PART 1: PARTICLES AND WAVES PREVIEW A photon is the smallest particle of light, and has an energy which is proportional to its frequency. The photon nature of light is the principle behind the photoelectric effect, in which the absorption of photons of a certain frequency causes electrons to be emitted from a metal surface. The Compton effect also verifies the photon nature of light by showing that momentum is conserved in a collision between a photon and an electron. The Heisenberg uncertainty principle states that both the position and momentum of a subatomic particle cannot be measured precisely. Since light waves exhibit particle (photon) properties, de Broglie suggested that particles, such as electrons, can exhibit wave properties. QUICK REFERENCE Important Terms blackbody radiation the radiation emitted by a blackbody, or perfect emitter and absorber of light, due to its temperature Compton effect the interaction of photons with electrons resulting in the increased wavelengths of the photons and kinetic energy of the electrons Compton wavelength of an electron half the maximum wavelength change of a photon in a Compton scattering with an electron de Broglie wavelength the wavelength associated with a moving particle with a momentum mv Heisenberg uncertainty principle the more accurately one determines the position of a subatomic particle, the less accurately its momentum is known photoelectric effect 1
Transcript
PART 1: PARTICLES AND WAVES
PREVIEW
A photon is the smallest particle of light, and has an energy which is proportional to its frequency. The photon nature of light is the principle behind the photoelectric effect, in which the absorption of photons of a certain frequency causes electrons to be emitted from a metal surface. The Compton effect also verifies the photon nature of light by showing that momentum is conserved in a collision between a photon and an electron. The Heisenberg uncertainty principle states that both the position and momentum of a subatomic particle cannot be measured precisely. Since light waves exhibit particle (photon) properties, de Broglie suggested that particles, such as electrons, can exhibit wave properties.
QUICK REFERENCE
Important Terms
blackbody radiationthe radiation emitted by a blackbody, or perfect emitter and absorber of light, due to its temperature
Compton effect the interaction of photons with electrons resulting in the increased wavelengths of the photons and kinetic energy of the electrons
Compton wavelength of an electronhalf the maximum wavelength change of a photon in a Compton scattering with an electron
de Broglie wavelength the wavelength associated with a moving particle with a momentum mv
Heisenberg uncertainty principle the more accurately one determines the position of a subatomic particle, the less accurately its momentum is known
photoelectric effect the ejection of electrons from certain metals when exposed to light of a minimum frequency
photon the smallest particle of light
Planck’s constant the quantity that results when the energy of a photon is divided by its frequency
quantized a quantity that cannot be divided into smaller increments forever, for which there exists a minimum, quantum increment
quantum mechanics the study of the properties of matter using its wave properties
wave-particle duality under certain circumstances, waves can behave like particles, and particles can behave like waves
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work function the minimum energy required to release an electron from a metal
Equations and Symbols
where
E = energy of a photonc = speed of light = 3 x 108 m/sf = frequency of lightλ = wavelength of lightWo = work function of a photoemissive surface (denoted by on the AP Physics exam)h = Planck’s constant = 6.63 x 10-34 J s = 4.14 x 10-15 eV sfo = threshold frequency of a photoemissive surfaceKEmax = maximum kinetic energy of electrons emitted in the photoelectric effecte = charge on one electronVstop = voltage needed to stop the emission of electrons
= wavelength of a photon after being scattered by a collision with an electronθ = angle between the scattered photon and electron after they collidep = momentum of a photonm = mass of a moving particlev = speed or velocity
The Wave-Particle Duality, Blackbody Radiation and Planck’s Constant and Photons and the Photoelectric Effect
In prior chapters we treated light as a wave. But there are circumstances when light behaves more like it is made up of individual bundles of energy, separate from each other, but sharing a wavelength, frequency, and speed. The quantum of light is called the photon.
In the late 19th century an effect was discovered by Heinrich Hertz which could not be explained by the wave model of light. He shined ultraviolet light on a piece of zinc metal,
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c
photonlight wavec
and the metal became positively charged. Although he did not know it at the time, the light was causing the metal to emit electrons. This effect of using light to cause electrons to be emitted from a metal is called the photoelectric effect. According to the theory of light at the time, light was considered a wave, and should not be able to “knock” electrons off of a metal surface. At the turn of the 20th century, Max Planck showed that light could be treated as tiny bundles of energy called photons, and the energy of a photon was proportional to its frequency. Thus, a graph of photon energy E vs. frequency f looks like this:
The slope of this line is a constant that occurs many times in the study of quantum phenomena called Planck’s constant. Its symbol is h, and its value is 6.62 x 10-34 J s (or J/Hz). The equation for the energy of a photon is
E = hf
or, since ,
The energy of a photon is proportional to its frequency, but inversely proportional to its wavelength. This means that a violet has a higher frequency and energy than a red photon.
Oftentimes when dealing with small amounts of energy like that of photons or electrons, we may prefer to use a very small unit of energy called the electron-volt (eV). The conversion between joules and electron-volts is
Planck’s constant can be expressed in terms of electron-volts as
In 1905, Albert Einstein used Planck’s idea of the photon to explain the photoelectric effect: one photon of energy which is higher than the energy (work function )which binds the electron to the metal is absorbed by one electron in the metal surface, giving the electron enough energy be released from the metal. Any energy left over from the photon after the work function has been met becomes the kinetic energy of the electron.
ΔE
Δf
Energy
frequency
hfEslope
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where fo is called the threshold frequency, which is the minimum frequency the incoming photonmust have to dig the electron out of the metal surface.
Example 1The metal sodium has a threshold frequency which corresponds to yellow light. Describe what will happen if (a) yellow light is shined on the sodium surface,(b) red light is shined on the metal surface,(c) green light is shined on the metal surface,(d) bright green light is shined on the metal surface.
Solution(a) If yellow light is shined on a sodium surface, the yellow photons will be absorbed by electrons in the metal, causing them to be released, but there will be no energy left over for the electrons to have any kinetic energy. (b) Red light has a lower frequency and energy than yellow light, therefore red photons do not have enough energy to release the electrons from the sodium surface. (c) Green light has a higher frequency and energy than yellow light, and therefore a green photon will be absorbed by a sodium electron and the electron will be released from the metal and have kinetic energy. (d) If a brighter (more photons) green light is shined on the surface, more electrons will be emitted, since one photon can be absorbed by one electron. If these electrons are funneled into a circuit, we can use them as current in an electrical device.
The photoelectric effect is the principle behind any process in which light produces electricity, such as a solar calculator or an auto-focus camera.
The graph of maximum kinetic energy of a photoelectron vs. frequency of light incident on a sodium surface would look like this:
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metale
Photon E
KE
frequency
hfEslope
KineticEnergy
R O Y G B V
Note that the electrons have no kinetic energy up to the threshold frequency (color), and then their kinetic energy is proportional to the frequency of the incoming light.
Example 2
Light is shined on a photoemissive surface of work function = 2.0 eV and electrons are released with a kinetic energy KEmax = 4.0 eV.
(a) What voltage, called the stopping voltage Vstop, would be necessary to stop the emission of electrons?
(b) Determine the energy of each of the incoming photons in eV and in Joules.
(c) Determine the frequency of the incoming photons in Hz.
(d) If photons of wavelength λ = 2.5x10-7m were shined on this photoemissive surface, would electrons be emitted from the surface? Justify your answer.
Solution(a) Stopping the emission of electrons requires work equal to the maximum kinetic energy of the electrons:
Thus, it would take 4.0 V to stop electrons with a kinetic energy of 4.0 eV, as we might expect.
(b)
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photoemissivesurface
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(c)
(d) The wavelength of these incoming photons corresponds to a frequency of
Now we need to check to see if this frequency is higher than the threshold frequency f0:
The incoming frequency is higher than the threshold frequency, so electrons will be emitted from the metal surface.
The Momentum of a Photon and the Compton Effect
Since a photon has energy, does it follow that it has momentum? Recall in an earlier chapter that we defined momentum as the product of mass and velocity. But a photon has no mass. It turns out that in quantum physics, photons do have momentum which is inversely proportional to its wavelength. The equation for the momentum of a photon is
Photons can and do impart momentum to sub-atomic particles in collisions that follow the law of conservation of momentum. This phenomena was experimentally verified by Arthur Compton in 1922. Compton aimed x-rays of a certain frequency at electrons, and when they collided and scattered, the x-rays were measured to have a lower frequency indicating less energy and momentum. The scattering of x-ray photons from an electron with a loss in energy of the x-ray photon is called the Compton effect. It is difficult to understand how a photon, having only energy and no mass, can collide with a particle like an electron and change its momentum, but this has been verified experimentally many times.
Example 3
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c
photonelectron
Before
c
photonelectron
After
A photon is fired at an electron which is initially at rest. The photon strikes the electron and reverses its direction, as shown in the diagram representing the photon and electron after the collision.
(a) Determine the shift in wavelength of the photon as a result of the collision.
The photon is an x-ray with a wavelength of 6.62 x 10-11 m. (b) Determine i. the initial momentum of the photon before the collision with the electron. ii. the final momentum of the photon after the collision with the electron.
(c) Find the final momentum of the electron after the collision.
Solution(a) The Compton equation gives the shift in wavelength:
(b) i.
ii. The momentum of the photon after the collision p΄corresponds to the new wavelength λ΄:
The momentum of the photon is given a negative sign, since it reverses direction after the collision.
(c) By the law of conservation of momentum, the momentum lost by the photon must have been gained by the electron.
lost by the photon. Thus, the momentum gained by the electron is 1.93 x 10-23 kgm/s.
The de Broglie Wavelength and the Wave Nature of Matter
In 1924, Louis de Broglie reasoned that if a wave such as light can behave like a particle, having momentum, then why couldn’t particles behave like waves? If the momentum of a
photon can be found by the equation , then the wavelength can be found by .
De Broglie suggested that for a particle with mass m and speed v, we could write the
equation as , and the wavelength of a moving particle could be calculated. This
hypothesis was initially met with a considerable amount of skepticism until it was shown by Davisson and Germer in 1927 that electrons passing through a nickel crystal were
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diffracted through the crystal, producing a diffraction pattern on a photographic plate. Thus, de Broglie’s hypothesis that particles could behave like waves was experimentally verified. Nuclear and particle physicists must take into account the wave behavior of subatomic particles in their experiments. We typically don’t notice the wave properties of objects moving around us because the masses are large in comparison to subatomic particles and the value for Planck’s constant h is extremely small. But the wavelength of any moving mass is inversely proportional to the momentum of the object.
The Heisenberg Uncertainty Principle
Since a photon is the smallest and most unobtrusive measuring device we have available to us, and even a photon has too large of a momentum to make accurate measurements of the speed and position of sub-atomic particles, we must admit to an uncertainty that will always exist in quantum measurements. This limit to accuracy at this level was formulated by Werner Heisenberg in 1928 and is called the Heisenberg uncertainty principle. It can be stated like this:
There is a limit to the accuracy of the measurement of the speed (or momentum) and position of any sub-atomic particle. The more accurately we measure the speed of a particular particle, the less accurately we can measure its position, and vice-versa.
REVIEW QUESTIONSFor each of the multiple choice questions below, choose the best answer.
1. The smallest, discrete value of any quantity in physics is called the (A) atom(B) molecule(C) proton(D) electron(E) quantum
2. The smallest discrete value of electromagnetic energy is called the (A) photon(B) proton(C) electron(D) neutron(E) quark
3. Which of the following photons has the highest energy?(A) x-ray(B) ultraviolet light(C) green light(D) microwave(E) radio wave
4. The threshold frequency of zinc for the photoelectric effect is in the ultraviolet range. Which of the following will occur if x-rays are shined on a zinc metal surface?(A) No electrons will be emitted from
the metal.(B) Electrons will be released from the
metal but have no kinetic energy.(C) Electrons will be released from the
metal and have kinetic energy.(D) Electrons will be released from the
metal but then will immediately be recaptured by the zinc atoms.
(E) Electrons will simply move from one zinc atom in the metal to another zinc atom in the metal.
5. A metal surface has a threshold frequency for the photoelectric effect which corresponds to green light. If blue light is shined on this metal, (A) no electrons will be emitted from the
metal.8
(B) the number of emitted electrons is proportional to the brightness (intensity) of the blue light.
(C) the electrons will have no kinetic energy.
(D) more electrons will be emitted than if green light were shined on the metal.
(E) electrons will be emitted from the metal, but since the light is not green, only a few electrons will be released.
6. Light is shined on a metal surface which exhibits the photoelectric effect according to the graph shown. What color(s) correspond to the threshold frequency of the metal?(A) red only(B) red and orange (C) red, orange, yellow, and green(D) blue only(E) blue, indigo, and violet
7. Which of the following is true of the momentum of a photon?
(A) It is proportional to the wavelength of the photon.
(B) It is inversely proportional to the wavelength of the photon.
(C) It is inversely proportional to the square of the wavelength of the photon.
(D) It is proportional to the mass of the photon.
(E) It is equal to the energy of the photon.
8. The Heisenberg uncertainty principle implies that(A) Electrons are too small to be studied.(B) Every photon is exactly the same
size.(C) The more you know about the
momentum of an electron, the less you can know about its position.
(D) The more you know about the energy of a photon, the less you can know about its frequency.
(E) You cannot state with accuracy the number of electrons in an atom.
9. Which of the following statements is true for the de Broglie wavelength of a moving particle?(A) It is never large enough to measure(B) It is proportional to the speed of the particle.(C) It is inversely proportional to the momentum of the particle.(D) It is equal to Planck’s constant.(E) It has no effect on the behavior of electrons.
10. When a photon transfers momentum to an electron, the wavelength of the photon(A) increases(B) decreases(C) remains the same(D) is equal to the wavelength of the electron(E) is always in the x-ray range
Free Response Question9
KE
R O Y G B V
Directions: Show all work in working the following question. The question is worth 15 points, and the suggested time for answering the question is about 15 minutes. The parts within a question may not have equal weight.
1. (15 points)
Light of a certain wavelength is shined on a photoemissive surface, ejecting electrons as shown above. The graph below show the maximum kinetic energy of each electron (x 10-20 J) vs. frequency of the incoming light (x 1014 Hz).
(a) On the graph below, draw the line that is your estimate of the best straight-line fit to the data points.
(b) Using your graph, find a value for Planck’s constant, and briefly explain how you found the value.
(c) From the graph, estimate the threshold frequency of the photoemissive surface.
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Adjustable Voltage
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(d) Photons of frequency 7.0 x 1014 Hz are shined on the metal surface. Determine the i. kinetic energy of the emitted electrons ii. speed of the emitted electrons iii. de Broglie wavelength of the emitted electrons.
ANSWERS AND EXPLANATIONS TO REVIEW QUESTIONS
Multiple Choice
1. EThe quantum is the smallest discrete value of any quantity, such as the electron for charge and the photon for light.
2. AA photon is the smallest bundle of light energy.
3. AThe x-ray has the highest frequency of the choices, and since energy is proportional to frequency, has the highest energy as well.
4. CSince the frequency of x-rays is higher than the ultraviolet threshold frequency, electrons will be emitted from the metal and have kinetic energy left over.
5. B
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After the threshold frequency is met, the number of photons (brightness) dictate how many electrons are emitted, since one photon can release one electron. Thus, a brighter light will release more electrons.
6. DThe electrons begin being released when blue light is shined on the metal, so blue has the threshold (minimum) frequency for this metal.
7. BSince the equation for the momentum of a photon is p = h/ , the momentum is inversely proportional to the wavelength of the photon, implying that a photon with a shorter wavelength has a higher momentum than one with a longer wavelength.
8. CHeisenberg’s uncertainty principle states that you have to sacrifice your knowledge of theposition of any subatomic particle to know its momentum accurately, and vice-versa.
9. CAccording to the equations for the de Broglie wavelength, the higher the momentum of the particle, the shorter its wavelength.
10. AA decrease in the photon’s momentum corresponds to an increase in the photon’s wavelength, since momentum and wavelength are inversely proportional to each other.
Free Response Question Solution
(a) 3 points
The best-fit straight line represents the average of the data points, and therefore there would be some data points above the line and some below the line. The best-fit line does not necessarily connect the first and last points, and does not necessarily pass through any particular data point. You should always use a straight-edge to draw a best-fit line that you know to be straight.
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(b) 4 pointsPlanck’s constant is equal to the slope of the graph. Let’s choose two convenient points on the line to find the slope, (4.0 x 1014 Hz, 0) and (8.0 x 1014 Hz, 20.0 x 10-20 J). Then the slope would be
This value for h is close to the actual value of h within a reasonable margin of error.
(c) 2 pointsThe threshold frequency can be found by marking the place where the graph crosses the frequency axis. From the graph, f0 = 4.5 x 1014 Hz.
(d) 6 pointsi. From the graph, the KE associated with the frequency of 7.0 x 1014 Hz is 14 x 10-20 J, or 1.4 x 10-19 J.
ii.
iii.
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PART 2: THE NATURE OF THE ATOM
PREVIEW
The atom is the smallest particle of an element that can be identified with that element. The atom consists of a nucleus surrounded by electrons which are in quantized, or discrete, energy levels. An electron can only change energy levels when it absorbs or emits energy. The energy emitted as a result of a downward energy level transition is typically in the form of a photon, the smallest particle of light, and the energy of the emitted photon is equal to the difference between the initial and final energy of the electron.
QUICK REFERENCE
Important Terms
atom the smallest particle of an element that can be identified with that element; the atom consists of protons and neutrons in the nucleus, and electrons in orbitals around the nucleus.
electron the smallest negatively charged particle; electrons orbit the nucleus of the atom
energy level amount of energy an electron has while in a particular orbit around the nucleus of an atom
excited state the energy level of an electron in an atom after it has absorbed energy
ground state the lowest energy level of an electron in an atom
ionization energy the energy needed to completely remove an electron from its orbital in an atom
line spectrum discrete lines which are emitted by a cool excited gas
principal quantum number an integer number n which determines the total energy of an atom
quantum model of the atom atomic model in which only the probability of locating an electron is known
x – rays high frequency and energy electromagnetic waves which are produced when high – energy electrons strike a metal target in an evacuated tube
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Equations and Symbols
where
E = energy of a photonc = speed of light = 3 x 108 m/sf = frequency of lightλ = wavelength of lightEf – Ei = difference between a final energy level of an electron in an atom and its initial energ
Line Spectra
The ancient Greeks were the first to document the concept of the atom. They believed that all matter is made up of tiny indivisible particles. In fact, the word atom comes from the Greek word atomos, meaning “uncuttable”. But a working model of the atom didn’t begin to take shape until J.J. Thomson’s discovery of the electron in 1897. He found that electrons are tiny negatively charged particles and that all atoms contain electrons. He also recognized that atoms are naturally neutral, containing equal amounts of positive and negative charge, although he was not correct in his theory of how the charge was arranged.
You may remember studying Thomson’s “plum-pudding” model of the atom, with electrons floating around in positive fluid. A significant improvement on this model of the atom was made by Ernest Rutherford around 1911, when he decided to shoot alpha particles (helium nuclei) at very thin gold foil to probe the inner structure of the atom. He discovered that the atom has a dense, positively charged nucleus with electrons orbiting around it.
In 1913, Niels Bohr made an important improvement to the Rutherford model of the atom. He observed that excited hydrogen gas gave off a spectrum of colors when viewed through a spectrosope. But the spectrum was not continuous, that is, the colors were bright, sharp lines which were separate from each other. It had long been known that every low pressure, excited gas emitted its own special spectrum in this way, but Bohr was the first to associate the bright-line spectra of these gases, particularly hydrogen, with a model of the atom. Section 30.2 in your textbook has excellent photographs of continuous and bright-line spectra.
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Nucleus
electron inorbit
He proposed that the electrons orbiting the nucleus of an atom do not radiate energy in the form of light while they are in a particular orbit, but only when they change orbits. Furthermore, an electron cannot orbit at just any radius around the nucleus, but only certain selected (quantized) orbits.
The Bohr Model of the Hydrogen Atom
The two postulates of the Bohr model of the atom are summarized below:
1. Electrons orbiting the nucleus of an atom can only orbit in certain quantized orbits, and no others. These orbits from the nucleus outward are designated n =1, 2, 3…, and the electron has energy in each of these orbits E1, E2, E3, and so on. The energies of electrons are typically measured in electron-volts (eV). The lowest energy (in the orbit nearest the nucleus) is called the ground state energy E1. (Fig. A)
2. Electrons can change orbits when they absorb or emit energy.(a) When an electron absorbs exactly enough energy to reach a higher energy level, it
jumps up to that level. If the energy offered to the electron is not exactly enough to raise it to a higher level, the electron will ignore the energy and let it pass. (Fig. B)
(b) When an electron is in a higher energy level, it can jump down to a lower energy level by releasing energy in the form of a photon of light. The energy of the emitted photon is exactly equal to the difference between the energy levels the electron moves between. (Fig. C)
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n = 1
n = 2
E1
E2
Fig. A Fig. B Fig. C
E1
E2
E3
photon
E1
E2
E3
photon E = E2-E1
Example 1Consider the energy level diagram for a particular atom shown below:
An electron begins in the ground state of this atom.
(a) How much energy must be absorbed by this electron to reach the 4th energy level?(b) How many possible photons can be emitted from this atom if the electron starts in the 4th energy level? Sketch the possible transitions on the diagram above using arrows to indicate a transition between levels.(c) The electron drops from E4 to E2 and emits a photon, then drops from E2 to E1 and emits a second photon. i. Calculate the frequency and wavelength of the photon emitted when the electron drops from E4 to E2. ii. Calculate the frequency and wavelength of the photon emitted when the electron drops from E2 to E1. (d) Are either, both, or neither of the photons emitted in part (c) above in the visible range? How can you tell?
Solution(a) E = E4 – E1 = 7 eV – 0 eV = 7 eV
(b) Six possible transitions
Energy above ground state
E1 = 0
E2 = 4 eV
E3 = 6 eV
E4 = 7 eV
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E1 = 0
E2 = 4 eV
E3 = 6 eV
E4 = 7 eV
(c) i. E42 = E4 – E2 = 7 eV – 4 eV = 3 eV
ii. E21 = E2 – E1 = 4 eV – 0 eV = 4 eV
(d) The range of visible wavelengths is about 4 x 10-7 m to 7 x 10-7 m. The photon emitted in the transition from E4 to E2 is in this visible range, but the photon emitted in the transition from E2 to E1 is not in this range.
REVIEW QUESTIONSFor each of the multiple choice questions below, choose the best answer.
1. An emission spectrum is produced when(A) electrons in an excited gas jump up
to a higher energy level and release photons.
(B) electrons in an excited gas jump down to a lower energy level and release photons.
(C) electrons are released from the outer orbitals of an excited gas.
(D) an unstable nucleus releases energy.(E) light is shined on a metal surface and
electrons are released.
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2. Consider the electron energy level diagram for a particular atom shown. An electron is in the ground state energy level. If a photon of energy 6 eV is given to the electron, which of the following will occur?(A) The electron will ignore the photon
since the photon’s energy does not match the energy levels.
(B) The electron will absorb the photon, jump up to the 5-eV level shown, and convert the remainder of the photon’s energy into kinetic energy, but will stay in the 5-eV energy level.
(C) The electron will absorb the photon, jump out of the atom completely, and convert the remainder of the photon’s energy into kinetic energy.
(D) The electron will absorb the photon, jump up to the 5-eV level, then back down to the 4 eV level.
(E) The electron will jump up to the 3-eV level, then immediately back down the ground state.
3. Consider the electron energy level diagram for hydrogen shown. An electron in the ground state of a hydrogen atom has an energy of- 13.6 eV. Which of the following energies is NOT a possible energyfor a photon emitted from hydrogen?(A) 1.9 eV(B) 13.6 eV(C) 0.65 eV(D) 11.1 eV(E) 10.2 eV
4. The reason why electrons can only orbit at certain circumferences is (A) some electrons are larger than others(B) the energy of electrons gets smaller as the circumference gets larger(C) electrons do not radiate energy when they are in a particular orbit(D) the atom is mostly empty space(E) a whole number of de Broglie wavelengths of the electron must fit into the orbit.
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Energy above ground state
E1 = 0
E2 = 2 eV
E3 = 3 eV
E4 = 4 eV
E5 = 5 eV
E1 = -13.6 eV
E2 = -3.4 eV
E3 = -1.5 eV
E4 = -0.85 eV
E5 = -0.54 eV
Free Response Question
Directions: Show all work in working the following question. The question is worth 10 points, and the suggested time for answering the question is about 10 minutes. The parts within a question may not have equal weight.
1. (10 points)
The energy level diagram for hydrogen is shown above. A free electron comes close enough to the hydrogen atom that it is captured and makes a transition to the third energy level of the atom. Then the electron makes a transition to the first energy level.
(a) Sketch arrows on the diagram above representing the two transitions made by the electron.(b) Calculate the wavelength of the photon emitted as the electron makes the transition to the third energy level.
While the electron is in the ground state it absorbs a 17-eV photon.
(c) Briefly describe what happens to the electron as a result of absorbing the 17-eV photon.(d) Calculate the de Broglie wavelength of the electron after absorbing the 17-eV photon.
ANSWERS AND EXPLANATIONS TO REVIEW QUESTIONS
Multiple Choice
1. BWhen electrons jump back to lower energy levels, they emit energy as photons.
2. CWhen an electron absorbs enough energy to completely escape the atom we say that the atom is ionized, and the energy remaining, in this case 1 eV, is converted to kinetic energy.
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E1 = -13.6 eV
E2 = -3.4 eV
E3 = -1.51 eV
E4 = - 0.85 eV
E = 0
3. DAn electron emits a photon of energy which corresponds exactly to the difference in two energy levels, and 11.1 eV does not correspond to any energy differences in the hydrogen atom.
4. EIf a whole number of electron wavelengths does not fit into a particular circumference, the electron wave would destructively interfere and could not exist in that orbit.
Free Response Question Solution
(a) 2 points
(b) 3 points
(c) 2 pointsIt takes 13.6 eV to release the electron from the ground state, and the remaining energy of 3.4 eV is the kinetic energy of the freed electron.
(d) 3 pointsThe speed of the ejected electron is
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E1 = -13.6 eV
E2 = -3.4 eV
E3 = -1.51 eV
E4 = - 0.85 eV
E = 0
photon
photon
PART 3: NUCLEAR PHYSICS AND RADIOACTIVITY
PREVIEW
The modern view of the atom includes electrons in energy levels around the nucleus of the atom. The nucleus contains positively-charged protons and neutral neutrons, each of which are made up of quarks. The nucleus is held together by the strong nuclear force, and the binding energy in the nucleus is a result of some of the mass of the particles (the mass defect) in the nucleus being converted into energy by the relationship E = mc2. The atomic number is the number of protons in the nucleus, and the atomic mass number is the number of nucleons (protons and neutrons) in the nucleus. Nuclear changes can take place, but the total amount of atomic mass in the process must remain constant.
QUICK REFERENCE
Important Terms
alpha particle positively charged particle consisting of two protons and two neutrons
atomic mass number (A) the number of protons and neutrons in the nucleus of an atom
atomic mass unit the unit of mass equal to 1/12 the mass of a carbon-12 nucleus; the atomic mass rounded to the nearest whole number is called the mass number
atomic number (Z) the number of protons in the nucleus of an atom
beta particle high speed electron emitted from a radioactive element when a neutron decays into a proton
binding energy the nuclear energy that binds protons and neutrons in the nucleus of the atom
element a substance made of only one kind of atom
isotope a form of an element which has a particular number of neutrons, that is, has the same atomic number but a different mass number than the other elements which occupy the same place on the periodic table
mass defect the mass equivalent of the binding energy in the nucleus of an atom by E = mc2
neutron an electrically neutral subatomic particle found in the nucleus of an atom
nuclear reaction any process in the nucleus of an atom that causes the number of
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protons and/or neutrons to changenucleons
protons or a neutronsstrong nuclear force
the force that binds protons and neutrons together in the nucleus of an atom
transmutation the changing of one element into another by a loss of gain of one or more protons
Equations and Symbols
where
ΔE = binding energy of the nucleusΔm = mass defect of the nucleusc = speed of light = 3 x 108 m/su = atomic mass unitX = element symbolA = atomic mass number (number of protons and neutrons)Z = atomic number (number of protons)
The Mass Defect of the Nucleus and Nuclear Binding Energy
The nucleus is made up of positively charged protons and neutrons, which have no charge. The proton has exactly the same charge as an electron, but is positive. The neutron is actually made up of a proton and an electron bound together to create the neutral particle. A proton is about 1800 times more massive than an electron, which makes a neutron only very slightly more massive than a proton. We say that a proton has a mass of approximately one atomic mass unit, u. The atomic number (Z) of an element is equal to the number of protons found in an atom of that element, and fundamentally is an indication of the charge on the nucleus of that element. All atoms of a given element have the same atomic number. In other words, the number of protons an atom has defines what kind of element it is. The total number of neutrons and protons in an atom is called the mass number (A) of that element. The symbol is used to show both the atomic number and the mass number of an X atom, where Z is the atomic number and A is the mass number.
Even though the number of protons must be the same for all atoms of a particular element, the number of neutrons, and thus the mass number, can be different. Atoms of the same element with different masses are known as isotopes of one another. For example, carbon-12 is a carbon atom with 6 protons and 6 neutrons, while carbon-14 is a carbon atom with 6 protons and 8 neutrons. We would write these two isotopes of carbon as and .
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The table below summarizes the basic features of protons, neutrons and electrons. Notice that we use an H to symbolize the proton, since the proton is a hydrogen nucleus.
Particle Symbol Relative mass Charge Locationproton 1 +1 nucleusneutron 1 0 nucleuselectron or e- 0 -1 electron orbitals
around the nucleus
Example 1Find the number of protons, electrons, and neutrons in a neutral atom of iron .
SolutionThis isotope of iron has an atomic number of 26 and a mass number of 56. Therefore, it will have 26 protons, 26 electrons, and 56 – 26 = 30 neutrons.
Since positive charges repel each other, one might wonder why protons stay together in the nucleus of the atom. There is a force holding the protons together which is greater than the electrostatic repulsion between them called the strong nuclear force, and it is a result of the binding energy of the nucleus.
According to Einstein’s famous equation E = mc2, mass and energy can be converted into one another. When an nucleus is assembled, each proton and neutron gives up a little of its mass to be converted into binding energy.
Example 2The nitrogen atom is composed of 7 protons and 7 neutrons, which gives a total of 14 atomic mass units (u). But if these particles are combined into a nitrogen nucleus, the resulting mass of the nitrogen nucleus is 14.003074 u. (a) Find the mass defect of the nitrogen nucleus. (b) What is the binding energy of the nitrogen nucleus?(c) What is the binding energy per nucleon?
Solution(a) The sum of the masses of the protons and neutrons is 7(1.007 825 u) = 7.054775 u 7(1.008 665 u) = 7.060655 u 14.115430 u
The mass defect is 14.115430 u- 14.003074 u 0.112356 u
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(b) The equivalence between mass in atomic mass units and energy in million electron- volts (MeV) is 1 u = 931 MeV. Then the binding energy of the nucleus is
At the end of the 19th century, there were elements discovered that continuously emitted mysterious rays. These elements were identified as being radioactive. A radioactive element spontaneously emits particles from its nucleus because the energy of the nucleus is unstable. Examples of naturally-occurring radioactive elements are uranium , radium , and carbon .
There are four types of particles that can be emitted when an element undergoes radioactive decay:
1. Alpha decay . Uranium, for example, undergoes alpha decay, meaning that it emits an alpha particle from its nucleus. An alpha particle is a helium nucleus, consisting of 2 protons and 2 neutrons. When an element emits an alpha particle, its nucleus loses 2 atomic numbers and 4 mass numbers, and thus changes into another element, called the daughter element. But what would this element be? We can write the nuclear equation for the radioactive decay of uranium as
where X is the daughter element and is the alpha particle. The atomic number on the left must equal the sum of the atomic numbers on the right, since charge and mass are conserved in this process. The same is true for the mass numbers on the left and right. So, the daughter element has an atomic number Z = 92-2 = 90 and a mass number A = 238 – 4 = 234. Uranium decays into the daughter element , thorium.
2. Beta decay. A beta particle is the name given to an electron emitted from the nucleus of a radioactive element. But what is an electron doing in the nucleus of an atom? Remember that we discussed the neutron in the nucleus of an atom as being a proton and an electron bound together. Beta decay is really just a neutron emitting an electron and becoming a proton. Thus, the daughter element resulting from beta decay is one atomic number higher than the parent nucleus, but the mass number essentially does not change. For example, carbon is a radioactive element that undergoes beta decay. The decay equation is
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We use the same symbol for a beta particle as we do for an electron. The daughter element must have an atomic number of 6 – (-1) = 7 and a mass number of 14 – 0 = 14. The daughter element is , nitrogen.
3. Gamma decay. Some radioactive elements emit a gamma ray, a very high energy electromagnetic wave which has no charge or mass, so only the energy of the nucleus changes, and neither Z nor A change.
4. Positron decay. A positron is exactly like an electron except for the fact that it is positively charged. A positron is not a proton, as their masses and other features are very different. Positron decay equations are typically not included on the AP Physics B exam.
CHAPTER REVIEW QUESTIONSFor each of the multiple choice questions below, choose the best answer.
1. The neutral element magnesiumhas (A) 12 protons, 12 electrons, and 24
neutrons.(B) 12 protons, 12 electrons, and 12
neutrons.(C) 24 protons, 24 electrons, and 12
neutrons.(D) 24 protons, 12 electrons, and 12
neutrons.(E) 12 protons, 24 electrons, and 24
neutrons.
2. All isotopes of uranium have (A) the same atomic number and the
same mass number.(B) different atomic numbers but the
same mass number.(C) different atomic numbers and
different mass numbers.(D) the same atomic number but
different mass numbers.(E) no electrons.
3. Six protons and six neutrons are brought together to form a carbon nucleus, but the mass of the carbon
nucleus is less than the sum of the masses of the individual particles that make up the nucleus. This missing mass, called the mass defect, has been(A) converted into the binding energy of
the nucleus.(B) given off in a radioactive decay
process.(C) converted into electrons.(D) converted into energy to hold the
electrons in orbit.(E) emitted as light.
4. The isotope of thorium undergoes alpha decay according to the equation
. The element X is(A)(B)(C)(D)(E)
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5. The isotope of cobalt undergoes beta decay according to the equation
. The element X is(A)(B)(C)(D)(E)
Free Response Question
Directions: Show all work in working the following question. The question is worth 10 points, and the suggested time for answering the question is about 10 minutes. The parts within a question may not have equal weight.
1. (10 points)
Particle or nucleus
Mass (u)
1.008 665 u1.007 825 u13.005738 u14.003074 u
A neutron is bound to a nitrogen nucleus, as shown in the equation above.(a) Find the mass defect and binding energy that holds a neutron to the nitrogen atom
(b) The U.S. uses about 1020 J of energy per year. How many nitrogen atoms would have to give up a neutron and release its binding energy to provide the energy needed in the U.S. for a year?
ANSWERS AND EXPLANATIONS TO CHAPTER REVIEW QUESTIONS
Multiple Choice
1. BThe atomic number 12 implies both 12 protons and 12 electrons, and the mass number 24 is the sum of protons and neutrons, giving 12 neutrons.
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2. DAll isotopes of a particular element must have the same atomic number (number of protons), since this number identifies the element, but can have a different mass number (number of neutrons).
3. AEach particle that makes up the nucleus gives up a little mass to be converted into energy by E = mc2 to bind the nucleus together.
4. BThe atomic number Z of element X is found by 90 = Z + 2, so Z = 88, and the mass number A is found by 234 = A + 4, so A = 230. The element X is radium.
5. CThe atomic number Z of element X is found by 27 = Z + (-1), so Z = 28, and the mass number A is found by 60 = A + 0, so A = 60. The element X is nickel.
Converting MeV to J:10.547299 MeV(1.6 x 10-13 J) = 1.69 x 10-12 J
Number of nitrogen atoms =
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PART 4: IONIZING RADIATION, NUCLEAR ENERGY, AND ELEMENTARY PARTICLES
PREVIEW
The nucleus of the atom can undergo changes, but the total amount of atomic mass in the process must remain constant. Nuclear reactions may be induced by bombardment of one nucleus with another, such as in the processes of fusion and fission. In general, when a nuclear change takes place, energy is released.
The content contained in sections 2, 3, 5, and 8 (Example 9) of chapter 32 of the textbook is included on the AP Physics B exam.
QUICK REFERENCE
Important Terms
chain reaction nuclear process producing more neutrons which in turn can create more nuclear processes, usually applied to fission
critical mass the minimum amount of mass of fissionable material necessary to sustain a nuclear chain reaction
elementary particles the particles (quarks and leptons) of which all matter is composed
neutron an electrically neutral subatomic particle found in the nucleus of an atom
nuclear fission the splitting of a heavy nucleus into two smaller ones
nuclear fusion the combining of two light nuclei into one larger one
nuclear reaction any process in the nucleus of an atom that causes the number of protons and/or neutrons to change
nuclear reactor device in which nuclear fission or fusion is used to generate electricity
quark one of the elementary particles of which all protons and neutrons are made
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Equations and Symbols
where
X = element symbolA = atomic mass number (number of protons and neutrons)Z = atomic number (number of protons)
DISCUSSION OF SELECTED SECTIONS
Nuclear Fission
Fission is a process in which a large nucleus splits in to smaller nuclei. Fission is usually caused artificially by shooting a slow neutron at a large atom such as uranium which absorbs the neutron and splits into two smaller atoms, along with the release of more neutrons and some energy.
For example, a fission reaction occurs when uranium absorbs a slow neutron and then splits into xenon and strontium, releasing three neutrons and some energy. The equation for this fission reaction is
Once again, the sum of the atomic and mass numbers on the left equal the sum of the atomic and mass numbers on the right. The three neutrons which are produced in this reaction can be used to split three more uranium atoms, which produce three more neutrons in each reaction, each of which can split three more uranium atoms, and so on. This is called a chain reaction, and is used to sustain the release of energy in fission reactions. However, before a chain reaction can be sustained, there must be a minimumamount of fissionable material, such as uranium or plutonium, present. The minimum amount of fissionable material that must be present to sustain a chain reaction is called the critical mass.
Nuclear Fusion
Fusion occurs when small nuclei combine into larger nuclei and energy is released. Many stars, including our sun, power themselves by fusing four hydrogen nuclei to make one helium nucleus. The fusion process is clean and a large amount of energy is released,
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which is why researchers here on earth are trying to find ways to use fusion as an alternative energy source.
For example, the element tritium is combined with , a hydrogen isotope called deuterium, to form helium and a neutron, along with the release of energy. The equation for this fusion reaction is
Note that the sum of the atomic numbers on the left must equal the sum of the atomic numbers on the right. The same is true of the mass numbers.
REVIEW QUESTIONSFor each of the multiple choice questions below, choose the best answer.
1. Consider the following nuclear fission equation: . The element which belongs in the blank is (A)(B)(C)(D)(E)
2. Consider the following nuclear fusion equation: . The element which belongs in the blank is (A) (B) (C) (D) (E)
3. Nitrogen is bombarded with an alpha particle, producing a nucleus of and (A) a neutron(B) an electron(C) a beta particle(D) a proton(E) an alpha particle
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