solut12 - CHAPTER TWELVE QUANTUM MECHANICS AND ATOMIC...

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344 CHAPTER TWELVE QUANTUM MECHANICS AND ATOMIC THEORY Light and Matter 21. Planck found that heated bodies only give off certain frequencies of light. Einstein’s analysis of the photoelectric effect used Planck&s concepts suggesting that electromagnetic radiation is ±quantized². 22. v = m 10 _ 780. m/s 10 _ 2.998 = c 9 - 8 λ = 3.84 × 10 14 s -1 ; E = hv = 2.54 × 10 -19 J where h = 6.626 × 10 -34 J s 23. v = m 10 _ 1.0 m/s 10 _ 3.00 = c 2 - 8 = 3.0 × 10 10 s -1 E = hv = 6.63 × 10 -34 J s × 3.0 × 10 10 s -1 = 2.0 × 10 -23 J/photon mol photons 10 _ 6.02 _ photon J 10 _ 2.0 23 -23 = 12 J/mol 24. Ionization energy = energy to remove an electron = E = J 10 _ 7.21 photon -19 E photon = hv and λ v = c. So, v = c and E = hc . λ = J 10 _ 7.21 m/s 10 _ 2.998 _ s J 10 _ 6.626 = E hc 19 - 8 -34 photon = 2.76 × 10 -7 m = 276 nm
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345 25. 99.5 MHz = 99.5 × 10 6 Hz = 99.5 × 10 6 s -1 ; λ = s 10 _ 99.5 m/s 10 _ 2.998 = c 1 - 6 8 ν = 3.01 m 26. E = hv = λ hc = nm 10 _ 1 m 1 _ nm 25 m/s 10 _ 3.00 _ s J 10 _ 6.63 9 8 -34 = 8.0 × 10 -18 J/photon mol photons 10 _ 6.02 _ photon J 10 _ 8.0 23 -18 = 4.8 × 10 6 J/mol 27. The energy needed to remove a single electron is: 10 _ 6.0221 mol 1 _ mol kJ 279.7 23 = 4.645 × 10 -22 kJ = 4.645 × 10 -19 J E = hc , λ = J 10 _ 4.645 m/s 10 _ 2.9979 _ s J 10 _ 6.6261 = E hc 19 - 8 -34 = 4.277 × 10 -7 m = 427.7 nm 28. atom J 10 _ 1.478 = atom kJ 10 _ 1.478 = atoms 10 _ 6.0221 mol 1 _ mol kJ 890.1 -18 -21 23 = ionization energy per atom J 10 _ 1.478 m/s 10 _ 2.9979 _ s J 10 _ 6.6261 = E hc = , hc = E 18 - 8 -34 = 1.344 × 10 -7 m = 134.4 nm No, it will take light with a wavelength of 134.4 nm or less to ionize gold. A photon of light with a wavelength of 225 nm is longer wavelength and, thus, lower energy than 134.4 nm light. 29. The energy to remove a single electron is:
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CHAPTER 12 QUANTUM MECHANICS AND ATOMIC THEORY 346 10 _ 6.022 mol 1 _ mol kJ 208.4 23 = 3.461 × 10 -22 kJ = 3.461 × 10 -19 J = E w Energy of 254 nm light is: m 10 _ 254 m/s) 10 _ (2.998 s) J 10 _ (6.626 = hc = E 9 - 8 -34 λ = 7.82 × 10 -19 J E photon = E K + E w , E K = 7.82 × 10 -19 J - 3.461 × 10 -19 J = 4.36 × 10 -19 J = maximum KE Hydrogen Atom: The Bohr Model 30. When something is quantized, it can only have certain discreet values. In the Bohr model of the H-atom, the energy of the electron is quantized. 31. For the H-atom (Z = 1): E n = -2.178 × 10 -18 J/n 2 ; For a spectral transition, E = E f - E i : E = -2.178 × 10 -18 J ÷ ø ö ç è æ n 1 - n 1 2 i 2 f where n i and n f are the levels of the initial and final states, respectively. A positive value of E always corresponds to an absorption of light and a negative value of E always corresponds to an emission of light. a. E = -2.178 × 10 -18 J ÷ ø ö ç è æ 3 1 - 2 1 2 2 = -2.178 × 10 -18 J ÷ ø ö ç è æ 9 1 - 4 1 E = -2.178 × 10 -18 J × (0.2500 - 0.1111) = -3.025 × 10 -19 J The photon of light must have precisely this energy (3.025 × 10 -19 J). E = E photon = hv = hc or λ = _ E _ hc = J 10 _ 3.025 m/s 10 _ 2.9979 _ s J 10 _ 6.6261 19 - 8 -34 = 6.567 × 10 -7 m = 656.7 nm
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CHAPTER 12 QUANTUM MECHANICS AND ATOMIC THEORY 347 b. E = -2.178 × 10 -18 J ÷ ø ö ç è æ 4 1 - 2 1 2 2 = -4.084 × 10 -19 J J 10 _ 4.084 m/s 10 _ 2.9979 _ s J 10 _ 6.6261 = _ E _ hc = 19 - 8 -34 λ = 4.864 × 10 -7 m = 486.4 nm c. E = -2.178 × 10 -18 J ÷ ø ö ç è æ 2 1 - 1 1 2 2 = -1.634 × 10 -18 J J 10 _ 1.634 m/s 10 _ 2.9979 _ s J 10 _ 6.6261 = 18 - 8 -34 = 1.216 × 10 -7 m = 121.6 nm d. E = -2.178 × 10 -18 J ÷ ø ö ç è æ 4 1 - 3 1 2 2 = -1.059 × 10 -19 J λ = J 10 _ 1.059 m/s 10 _ 2.9979 _ s J 10 _ 6.6261 = _ E _ hc 19 - 8 -34 = 1.876 × 10 -6 m or 1876 nm 32. Ionization from n = 1 corresponds to the transition n i = 1 n f = where E = 0.
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solut12 - CHAPTER TWELVE QUANTUM MECHANICS AND ATOMIC...

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