Answers to Quick Quizzes, Odd-Numbered Conceptual Questions and Problems A.29light will generally increase the size of the current, butwill not change the energy of the individual electronsthat are ejected. Thus, the stopping potential remainsconstant.11. Wave theory predicts that the photoelectric effect shouldoccur at any frequency, provided that the light intensity ishigh enough. However, as seen in photoelectric experiments,the light must have sufficiently high frequency forthe effect to occur.13. (a) Electrons are emitted only if the photon frequency isgreater than the cutoff frequency.15. No. Suppose that the incident light frequency at whichyou first observed the photoelectric effect is above the cutofffrequency of the first metal, but less than the cutofffrequency of the second metal. In that case, the photoelectriceffect would not be observed at all in the secondmetal.17. The frequency of the scattered photon must decrease,because some of its energy is transferred to the electron.PROBLEMS1. (a) 3 000 K (b) 20 000 K3. 500 nm5. (a) 2.49 10 5 eV (b) 2.49 eV (c) 249 eV7. 2.27 10 30 photons/s9. (a) 2.3 10 31 (b) E/E 4.3 10 3211. (a) 2.24 eV (b) 555 nm (c) 5.41 10 14 Hz13. 234 nm15. 148 days, incompatible with observation17. 4.8 10 14 Hz, 2.0 eV19. 1.2 10 2 V and 1.2 10 7 V, respectively21. 41.4 kV23. 0.078 nm25. 0.281 nm27. 1.78 eV, 9.47 10 28 kg m/s29. 70°31. 1.18 10 23 kg m/s, 478 eV33. (a) 1.2 eV (b) 6.5 10 5 m/s35. (a) 1.46 km/s (b) 7.28 10 11 m37. (a) 10 2 MeV (b) No. With kinetic energy much largerthan the magnitude of the negative potential energy, theelectron would immediately escape.39. 3.58 10 13 m41. (a) 15 keV (b) 1.2 10 2 keV43. 10 6 m/s45. 116 m/s47. 5 200 K; clearly, a firefly is not at that temperature, sothis cannot be blackbody radiation.49. 18.2°51. 1.36 eV53. 2.00 eV55. (a) 0.022 0c (b) 0.999 2c57. (b) 3.72 km/s59. (b) 5.19 10 16 m61. (a) 0.263 kg (b) 1.81 W(c) 0.015 3°C/s 0.919°C/min (d) 9.89 m(e) 2.01 10 20 J (f) 8.98 10 19 photon/sChapter 28QUICK QUIZZES1. (b)2. (a)3. (a) 5 (b) 9 (c) 254. (d)CONCEPTUAL QUESTIONS1. If the energy of the hydrogen atom were proportional to n(or any power of n), then the energy would become infiniteas n grew to infinity. But the energy of the atom is inverselyproportional to n 2 . Thus, as n grows to infinity, the energyof the atom approaches a value that is above the groundstate by a finite amount, namely, the ionization energy13.6 eV. As the electron falls from one bound state to another,its energy loss is always less than the ionization energy.The energy and frequency of any emitted photon are finite.3. The characteristic x-rays originate from transitions withinthe atoms of the target, such as an electron from the Lshell making a transition to a vacancy in the K shell. The vacancyis caused when an accelerated electron in the x-raytube supplies energy to the K shell electron to eject it fromthe atom. If the energy of the bombarding electrons wereto be increased, the K shell electron will be ejected fromthe atom with more remaining kinetic energy. But the energydifference between the K and L shell has not changed,so the emitted x-ray has exactly the same wavelength.5. A continuous spectrum without characteristic x-rays is possible.At a low accelerating potential difference for theelectron, the electron may not have enough energy toeject an electron from a target atom. As a result, there willbe no characteristic x-rays. The change in speed of theelectron as it enters the target will result in the continuousspectrum.7. The hologram is an interference pattern between lightscattered from the object and the reference beam. If anythingmoves by a distance comparable to the wavelengthof the light (or more), the pattern will wash out. Theeffect is just like making the slits vibrate in Young’s experiment,to make the interference fringes vibrate wildly sothat a photograph of the screen displays only the averageintensity everywhere.9. If the Pauli exclusion principle were not valid, theelements and their chemical behavior would be grosslydifferent, because every electron would end up in the lowestenergy level of the atom. All matter would thereforebe nearly alike in its chemistry and composition, since theshell structures of each element would be identical. Mostmaterials would have a much higher density, and the spectraof atoms and molecules would be very simple, resultingin the existence of less color in the world.11. The three elements have similar electronic configurations,with filled inner shells plus a single electron in an sorbital. Since atoms typically interact through their unfilledouter shells, and since the outer shells of theseatoms are similar, the chemical interactions of the threeatoms are also similar.13. Each of the eight electrons must have at least one quantumnumber different from each of the others. They candiffer (in m s ) by being spin-up or spin-down. They can differ(in ) in angular momentum and in the general shapeof the wave function. Those electrons with 1 can differ(in m ) in orientation of angular momentum.15. Stimulated emission is the reason laser light is coherentand tends to travel in a well-defined parallel beam. Whena photon passing by an excited atom stimulates that atomto emit a photon, the emitted photon is in phase with the
A.30 Answers to Quick Quizzes, Odd-Numbered Conceptual Questions and ProblemsCONCEPTUAL QUESTIONS1. Isotopes of a given element correspond to nuclei with differentnumbers of neutrons. This will result in a variety ofdifferent physical properties for the nuclei, including theobvious one of mass. The chemical behavior, however, isgoverned by the element’s electrons. All isotopes of agiven element have the same number of electrons and,therefore, the same chemical behavior.3. An alpha particle contains two protons and two neutrons.Because a hydrogen nucleus contains only one proton, itcannot emit an alpha particle.5. In alpha decay, there are only two final particles: thealpha particle and the daughter nucleus. There are alsotwo conservation principles: of energy and of momentum.As a result, the alpha particle must be ejected with a discreteenergy to satisfy both conservation principles. However,beta decay is a three-particle decay: the beta particle,the neutrino (or antineutron), and the daughter nucleus.As a result, the energy and momentum can be shared in avariety of ways among the three particles while still satisfyingthe two conservation principles. This allows a continuousrange of energies for the beta particle.7. The larger rest energy of the neutron means that a freeproton in space will not spontaneously decay into a neutronand a positron. When the proton is in the nucleus,however, the important question is that of the total restenergy of the nucleus. If it is energetically favorable for thenucleus to have one less proton and one more neutron,then the decay process will occur to achieve this lower energy.9. Carbon dating cannot generally be used to estimate theage of a stone, because the stone was not alive to take upcarbon from the environment. Only the ages of artifactsthat were once alive can be estimated with carbon dating.11. The protons, although held together by the nuclear force,are repelled by the electrostatic force. If enoughprotons were placed together in a nucleus, the electrostaticforce would overcome the nuclear force, which isbased on the number of particles, and cause the nucleusto fission.The addition of neutrons prevents such fission. Theneutron does not increase the electrical force, being electricallyneutral, but does contribute to the nuclear force.13. The photon and the neutrino are similar in that both particleshave zero charge and very little mass. (The photonhas zero mass, but recent evidence suggests that certainkinds of neutrinos have a very small mass.) Both musttravel at the speed of light and are capable of transferringboth energy and momentum. They differ in that the phooriginalphoton and travels in the same direction. As thisprocess is repeated many times, an intense, parallel beamof coherent light is produced. Without stimulated emission,the excited atoms would return to the ground stateby emitting photons at random times and in random directions.The resulting light would not have the usefulproperties of laser light.17. The atom is a bound system. The atomic electron doesnot have enough kinetic energy to escape from its electricalattraction to the nucleus. The electrical potential energyof the atom is negative and is greater than the kineticenergy, so the total energy of the atom is negative.19. (a) The wavelength of photon A is greater. (b) Theenergy of photon B is greater.PROBLEMS1. 656 nm, 486 nm, and 434 nm3. (a) 2.3 10 8 N (b) 14 eV5. (a) 1.6 10 6 m/s (b) No, v/c 5.3 10 3 1(c) 0.46 nm (d) Yes. The wavelength is roughly the samesize as the atom.7. (a) 0.212 nm (b) 9.95 10 25 kg m/s(c) 2.11 10 34 J s(d) 3.40 eV (e) 6.80 eV (f ) 3.40 eV11. E 1.51 eV (n 3) to E 3.40 eV (n 2)13. (a) 0.967 eV (b) 0.266 eV15. (a) 122 nm, 91.1 nm (b) 1.87 10 3 nm, 820 nm17. 97.2 nm19. (a) 488 nm (b) 0.814 m/s21. (d) n 2.53 10 74 (e) No. At such large quantumnumbers, the allowed energies are essentiallycontinuous.23. (a) 2.47 10 14 Hz, f orb 8.23 10 14 Hz(b) 6.59 10 3 Hz, f orb 6.59 10 3 Hz. For large n,classical theory and quantum theory approach each otherin their results.25. 4.42 10 4 m/s27. (a) 122 eV (b) 1.76 10 11 m29. (a) 0.026 5 nm (b) 0.017 6 nm (c) 0.013 2 nm31. 1.33 nm33. n 3, 1, m 1, m s 1/2; n 3, 1, m 0,m s 1/2; n 3, 1, m 1, m s 1/235. Fifteen possible states, as summarized in the followingtable:n 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2m 2 2 2 1 1 1 0 0 0 1 1 1 2 2 2m s 1 0 1 1 0 1 1 0 1 1 0 1 1 0 137. (a) 30 possible states (b) 3639. (a) n 4 and 2 (b) m (0, 1, 2), m s 1/2(c) 1s 2 2s 2 2p 6 3s 2 3p 6 3d 10 4s 2 4p 6 4d 2 5s 2 [Kr] 4d 2 5s 241. 0.160 nm43. L shell: 11.7 keV; M shell: 10.0 keV; N shell: 2.30 keV45. (a) 10.2 eV (b) 7.88 10 4 K47. (a) 8.18 eV, 2.04 eV, 0.904 eV, 0.510 eV, 0.325 eV(b) 1.09 10 3 nm and 609 nm49. The four lowest energies are 10.39 eV, 5.502 eV, 3.687 eV, and 2.567 eV (b) The wavelengths of theemission lines are 158.5 nm, 185.0 nm, 253.7 nm,422.5 nm, 683.2 nm, and 1 107 nm(c) 1.31 10 6 m/s51. (a) 4.24 10 15 W/m 2 (b) 1.20 10 12 J55. (a) E n ( 1.49 10 4 eV)/n 2 (b) n 4 : n 157. (a) 9.03 10 22 m/s 2 (b) 4.63 10 8 W(c) 10 11 sChapter 29QUICK QUIZZES1. (c)2. (c)3. (a)4. (a) and (b)5. (b)
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Summary 965Photo Researchers, Inc./
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