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eng 45 homework1_sol - 2.10 The orbital electrons of an...

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Unformatted text preview: 2.10. The orbital electrons of an atom can be ejected by exposure to a beam of electromagnetic radiation. Specifically, an electron can be ejected by a photon with energy greater than or equal to the electron’s binding energy. Given that the photon energy (E) is equal to hc/A, where h is Planck’s constant. c the speed of light, and A the wavelength, calculate the maximum wavelength of radiation (correspond-' ing to the minimum energy) necessary to eject a Is electron from a C12 atom. (See Figure 2—3.) 5"“ Fgww 4—3, /E/= 2:3.9¢M = (0.“ .74 x10" 3J5X0.247Ix10 RA) N (.783. 9c V)(.t 1/6. 242nm": V) = 4.37x/0' M x Jun/Ia'a.‘ =’ 4-37HM ”aft: [0! m1 741 ma’réfl 0/7/11 1.42761» Lin/.27 M £4,461 {ll ar‘fi‘ «24% . '74 76 WA! i%" fix: a P“ a 2g far/rib! Ln . 2.12 The mechanism for producing a photon of specific en- ergy is outlined in Problem 2.11. The magnitude of photon energy increases with the atomic number of the atom from which emission occurs. (This is due to the stronger binding forces between the negative electrons and the positive nucleus as the numbers of protons and electrons increase with atomic number.) As noted in Problem 2.10. E = Inc/1. which means that a higher- energy photon will have a shorter wavelength. Verify that higher atomic number materials will emit higher-energy. shorter-wavelength photons by calculating E and A. for emission from iron (atomic number 26 compared to 6 for carbon). given that the energy levels for the first two elecn'on orbitals in iron are at —7.112eV and —708eV. /4£/= [-7112 —(—7o:)/ev .—. 4404” or = 5—f- ; (0.601bx10'331- S)(O- “WM 7W5 45 (4404: v)(1.T/6..74.1 no” e V) = £74110",th bun/[0-1M = O./94- him 2.15 So far. we have concentrated on the coulombic force of attraction between ions. But like ions repel euh other. A nearest-neighbor pair of Na“ ions in Figure 2-5 are separated by a dismnce of Jim. where no is defined in Figure 2-7. Calculate the coulombic force of repulsion between such a pair of like ions. [El 5:- - way, > a _._ __ (fix/0’ v... Ago/Mm x/od’c)(+/)(a./6 x/o"’c) (2)(o.27a x/a‘9~)2 = —/.47x/o-’/v ll 2.16 Calculgte the coulombic force of attraction between Ca2+ and O" in C30. which has the N aCl-type structure. Fm flfflnakfpl ”a”: 0.106 m. ; rga'50-l32hm Than, 4 = roar +r’a,_ = o./o£.-+a./3.?..~= 0.235,. I ‘ (0.23: urn-9...)-1 = law/041w ...
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