ANSWERS TO QUESTIONS 1.

The energy of an object or a system is ‘quantized’ if it can take on only a certain set of prescribed, discrete values instead of ones which are continuously distributed. In a quantized system, energy is available only in individual chunks having certain sizes, not in any arbitrary amount or quantity. 2. Oil, tires, antifreeze, transmission fluid, and windshield washer fluid are normally sold quantized. Gasoline, diesel fuel, and air for tires are usually not quantized. 3. Planck was able to account for the observed spectra of blackbodies by assuming that the allowed energies of an oscillating atom in a blackbody were quantized—that is, that these atoms could have only certain fixed values of energy, ranging from 0 on up, in steps of some fundamental quantum of energy, E. (See page 391.) 4. The steering wheel controls the orientation of the car's front wheels in a continuous fashion. Except in old cars, the station selector on the radio uses a quantized process, meaning the frequency can only be certain discrete values, like 91.3 MHz, 91.5 MHz, 91.7 MHz, etc. 5. A photon is a particle or quantum of electromagnetic radiation. The energy of a photon is directly proportional to its frequency and inversely proportional to its wavelength: hc Ephoton = hf = λ The proportionality constant, h, is Planck’s constant. (See page 391.) 6. See Section 10.2 beginning on page 392. 7. The quantum of energy for atomic oscillators and photons would be larger in proportion to the size of Planck’s constant, as would the de Broglie wavelength of particles and the allowed values of angular momentum of an electron in an atom. Assuming nothing else changed, solar cells would put out more energy for the same number of photons absorbed, the Bohr orbits would be larger, the electron energies in atomic energy levels would have smaller magnitudes so the frequencies of light in absorption and emission spectra would be lower, the frequencies of lasers would be shifted, the diffraction of particles would be greater for the same speed, and so on. 8. The electron absorbing the ultraviolet photon will be ejected with a higher energy than the one absorbing the violet photon. 9. It makes sense that electromagnetic waves can give energy to electrons—the electrons are charged, and the electric field in the EM wave will therefore accelerate them and give them energy. This would be expected even without photons. But the fact that even very dim light of high enough frequency can eject some electrons immediately from a surface implies that the electron absorbed energy very quickly, as if it were concentrated in a packet that the electron absorbed all at once—a photon. Also, the observation that EM waves of too low a frequency won’t eject electrons at all shows that the electron has absorbs energy in sort of an “all or nothing” fashion, i.e., it doesn’t absorb energy gradually and then leave the surface, which one would expect classically. 10. Use lenses that produce an enlarged or reduced image on the photoconductive surface.

11. Our bodies are simply too massive to be noticeably affected by the tiny amount of momentum the photons deliver. 12. An emission line spectrum shows emission only at a few particular frequencies, with no emission at frequencies in between. A continuous spectrum has emission over all the frequencies in a frequency range. 13. The emission spectra of both neon and hydrogen would be seen. See Figure 10.12 on page 397 (Transparency 160). 14. The basic assumptions of the Bohr model of the atom are: (i) Electrons move about the nucleus of an atom in well-defined circular orbits much like the orbits of planets about the sun. (ii) The electron orbits are quantized—each orbit has a well-defined energy and angular momentum associated with it, such that the larger the orbit, the greater the energy. The  = h 2 π ). angular momentum of the electron must be an integer multiple of  . (  (iii) Electrons may make transitions between the allowed orbits. Transitions to larger orbits require that the electrons gain energy, while transitions to smaller orbits require that they lose energy. The Bohr model accounts for the observed emission line spectra of the elements because when making ‘jumps’ to smaller orbits, the atom may give up energy in the form of photons whose energies are equal to the differences in the energies of the orbits between which the ‘jumps’ occur. Since the photon wavelengths are inversely proportional to their energies, the larger the energy differences of the orbits involved in the transitions, the smaller the wavelengths of the photons released. Thus, the spectrum of emitted photons depends on the number and energies of the allowed electron ‘jumps’ in the atom and, hence, on the structure of the atomic orbits. Since each atom has its own individual set of atomic orbits, each will have its own individual set of emission lines. 15. Ionization is a process whereby an atom acquires enough energy so that one or more of its electrons become unbound or free. The freed electron(s) generally move(s) off with some kinetic energy, leaving behind an ion, that is, an atom with a net positive charge. Ionization may occur as a result of a collision between the atom and another particle, during which the energy needed to overcome the natural attraction between the electron(s) and the nucleus is transferred to the atom by the colliding particle, or as a result of the absorption of a photon by the atom. In the latter instance, the energy of the photon must be larger than the difference between the energy associated with the electron’s orbit and that of the largest allowed orbit in the atoms. 16. The red line in the helium spectrum is closer to orange than the red line in the hydrogen spectrum indicating that it has a higher frequency. Therefore the photons of red light emitted by helium have higher energy than those emitted by hydrogen. 17. Because of the Doppler effect, the frequencies would be shifted to lower values. 18. The light from the star passes through the earth’s atmosphere in reaching the earth’s surface, so the ground-based telescope will see the effects of selective absorption by the atmosphere, whereas the HST will not.

19. The photon’s energy is reduced, lowering its frequency and increasing its wavelength. 20. De Broglie proposed that particles also have wavelike properties. The de Broglie wavelength is the wavelength of the wave associated with any moving particle. Since the de Broglie wavelength is inversely proportional to the momentum of a particle, the wavelength of an electron decreases when its speed is increased. 21. Because the proton’s mass is larger than the electron’s mass it will have a larger momentum. Consequently the electron’s de Broglie wavelength is longer. 22. Electrons in an electron microscope behave like waves, and therefore act analogously to light in an optical microscope. They have a much shorter wavelength, though, and can do a much better job of showing details that are beyond the diffraction limitations of light microscopes. 23. Since particles possess wavelike properties, it is not possible to specify both the exact position and momentum of a particle. The uncertainty principle states that the uncertainty in the position of a particle multiplied by the uncertainty in its momentum is at best equal to Planck’s constant. (Actually, the “best” value is really Planck’s constant divided by 4π.) 24. The uncertainty principle cannot accommodate the Bohr model’s well-defined circular paths for the electrons in an atom. The electrons are actually spread out in space and act like an “electron cloud”. (See page 407) 25. The electron is in the n = 1 energy level, the lowest one. 26. A hydrogen atom with its electron in the ground state will only absorb a photon whose energy equals the difference between the energy of the ground state (n=1 ) and that of the second excited state (n=3) if it is to make a transition to the n = 3 state. It can do nothing else because the permitted electron energy levels are quantized, and electrons in those levels can have only certain, specific amounts of energy. Photons having slightly more or slightly less energy than the amount required for the transitions will be “ignored” because they would produce electron jumps to impermissible, non-existent energy levels. 27. Since all the electrons have energies higher than the ionization level and these energies are not quantized, the spectrum should be a continuous spectrum. 28. Ionizing a hydrogen atom from the ground state requires a photon whose energy is larger than that of a photon capable of producing ionization from the first excited state. The energy-level diagram for hydrogen (Figure 10.30 on page 408) shows that a photon of energy 13.6 eV is required to ionize a hydrogen atom from the ground state, but a photon with energy 3.4 eV will ionize the atom from the n=2 state. 29.

Energy 3 mgd 2 mgd mgd

d = height of each step

0

30. The Lyman emission series of hydrogen lies in the ultraviolet part of the electromagnetic spectrum—the Balmer series falls in the visible part, and the Paschen series is in the infrared region. Each series consists of the set of lines produced by all possible downward oneelectron transitions between allowed upper states and the ground state, (n=1), first excited state (n=2), and second excited state (n=3), respectively. (See Figure 10.32, Transparency #168). 31. Sr and Ca are both in the same group (IIA) in the periodic table and therefore have similar ground state electron configurations, leading to their similar chemical behavior. 32. The x-ray spectrum of heavy elements typically consists of a characteristic spectrum of discrete x-ray lines or peaks superimposed on a continuous bremsstrahlung spectrum. The characteristic spectra are produced when electrons in higher allowed energy levels cascade downward to fill vacancies in lower levels left when electrons are ejected from target metal atoms by an incoming stream of high speed bombarding electrons. The bremsstrahlung spectrum represents radiation which is emitted continuously by the bombarding electrons as they are gradually decelerated (“braked”) upon passing into the metallic targets.

33. The frequency of Kα radiation in the x-ray spectrum of Cu will be lower than that of Kα radiation for W because the atomic number of Cu (Z=29) is smaller than the atomic number of W (Z=74). According to Moseley’s law (See Figure 10.36) the frequency of the emitted characteristic radiation is proportional to Z . As Z increases, so too does the frequency. 34. Characteristic x-ray spectra result from transitions between allowed atomic energy levels in the same way that optical emission line spectra, as described using the Bohr model, do. The only difference is in the energies of the photon released when the electrons jump from a higher energy state to a lower one. In the case of the x-ray line spectra of heavy elements, the photon energies are large (the wavelengths are small) because the energy differences between the states involved in the jumps are large—this is so because these states are generally low-lying ones which are tightly bound to the nucleus by its large positive charge. Optical spectra in heavy elements generally involve other electronic states where the electrons are more loosely bound and the energy level differences are smaller—these yield lower energy (longer wavelength) photons during transitions. In light atoms like hydrogen with only a single positive charge in the nucleus, even the innermost electrons are not strongly enough bound to produce x-ray photons when transitions involving these levels occur. 35. Laser is an acronym for the phrase ‘light amplification by stimulated emission of radiation.’ 36. Laser light is coherent and monochromatic and usually of much higher intensity than the light from the light bulb. 37. A metastable state is a state in which an electron, before spontaneously jumping to a lower state, remains for a time which is long compared to that typically associated with its decay from most other states in the atom (a few nanoseconds). The existence of metastable states or levels is required for the operation of a laser because they are instrumental in producing a population inversion within the laser system. (See Question 38.) 38. A population inversion is a circumstance within a lasing system in which there are more atoms with their electrons in an upper excited state than in a specific preferred lower state. This condition must be fulfilled in a successfully functioning laser in order for there to be more photons released in downward transitions induced in the excited atoms by the stimulating radiation than are absorbed from the stimulating radiation in upward transitions by unexcited atoms. Otherwise there will not be a net output of photons having the characteristics of the laser light. 39. Getting UV laser light out using yellow light to pump would imply getting more energy out that you put in. The yellow pump photons should not have enough energy to drive the lasing atoms into a state from which they could decay and emit the larger energy UV photons. Extra energy would have to be coming from somewhere. If the yellow light were pumping the atoms from an already excited state instead of the ground state, then a transition from a higher excited state all the way down to the ground state could generate ultraviolet laser light, but there would have to be some other mechanism to get the atoms into the original excited state to begin with. 40. (See Figure 10.40 and related discussion in Section 10.8.) 41. The Solvay Conferences are international meetings held from time to time to discuss important topics in physics and physical chemistry. They are organized and supported by the Solvay Foundation which is funded by monies provided through the legacy of the late

Belgian industrialist Ernest Solvay. These meetings permitted the free and vigorous exchange of ideas between the most renowned physicists in the world between 1911 and 1933 and led to a deepening and a refining of our understanding of the world of microscopic physics through the development and nurturing of the subject of quantum mechanics. ANSWERS TO EVEN NUMBERED PROBLEMS 2. 4. 6. 8. 10. 12. 14. 16. 18. 20.

E = 0.125 eV f = 1.24 × 1020 Hz λ = 4.1 × 10–12 m λ = 5.0 × 10–15 m a) E = 12.1 eV b) E = 12.1 eV Four levels are required. Two electrons are in the n =1 level, 8 in the n = 2 level, 18 in the n = 3 level, and 2 in the n = 4 level. 1.108 E = 17,200 eV = 17.2 keV a) f = 2.83 × 1013 Hz b) infrared a) iii) Indium b) ii) Calcium

ANSWERS TO CHALLENGES 1.

2.

Some of the electrons in metals are not tightly bound to individual atoms. This is why metals are good conductors of electricity—these loosely bound electrons flow readily. Also these electrons only have to gain relatively little energy to break free from the metal itself. That is why metals exhibit the photoelectric effect so easily. The speed of light in air increases as it is heated because heating causes air to expand and consequently lowers its density. Light can travel faster in thinner air. The air along the center of the path of an intense laser beam will be hotter than the air near the edges of the beam. An individual ray that, for whatever reason, strays from the center of the beam passes from hotter air into cooler air. In so doing it goes from a region of higher speed of light to a region of lower speed. From Section 9.3 we know that this causes the ray to bend toward the normal to the boundary between the two regions, which means that it diverges even more rapidly away from the center of the beam. The net result is that the beam spreads out.

3.

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6.

Even though we can think of the price of something as having a continuous range of possible values, including numbers like 54.678593 cents and 6778.09846657 dollars, in reality prices and money are quantized. Money comes in units of cents. The cent is the quantum of monetary value. The muon orbits would follow Bohr’s condition that the orbital angular momentum, mvr, must equal an integer multiple of 2hπ . Since the muon’s mass is larger than the electron’s, the product vr must be smaller. Since the muon has the same charge as the electron, the force acting on the muon for a given radius is the same as that on an electron. From Chapter 2 we know that in circular orbits v is higher when r is smaller. Equating the centripetal force and the Coulomb force: 9 × 10 9 q2 mv 2 =F= r r2 mv 2 r = 9 × 109 q 2 , a constant. For each orbit, mvr is a constant and mv2r is a constant, and these constants are the same for the muonic atom and the ordinary atom. So we have mev ere = mµ vµ rµ and 2 mev e re = mµ vµ 2 rµ This means that the speed v of the muon must be the same as that of the electron. Consequently the radius r of each Bohr orbit is smaller by a factor of about 200. In each orbit both the kinetic energy and the potential energy of the muon will be about 200 times larger than the KE and PE of an electron. Consequently the energy levels will be about 200 times farther apart and the energies of all of the photons that can be emitted during energy level transitions will be about 200 times larger. The emission spectrum of the muonic atom is shifted to higher frequencies by this same factor compared to the emission spectrum of ordinary hydrogen. The aurora borealis consists of the emission spectra of excited atoms, molecules, and ions in the upper atmosphere. Consequently it exhibits a line spectrum. One of the important energy level transitions starts from an excited metastable state with a lifetime of about 200 seconds. Consequently, many photons are emitted long after the solar wind particles which caused the excitation have ceased to be available in large numbers. (For more on the mechanism of the aurora, see: Akasofu, Syun-Ichi “The Dynamic Aurora,” Scientific American , May 1989.) The wavelength of an “average” photon emitted by a filament at 3000K is given by 0.0029 0.0029 −7 = = 9.67⋅10 m . λ max = 3000K T 3⋅10 8 m / s c = 3.10 ⋅1014 Hz . This makes the = f = The frequency of such a photon is −7 λ 9.67 ⋅10 m −34 14 −19 photon’s energy E = hf = 6.63 ⋅10 J ⋅ s ⋅3.10 ⋅10 Hz = 2.06 ⋅10 J . We can express the power in terms of a photon rate: J 1 photon 20 P = 100W = 100 ⋅ = 4.86 ⋅10 photons/ s . −19 s 2.06 ⋅10 J

7.

n = 100 orbit circumference = 2π (100) ⋅ 0.0529nm = 3323.8nm . ∆E 1  13.6eV 13.6eV  9 = 6.48⋅10 Hz − f = =  2 2 101  h h 100 1 −10 orbital period = = 1.54 ⋅10 s f circumference 3323.8 ⋅10 −9 m = 21500 m / s , far less than the speed of = orbital speed = −10 1.54⋅10 s period light. 2