chapter07_part1 - Chapter 7: Electron Configurations and...

Info iconThis preview shows pages 1–3. Sign up to view the full content.

View Full Document Right Arrow Icon

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

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: = 2.463 10 15 s 1 , E = 1.632 10 18 J; 3 1: = 2.922 10 15 s 1 , E = 1.936 10 18 J; 5 2: = 6.903 10 14 s 1 , E = 4.574 10 19 J; 4 3: = 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, = 121.7 nm in the ultraviolet region Second transition: 3 1, = 102.6 nm in the ultraviolet region Third transition: 5 2, = 434.3 nm in the visible region (blue) Fourth transition: 4 3, = 1876 nm in the infrared region Determine the frequency ( ) 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 = (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 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 metals surface). It is not necessary to calculate the actual energies of the photons eject electrons from that metals surface)....
View Full Document

This note was uploaded on 12/17/2009 for the course CHE 129/130 taught by Professor Hanson during the Spring '09 term at SUNY Stony Brook.

Page1 / 18

chapter07_part1 - Chapter 7: Electron Configurations and...

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online