Quantum Physics

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)