1111_PartUniversity Physics Solution

1111_PartUniversity Physics Solution - Atomic Structure...

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Atomic Structure 41-5 (d) eff 15 12 3 Z =−= for the 3 p level. 22 eff 3 (13.6 eV) (13.6 eV) 13.6 eV 3 n Z E n ⎛⎞ =− ⎜⎟ ⎝⎠ EVALUATE: In these ions there is one electron outside filled subshells, so it is a reasonable approximation to assume full screening by these inner-subshell electrons. 41.30. (a) 2 2e f f e f f 13.6 eV , so 1.26. 4 EZ Z = (b) Similarly, eff 2.26. Z = (c) eff Z becomes larger going down the columns in the periodic table. 41.31. IDENTIFY and SET UP: Estimate eff Z by considering electron screening and use Eq.(41.27) to calculate the energy. eff Z is calculated as in Example 41.8. EXECUTE: (a) The element Be has nuclear charge 4. Z = The ion + Be has 3 electrons. The outermost electron sees the nuclear charge screened by the other two electrons so eff 422 . Z 2 eff 2 (13.6 eV) n Z E n so 2 2 2 2 (13.6 eV) 13.6 eV 2 E (b) The outermost electron in + Ca sees a eff 2. Z = 2 4 2 2 (13.6 eV) 3.4 eV 4 E EVALUATE: For the electron in the highest l -state it is reasonable to assume full screening by the other electrons, as in Example 41.8. The highest l -states of + Be , ++ Mg , Ca , etc. all have a eff 2. Z = But the energies are different because for each ion the outermost sublevel has a different n quantum number. 41.32. 2 (1 ) ( 1 0 . 2 e V ) kx ≅− . 3 7.46 10 eV 12 8 . 0 , 10.2 eV Z × ≈+ = which corresponds to the element Nickel (Ni). 41.33. (a) 15 2 17 20: (2.48 10 Hz)(20 1) 8.95 10 Hz Zf == × = × . 8 15 17 10 17 3.00 10 m s (4.14 10 eV s) (8.95 10 Hz) 3.71 keV. 3.35 10 m. 8.95 10 Hz c Eh f f λ × × × = = × × (b) Z = 27: 18 10 1.68 10 Hz. 6.96 keV. 1.79 10 m. fE = (c) 18 11 48: 5.48 10 Hz, 22.7 keV, 5.47 10 m. E × = = × 41.34. IDENTIFY: The orbital angular momentum is limited by the shell the electron is in. SET UP: For an electron in the n shell, its orbital angular momentum quantum number l is limited by 0 l < n , and its orbital angular momentum is given by ) Ll l =+ = . The z- component of its angular momentum is , zl Lm = = where m l = 0, ±1, … , ± l , and its spin angular momentum is 3/4 S = = for all electrons. Its energy in the n th shell is 2 (13.6 eV)/ n En . EXECUTE: (a) ) 1 2 3 l l = = . Therefore the smallest that n can be is 4, so E n = – (13.6 eV)/ n 2 = – (13.6 eV)/4 2 = –0.8500 eV. (b) For l = 3, m l = ±3, ±2, ±1, 0. Since , = = the largest L z can be is 3 = and the smallest it can be is –3 = . (c) S = = for all electrons. (d) In this case, n = 3, so l = 2, 1, 0. Therefore the maximum that L can be is max 2(2 1) 6 L = . The minimum L can be is zero when l = 0. EVALUATE: At the quantum level, electrons in atoms can have only certain allowed values of their angular momentum. 41.35. IDENTIFY: The total energy determines what shell the electron is in, which limits its angular momentum. SET UP: The electron’s orbital angular momentum is given by ) l = , and its total energy in the n th shell is 2 (13.6 eV)/ n . EXECUTE: (a) First find n : 2 (13.6 eV)/ n = 0.5440 eV which gives n = 5, so l = 4, 3, 2, 1, 0. Therefore the possible values of L are given by ) l = , giving L = 0, 2, = 6, = 12 , = 20 . = (b) E 6 = – (13.6 eV)/6 2 = –0.3778 eV. Δ E = E 6 E 5 = –0.3778 eV – (–0.5440 eV) = +0.1662 eV This must be the energy of the photon, so Δ E = hc/ , which gives = hc / Δ E = (4.136 × 10 –15 eV s )(3.00 × 10 8 m/s)/(0.1662 eV) = 7.47 × 10 –6 m = 7470 nm, which is in the infrared and hence not visible.
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1111_PartUniversity Physics Solution - Atomic Structure...

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