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chapter07_part1

# chapter07_part1 - Chapter 7 Electron Configurations and the...

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Unformatted text preview: Chapter 7: Electron Configurations and the Periodic Table 297 End-of-Chapter Solutions for Chapter 7 Summary Problem Answer: (a) See diagram below; 2 → 1: ν = 2.463 × 10 15 s –1 , E = 1.632 × 10 –18 J; 3 → 1: ν = 2.922 × 10 15 s –1 , E = 1.936 × 10 –18 J; 5 → 2: ν = 6.903 × 10 14 s –1 , E = 4.574 × 10 –19 J; 4 → 3: ν = 1.598 × 10 14 s –1 , E = 1.059 × 10 –19 J (b) All three metals, Li, K, and Cs, can exhibit photoelectric effect when photons from 2 → 1 (E = 1.936 × 10 –18 J), 3 → 1 (E = 4.574 × 10 –19 J), and 5 → 2 (E = 1.059 × 10 –19 J) strike their surface. (b) (i) 1s 2 2s 2 2p 6 3s 2 3p 6 3d 3 4s 2 (ii) [Ar]3d 3 (iii) (iv) one unpaired electron (v) n = 3, l = 2, m l = 2, m s = + 1 2 ; n = 3, l = 2, m l = 1, m s = + 1 2 ; n = 3, l = 2, m l = 0, m s = + 1 2 (vi) not paramagnetic, because there are no unpaired electrons (d) Metalloid; the other three options can be eliminated: a transition metal would be a good electrical conductor, an alkali metal chloride would be a solid, and the known halogens are not gray. (e) many different responses can be given (see examples below) Strategy and Explanation: (a) Draw and label the first five energy levels of a hydrogen atom, without referring back to the chapter, then indicate several given transition on that diagram. The levels get increasingly closer in energy as n increases. Draw lines from one level to another to show the emission lines. Look up the wavelengths of light emitted and the spectral region for transition of electrons from one level to another in Table 7.2. First transition: 2 → 1, λ = 121.7 nm in the ultraviolet region Second transition: 3 → 1, λ = 102.6 nm in the ultraviolet region Third transition: 5 → 2, λ = 434.3 nm in the visible region (blue) Fourth transition: 4 → 3, λ = 1876 nm in the infrared region Determine the frequency ( ν ) of light for each transition and the energy (E) of each photon. Use appropriate length conversions and the equations in Section 7.1 and Section 7.2. 1 Hz is defined as 1 s –1 . First Transition: 2 → 1: " = c # = 2.998 \$ 10 8 m/s 121.7 nm \$ 1 \$ 10 % 9 m 1 nm = 2.463 \$ 10 15 s % 1 E = h ν = (6.626 × 10 –34 J ⋅ s) × (2.463 × 10 15 s –1 ) = 1.632 × 10 –18 J for one photon Chapter 7: Electron Configurations and the Periodic Table 298 Use the same method to determine the frequency and photon energy for the other transitions. Transition λ ν E 2 → 1 121.7 nm 2.463 × 10 15 s –1 1.632 × 10 –18 J 3 → 1 102.6 nm 2.922 × 10 15 s –1 1.936 × 10 –18 J 5 → 2 434.3 nm 6.903 × 10 14 s –1 4.574 × 10 –19 J 4 → 3 1876 nm 1.598 × 10 14 s –1 1.059 × 10 –19 J (b) Given the wavelengths of light for the photoelectric threshold for ejecting electrons from the surface of three metals, determine which photons of light from (a) have sufficient energy to cause the photoelectric effect (i.e., eject electrons from that metal’s surface). It is not necessary to calculate the actual energies of the photons eject electrons from that metal’s surface)....
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chapter07_part1 - Chapter 7 Electron Configurations and the...

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