Solve the electron configuration in the ground state

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Solve: The electron configuration in the ground state of Ca ( Z = 20) is 1 s 2 2 s 2 2 p 6 3 s 2 3 p 6 4 s 2 . The configuration for V ( Z = 23) is 1 s 2 2 s 2 2 p 6 3 s 2 3 p 6 4 s 2 3 d 3 . Ni ( Z = 28), As ( Z = 33), and Kr ( Z = 36) are 1 s 2 2 s 2 2 p 6 3 s 2 3 p 6 4 s 2 3 d 8 1 s 2 2 s 2 2 p 6 3 s 2 3 p 6 4 s 2 3 d 10 4 p 3 1 s 2 2 s 2 2 p 6 3 s 2 3 p 6 4 s 2 3 d 10 4 p 6
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42.37. Visualize: Please refer to Figure 42.25. Solve: (a) The allowed transitions are those with l = ± 1. Starting from an s -state, the only allowed transitions are to p -states. (b) Once the states are identified, the wavelength is λ = hc / E = 1240 eV nm/ E . Transition E λ 6 s 5 p 6 s 4 p 6 s 3 p 0.17 eV 0.76 eV 2.41 eV 7290 nm 1630 nm 515 nm
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42.38. Visualize: Please refer to Figure 42.25. Solve: For sodium, the energy of the 3 p state (which is the first excited state) is E 3 p = 2.104 eV, compared to the ground state at E = 0 eV. We need to add the transition energy E 5 d 3 p to E 3 p to obtain the energy of the 5 d state. The transition energy is 5 3 5 3 5 3 1240 eV nm 2.485 eV 2.104 eV 2.485 eV 4.59 eV 499 nm d p d p d p hc E E E E λ = = = = + = + = Assess: E 5 d is larger than E 6 s , E 5 p , and E 4 d and less than the ionization energy limit of 5.14 eV. This is reasonable.
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42.39. Visualize: Please refer to Figure 42.25. Solve: A photon wavelength of 818 nm corresponds to an energy of 1240 eV nm 1.516 eV 818 nm hc E λ = = = From Figure 42.25, the transition that obeys the selection rule l = 1 and has a magnitude around 1.5 eV is 3 p 3 d . Note that E 3 d E 3 p = 3.620 eV – 2.104 eV = 1.516 eV, which is exactly equal to the photon energy. The atom was excited to the 3 d state from the ground state. Thus the minimum kinetic energy of the electron was ( ) ( ) 19 2 6 31 2 3.62 eV 1.60 10 J/eV 1 3.62 eV 1.13 10 m/s 2 9.11 10 kg mv v × = = = × ×
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42.40. Model: Allowed transitions are those with l = ± 1. Visualize: For the fictitious atom, our task is to draw an energy level diagram that will have four emission wavelengths 310.0 nm, 354.3 nm, 826.7 nm, and 1240.0 nm. These wavelengths correspond to energies of 4.00 eV, 3.50 eV, 1.50 eV, and 1.00 eV. A ground p -state ( l = 1), an excited s -state ( l = 0) at 3.50 eV, an excited d -state ( l = 2) at 4.0 eV, and an excited p -state ( l = 1) at 5.0 eV will lead to the four energies or transitions. The excited p -state can decay only to the s and d states, and s and d states can decay only to the ground state. So, the four transitions make a diamond pattern on the diagram.
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42.41. Visualize: Please refer to Figure P42.41. Solve: We need to use the condition l = 1 to determine the allowed transitions of the emission spectrum, and the equation 1240 eV nm hc E λ λ = = to find the wavelengths. Visible wavelengths are in the range 400–700 nm, with infrared being greater than 700 nm and ultraviolet being less than 400 nm. The absorption spectrum has only transitions from the ground state, so the wavelengths of the emission spectrum that are in the absorption spectrum belong to transitions that end on the 2 s ground state.
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