Chapters_9%2c10%2c11_Problem_Set_Answers

Chapters_9%2c10%2c11_Problem_Set_Answers - Chapters 9 &...

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Chapters 9 & 10 Problems 25 . (a) TRUE Frequency and wavelength are inversely proportional so radiation of shorter wavelength has the higher frequency (b) FALSE Light of wavelengths between 390nm and 790nm is visible to the naked eye (c) FALSE All electromagnetic radiation has the same speed in a vacuum (c) (d) TRUE The wavelength of an X-ray is approximately 0.1nm (a) = c = 2.9979 x 10 8 m s -1 x 10 9 nm 418.7 nm 1m = 7.16 x 10 14 Hz (b) Light of wavelength 418 nm is in the visible part of the spectrum (c) The light will have a violet colour
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31. E = -R H z 2 n 2 E = -R H ( 1 -1 ) n i 2 n f 2 z=1 for H = 3.2881 x 10 15 s -1 ( 1 ) 2 2 3 2 E = h = R H ( 1 ) h n f 2 n i 2 = 4.5668 x 10 14 s -1 = c = 2.9979 x 10 8 m s -1 = 6.5646 x 10 -7 m 4.5668 x 10 14 s -1 = 656.46 nm n i =4 n f =2 = 486.26 nm n i =5 n f =2 = 434.14 nm n i =6 n f =2 = 410.28 nm The longest wavelength has the lowest frequency (and smallest energy) 34. (a) The maximum wavelength occurs when n=2 and the shortest when n is at a maximum ie; n = and 1/n 2 = 0 = 3.281 x 10 15 s -1 (1 ) 1 2 2 2 = 2.4661 x 10 15 s -1 ..... 121.56nm = 3.281 x 10 15 s -1 (1 -0 ) 1 2 = 3.2881 x 10 15 s -1 ..... 91.174nm (b) = 95.0 nm =c = 2.9979 x 10 8 ms -1 = 3.16 x 10 15 s -1 95.0 x 10 -9 m = 3.2881 x 10 15 s -1 ( 1 ) 1 2 n 2 ( 1 ) = 0.960 1 2 n 2 1 = 0.041 n = 5 n 2 (c) here 1 = 0.1597 so n = 2.505 n 2 As n is not an integer there is no atomic line at 108.5nm
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(a) = 2.197 x 10 -18 J (1 -1 ) 6.626 x 10 -34 J s 4 2 7 2 = 1.384 x 10 14 s -1 (b) = c = 2.998 x 10 8 ms -1 = 2.166x 10 -6 m 1.384 x 10 14 s -1 = 2166 nm (c) This is infrared radiation 49. = c = 3.281 x 10 15 s -1 ( 1 ) n 2 7 2 = 2.997 x 10 8 ms -1 2.170 x 10 -6 m ( 1 ) = 0.04196 n 2 7 2 ( 1 )= 0.06237 n 2 n = 4 69. (a) n=5 l=1 m l =0 5p ( l=1 for all p orbitals) (b) n=4 l=2 m l =-2 4d ( l=2 for all d orbitals) (c) n=2 l=0 m l =0 2s ( l=0 for all s orbitals)
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2 electrons in 3s must have opposite spins Correct Each 3p orbital must contain one electron before they can be paired The 3 unpaired electrons must have the same spin, either all up or all down (a) P from question 83 must have 3 unpaired electrons (b) Br: [Ar]4s 2 3d 10 4p 5 10 d-electrons in Br (c) Ge: [Ar]4s 2 3d 10 4p 2 2 4p electrons in Ge (d) Ba: [Xe]6s 2 2 6s electrons in Ba (e) Au: [Xe]6s 1 4f 14 5d 10 14 4f electrons solution manual incorrect: anomaly like Cu 87. (a) N 1s 2 2s 2 2p 3 3 p electrons (b) Rb: [Kr]5s 1
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This note was uploaded on 05/19/2011 for the course PHYS PHYS 100 taught by Professor Lioudmila during the Winter '10 term at The University of British Columbia.

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Chapters_9%2c10%2c11_Problem_Set_Answers - Chapters 9 &...

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