08_Petrucci10e_SSM

# 08_Petrucci10e_SSM - CHAPTER 8 ELECTRONS IN ATOMS PRACTICE...

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177 CHAPTER 8 ELECTRONS IN ATOMS PRACTICE EXAMPLES 1A Use c =  , solve for frequency. = 2.9979 10 690 10 1 = 4.34 10 89 14  m/s nm m Hz 1B Wavelength and frequency are related through the equation c = , which can be solved for either one. == 2.9979 10 91.5 10 = 3.28 8 61 c s m Note that Hz s = 1 2A The relationship =/ c can be substituted into the equation Eh = to obtain c . This energy, in J/photon, can then be converted to kJ/mol. 34 1 8 1 23 9 6.626 10 Jsphoton 2.998 10 ms 6.022 10 photons 1kJ 520 kJ/mol 1m 1mol 1000J 230nm 10 nm hc E   With a similar calculation one finds that 290 nm corresponds to 410 kJ/mol. Thus, the energy range is from 410 to 520 kJ/mol, respectively. 2B The equation = is solved for frequency and the two frequencies are calculated. 3.056 10 6.626 10 19 34 E h  J / photon J s/ photon E h 4.414 10 J / photon 6.626 10 J s/ photon 19 34 = 4.612 10 14 6 662 10 14 .H z To determine color, we calculate the wavelength of each frequency and compare it with text Figure 8-3. 2.9979 10 4.612 10 10 1 = 650 8 14 9 c Hz nm m nm orange 2.9979 10 6.662 10 10 1 = 450 8 14 9 c Hz nm m nm indigo The colors of the spectrum that are not absorbed are what we see when we look at a plant, namely in this case blue, green, and yellow. The plant appears green. 3A We solve the Rydberg equation for n to see if we obtain an integer. nn R E H n = 2.179 10 J 2.69 10 J 81.00 9.00 18 20 2  This is E 9 for n = 9. 3B H n 2 18 20 2 2 -R E n 2.179 10 J 4.45 10 J = n n 48.97 n 7 

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Chapter 8: Electrons in Atoms 178 4A We first determine the energy difference, and then the wavelength of light for that energy. 18 19 22 11 = = 2.179 10 J = 4.086 10 J 24 H ER      34 8 1 7 19 6.626 10 J s 2.998 10 m s = = = 4.862 10 m or 486.2 nm 4.086 10 J hc E  4B The longest wavelength light results from the transition that spans the smallest difference in energy. Since all Lyman series emissions end with n f =1, the smallest energy transition has n i = 2 . From this, we obtain the value of E . 18 18 1 1 = = 2.179 10 J = 1.634 10 J 21 H if nn From this energy emitted, we can obtain the wavelength of the emitted light: Eh c =/ 34 8 1 7 18 6.626 10 = = = 1.216 10 m or 121.6 nm (1216 angstroms) 1.634 10 J hc E 5A E f - 1 8 H 2 2 f - -4 2.179 10 J == 3 ZR n E f = -3.874 10 -18 J E i - 1 8 H i - -4 2.179 10 J 5 n E i = -1.395 10 -18 J E = E f E i E = (-3.874 10 -18 J) – (-1.395 10 -18 J) E = -2.479 10 -18 J To determine the wavelength, use E = h = hc ; Rearrange for : = 34 8 18 (6.626 10 Js) 2.998 10 2.479 10 m hc s Ex J = 8.013 10 -8 m or 80.13 nm 5B Since E 2 H 2 - = Z R n , the transitions are related to Z 2 , hence, if the frequency is 16 times greater, then the value of the ratio 2 2 ? (?-atom) 16. (H-atom) 1 Z Z Z We can see Z 2 = 16 or Z = 4 This is a Be nucleus. The hydrogen-like ion must be Be 3+ .
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## This note was uploaded on 11/07/2011 for the course CHEMISTRY 1500 taught by Professor Hammedmirza during the Fall '11 term at York University.

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08_Petrucci10e_SSM - CHAPTER 8 ELECTRONS IN ATOMS PRACTICE...

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