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423 Chapter 28 Atomic Physics Answers to Even Numbered Conceptual Questions 6. Classically, the electron can occupy any energy state. That is, all energies would be allowed. Therefore, if the electron obeyed classical mechanics, its spectrum, which originates from transitions between states, would be continuous rather than discrete. 8. The de Broglie wavelength of macroscopic objects such as a baseball moving with a typical speed such as 30 m/s is very small and impossible to measure. That is, hmvλ= , is a very small number for macroscopic objects. We are not able to observe diffraction effects because the wavelength is much smaller than any aperture through which the object could pass. 10. In both cases the answer is yes. Recall that the ionization energy of hydrogen is 13.6 eV. The electron can absorb a photon of energy less than 13.6 eV by making a transition to some intermediate state such as one with 2n=. It can also absorb a photon of energy greater than 13.6 eV, but in doing so, the electron would be separated from the proton and have some residual kinetic energy. Problem Solutions 28.1 The Balmer equation is H221112Rnλ⎛⎞=−⎜⎟⎝⎠, or 22H44nRnλ⎛⎞=⎜⎟−⎝⎠ When 3n =, 771496.56 10 m656 nm1.097 37 10 m94λ−−⎛⎞==×=⎜⎟×−⎝⎠ When 4n=, 7714164.86 10 m486 nm1.097 37 10 m164λ−−⎛⎞==×=⎜⎟×−⎝⎠ When 5n =, 7714254.34 10 m434 nm1.097 37 10 m254λ−−⎛⎞==×=⎜⎟×−⎝⎠424 CHAPTER 28 28.3 (a) From Coulomb’s law, ()()()2922 1912822108.99 10 N m C 1.60 10 C=2.310 N1.010 mekqqFr−−−×⋅ ×==×× (b) The electrical potential energy is ()()()922 19 19121018-198.99 10 N m C 1.60 10 C1.60 10 C1.010 m1 eV2.310 J14 eV1.60 10 JekqqPEr−−−−×⋅ −× ×==×⎛⎞=− × = −⎜⎟×⎝⎠ 28.7 (a) 20nrna= yields ()240.052 9 nm 0.212 nmr == (b) With the electrical force supplying the centripetal acceleration, 222en ennmv kerr=, giving 2enenkevmr= and 2eenennmkepmvr== Thus, ()()()231 9 2 2 192292259.11 10 kg 8.99 10 N m C 1.610 C0.212 10 m9.95 10 kg m seemkepr−−−−××⋅×==×=× ⋅ (c) 343426.63 10 Js 22.11 10 Js22nhLn Lππ−−⎛⎞ ⎛ ⎞×⋅=→= =×⋅⎜⎟ ⎜ ⎟⎝⎠ ⎝ ⎠ (d) ()()22522 19222319.95 10 kg m s15.44 10 J3.40 eV2229.11 10 kgeepKE m vm−−−×⋅=== =×=× (e) ()()()()2922 19292188.99 10 N m C 1.60 10 C0.212 10 m1.09 10 J6.80 eVekeePEr−−−×⋅ ×−==−×=− × = − (f) 2223.40 eV 6.80 eV = 3.40 eVEKEPE=+= − −Atomic Physics 425 28.10 (b) From H22111fiRnnλ⎛⎞=−⎜⎟⎜⎟⎝⎠ or 2222H1ififnnRnnλ⎛⎞=⎜⎟⎜⎟−⎝⎠ with 6 and 2ifnn== ()()77136 414.10 10 m410 nm1.097 37 10 m364λ−−⎡⎤==×=⎢⎥×−⎣⎦ (a) ()()34 81996.63 10 Js 3.00 10 m s4.85 10 J3.03 eV410 10 mhcEλ−−−×⋅ ×== = × =× (c) 81493.00 10 m s7.32 10 Hz410 10 mcfλ−×== = ×× 28.15 From H22111fiRnnλ⎛⎞=−⎜⎟⎜⎟⎝⎠, it is seen that (for a fixed value of fn) maxλ occurs when 1ifnn=+ and minλ occurs when in →∞. (a) For the Lyman series ()1fn=, ()71 7max22max1111.097 37 10 m 1.22 10 m122 nm12λλ−−⎛⎞=× −→=×=⎜⎟⎝⎠ and ()71 8min2min1111.097 37 10 m 9.11 10 m91.1 nm1λλ−−⎛⎞=× −→=×=⎜⎟⎝⎠∞ (b) For the Paschen series ()3fn=, ()71 6 3max22max1111.097 37 10 m1.87 10 m1.87 10 nm34λλ−−⎛⎞=× −→=×=×⎜⎟⎝⎠ and ()71 7min2min1111.097 37 10 m 8.20 10 m 820 nm3λλ−−⎛⎞=× −→=×=⎜⎟⎝⎠∞426 CHAPTER 28 28.17 The batch of excited atoms must make these six transitions to get back to the ground state: 21ifnn=→ =, also 32ifnn=→= and 31ifnn=→=, and also 43ifnn=→ = and 42ifnn=→ = and 41ifnn=→ =. Thus, the incoming light must have just enough energy to produce the 14ifnn=→ = transition. It must be the third line of the Lyman series in the absorption spectrum of hydrogen. The incoming photons must have wavelength given by HH221511114 16RRλ⎛⎞=−=⎜⎟⎝⎠ or ()71H16 1697.2 nm1515 1.097 37 10 mRλ−== =× 28.27 (a) From ()2213.6 eVnZEn=−, ()( )()2123 13.6 eV122 eV1E =− = − (b) Using 20nnarZ= gives ()29110110.052 9 10 m1.76 10 m33ar−−×== =× 28.33 In the 3p subshell, 3n = and 1=A . The 6 possible quantum states are 3n = 1=A 1m=+A 12sm=± 3n = 1=A 0m=A 12sm=± 3n = 1=A 1m=−A 12sm=± 28.34 (a) For a given value of the principle quantum number n, the orbital quantum number l varies from 0 to 1n− in integer steps. Thus, if 4n=, there are 4 possible values of l : 0,1,2, and 3=l (b) For each possible value of the orbital quantum number l, the orbital magnetic quantum number ml ranges from to−+ll in integer steps. When the principle quantum number is 4n= and the largest allowed value of the orbital quantum number is 3=l , there are 7 distinct possible values for ml. These values are: 3, 2, 1,0, 1, 2, and 3m =− − − + + +l 28.36 (a) The electronic configuration for oxygen ()8Z= is 22 4122ssp (b) The quantum numbers for the 8 electrons can be: 1 statess 1n= 0=A 0m=A 12sm=± 2 statess 2n = 0=A 0m=A 12sm=± 2 statesp 2n = 1=A 0m=A 1m=A 12sm=± 12sm=±Atomic Physics 427 28.37 (a) For Electron #1 and also for Electron #2, 3n= and 1=A . The other quantum numbers for each of the 30 allowed states are listed in the tables below. mA sm mA sm mA sm mA sm mA sm mA sm Electron #1 +1 12+ +1 12+ +1 12+ +1 12− +1 12− +1 12− Electron #2 +1 12− 0 12± -1 12± +1 12+ 0 12± -1 12± mA sm mA sm mA sm mA sm mA sm mA sm Electron #1 0 12+ 0 12+ 0 12+ 0 12− 0 12− 0 12− Electron #2 +1 12± 0 12− -1 12± +1 12± 0 12+ -1 12± mA sm mA sm mA sm mA sm mA sm mA sm Electron #1 -1 12+ -1 12+ -1 12+ -1 12− -1 12− -1 12− Electron #2 +1 12± 0 12± -1 12− +1 12± 0 12± -1 12+ There are 30 allow ed states , since Electron #1 can have any of three possible values of mA for both spin up and spin down, totaling six possible states. For each of these states, Electron #2 can be in either of the remaining five states. (b) Were it not for the exclusion principle, there would be 36 possible states, six for each electron independently. 28.41 For nickel, 28Z = and ()()()( )223213.6 eV1 27 13.6 eV 9.91 10 eV1KEZ≈− − =− =− × ()()()()223213.6 eV13.6 eV325 2.13 10 eV42LEZ≈− − =− =− × Thus, ()2.13 keV


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UCSD PHYS 1C - Atomic Physics

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