n=5
n=4
n=3 2
IONIZATION n n n n
= = = =
∞ 5 4 3
n=2
ev 0.0 -0.54 -0.85 -1.5
-3.4
The final assumption that Bohr made in his model concerned these orbit changes. Postulate Three: a. No radiation is emitted by an electron while it is in one of its allowed orbits. b. Electromagnetic radiation is emitted only when an electron jumps from one orbit to another. c. The frequency of this radiation is determined by the energy difference between the two orbits.
GROUND STATE
n=1
ENERGY LEVELS
FOR
-13.6
HYDROGEN
The success of Bohr’s theory is not that it gives us the “right answer,” for we now know that the theory is much too simplified. The model gave rise to an equation that agrees exactly with experimental results, and this equation is the significant contribution.
hv = E1 - E2 E1 is the energy of the electron in the initial orbit and E2 is its energy in the final orbit. Of course, hv is the energy of the radiated photons.
Quantum numbers. A man named Stoner worked with Bohr to extend the orbital model of the hydrogen atom to other elements. The helium atom has an atomic number Z = 2: Its nucleus contains two protons, which are electrically balanced by two orbiting electrons. Lithium, Z = 3, has three protons and three electrons. The allowed orbits are called shells and are the orbits for the quantum number n = 1, 2, 3 and so on. To explain all the experimental data, Bohr and Stoner had to limit the number of electrons in each shell. The maximum number of electrons per shell is 2n2.
Atomic excitation. Obviously, radiation is emitted only when an electron goes from the higher orbit to the lower energy orbit. As with a gravitational system the electron loses potential energy as it falls into inner orbits; hence the photons are emitted as the electrons move to orbits represented by smaller quantum numbers. In a sample of hydrogen gas, most of the electrons are in the n = 1 orbit. Some of the electrons, however, are knocked into higher orbits by collisions with other atoms or with other electrons. As the temperature of the hydrogen increases, this excitation is more likely to happen.
16
Quantum Number n
Maximum Number of Electrons
1
2
•
12 = 2
2
2
•
22 = 8
3
2
•
32 = 18
4
2
•
42 = 32
5
2
•
52 = 50
6
2
•
62 = 72
The shells are generally filled in order as Z increases, but exceptions occur for some of the heavier elements. For some reason the number of electrons that fill a shell is a “preferred number.” For example, mercury with Z = 80 has the following configuration: n = 1 shell
2 electrons
n = 2 shell
8 electrons
n = 3 shell
18 electrons
n = 4 shell
32 electrons
n = 5 shell
18 electrons
n = 6 shell
2 electrons
electrons enter the next shell before the incomplete shell (n = 5 in this case) finishes filling. The equation for orbit size is valid only for hydrogen. As Z gets larger, the electrons are attracted more strongly; consequently, the orbits get smaller. Therefore, all other atoms are smaller than the hydrogen model would predict. For instance, the helium (Z = 2) atom is only about 70 percent the size of the hydrogen atom. Lithium (Z = 3) is slightly larger than hydrogen; but since only two electrons can be in the n = 1 shell, one must be in the n = 2 shell. Beryllium (Z = 4) is again smaller than hydrogen. In fact, lithium (Z = 3), sodium (Z = 11) and mercury (Z = 80) are all about the same size. As a result the heaviest elements are not much larger in diameter than the lightest elements.
The n = 5 shell stops filling at eighteen, the number for a filled n = 3 shell, and two electrons go to the n = 6 shell even though the n = 5 shell has room for thirty-two more electrons. Not many
✍
Complete these activities.
1.33
Calculate the orbital radii of the hydrogen atom for the first six principal quantum numbers.
1.34
Assume that the radius of the hydrogen nucleus is 1.4 • 10-15 meters. How much larger than the nucleus is the entire hydrogen atom? (Calculate the atomic radius for n = 1.)
1.35
Assume that the radius of the sun is 865,000 miles. According to the atomic scale of distances for hydrogen, considering the earth to be an orbiting electron, calculate the earth’s orbital radius.
17
1.36
An atom that has six principal quantum numbers can yield several emission lines. An excited electron that jumps from the n = 1 to the n = 2 can only drop back to the n = 1. But, an excited electron that jumps from the n = 1 to the n = 6 can jump back 1, 2, 3, 4, or 5 orbits to drop back to the n = 1. a.
Write the number of possible electron shell transitions for which energy is radiated. (Hint: if an electron has jumped from n = 1 to n = 6, it can have these transitions: n = 6 to n = 5, n = 6 to n = 4, n = 6 to n = 3, n = 6 to n = 2, n = 6 to n = 1. That's 5 of them – can you find the rest? Remember, the electron didn't have to jump all the way to the n=6 level to start out with.)
b.
Use the Energy Levels for Hydrogen Chart to calculate the wavelength corresponding to each electron transition. Note that energy is given in electron volts.
TRANSITION
ENERGY
18
IN EV’S
EMITTED WAVELENGTHS
✍
Answer these questions.
1.37
What is a quantum number?
1.38
What are the Bohr postulates? a. b. c.
1.39
Why does a neon sign light up?
1.40
What forces must be balanced in the Bohr model?
✍
Solve these problems.
1.41
Calculate the speed of an electron in the innermost orbit of a hydrogen atom.
1.42
If an electron spends 10-8 seconds in the n = 2 orbit, calculate how many revolutions it makes around the nucleus.
Review the material in this section in preparation for the Self Test. This Self Test will check your mastery of this particular section. The items missed on this Self Test will indicate specific areas where restudy is needed for mastery.
19
SELF TEST 1 Match these items (each answer, 2 points). 1.01
small unit of energy
a. electron volt
1.02
lowest energy level
b. de Broglie wave
1.03
characterizes an electron orbit
c. ground state
1.04
smallest amount of light
d. uncertainty principle
1.05
wave associated with moving matter
e. quantum number
1.06
statement about limit of position measurement
f. continuous spectrum
1.07
light from a heated metal
g. photon
1.08
light from a heated gas
h. line spectrum
1.09
radiation from decelerated electrons
i. wave theory
1.010
6.63
j. X-ray
•
10
-34
joule
•
sec.
k. Planck’s constant Complete these sentences (each answer, 3 points). 1.011
When a photon collides with an electron and is deflected, the photon’s
1.012
According to Einstein’s interpretation, the maximum photoelectron energy is equal to the energy of the incident photon minus the energy required to .
1.013
When the number of electrons striking the anode of an X-ray tube is increased, the of the emitted X-rays increases.
1.014
When the speed of the electrons striking the anode is increased, the the emitted X-rays increases.
1.015
The lowest energy level in an atom is called its
1.016
The energy of a photon is equal to a.
1.017
The wavelength of a moving particle is equal to a. by b. .
1.018
De Broglie’s hypothesis was verified when electrons scattered by certain crystals were observed to experience the wave phenomenon called .
1.019
The orbit of an electron in a hydrogen atom is always an integral number of in circumference.
decreases.
of state.
times b.
. divided
Choose the correct answer (each answer, 2 points). 1.020
In the Rutherford model of the atom, the positive charge in an atom is a.
concentrated at its center
b.
spread uniformly throughout its volume
c.
in the form of positive electrons at some distance from its center
d. readily deflected by an incident alpha particle 1.021
Modern physics theories indicate that
.
a. all particles exhibit wave behavior b. only moving particles exhibit wave behavior c. only charged particles exhibit wave behavior d. only uncharged Particles exhibit wave behavior
20
.
1.022
1.023
1.024
1.025
Photoelectrons are emitted by a metal surface only when the light shining on the surface exceeds a certain minimum . a. wavelength
c. velocity
b. frequency
d. charge .
The transition in a hydrogen atom that absorbs the photon of highest frequency is a. n = 1 to n = 2
c. n = 2 to n = 6
b. n = 2 to n = 1
d. n = 6 to n = 2
When light shines on a metal surface, the maximum energies of emitted electrons a. vary with the intensity of the light
c. vary with the speed of the light
b. vary with the frequency of the light
d. are random
.
An electron can rotate around an atomic nucleus indefinitely without radiating energy if its . orbit a. is a perfect circle b. is sufficiently far from the nucleus c. is less than a de Broglie wavelength in circumference d. contains an integral number of de Broglie wavelengths
1.026
1.027
A neon sign does not produce
.
a. a line spectrum
c. an absorption spectrum
b. an emission spectrum
d. photons
The de Broglie wavelength of a particle is
.
a. proportional to its momentum
c. inversely proportional to its momentum
b. proportional to its energy
d. inversely proportional to its energy
Complete these items (each answer, 5 points). 1.028
If the electron of a hydrogen atom is in the ninth circular orbit of diameter 85.9 A, the frequency of revolution is .
1.029
Make a diagram of an arsenic atom (Z = 33) according to the Bohr-Stoner scheme of atomic structure. Show orbits with numbers indicating electrons in each.
Score Adult check
63 79
______________________ Initial
21
Date
