ch01_odd _final - CHAPTER 1 ATOMS: THE QUANTUM WORLD 1.1...

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SM-46 CHAPTER 1 ATOMS: THE QUANTUM WORLD 1.1 microwaves < visible light < ultraviolet light < x-rays < γ -rays 1.3 (a) If wavelength is known, the frequency can be obtained from the relation 81 9 c: 2.997 92 10 m s ( ) (925 10 m) νλ ν −− = ×⋅ = × 9 14 1 2.997 92 10 m s 925 10 m 3.24 10 s = × (b) 3 2.997 92 10 m s ( ) (4.15 10 m) = × 3 10 1 2.997 92 10 m s 4.15 10 m 7.22 10 s = × 1.5 3 max Wien’s law states that constant 2.88 10 K m. T λ = 3 max 6 max 2.88 10 K m If /K 1540 C 273 C 1813 K, then 1813 K 1.59 10 m, or 1590 nm T × + ° = = 1.7 (a) From and , we can write cE h == nn 1 34 8 9 1 19 (6.626 08 10 J s) (2.997 92 10 ) (589 10 m) 3.37 10 J Eh c = ××
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SM-47 (b) (c) 23 1 19 1 5 (6.022 10 atoms mol )(3.37 10 J atom ) 2.03 10 J or 203 KJ E −− ×⋅ 1.9 The energy is first converted from eV to joules: 31 9 1 1 4 (140.511 10 eV) (1.6022 10 J eV ) 2.2513 10 J E × = × From Eh v = and cv λ = we can write 34 8 1 14 12 (6.626 09 10 J s) (2.997 92 10 m s ) 2.2513 10 J 8.8236 10 m or 8.8236 pm hc E = = × 1.11 (a) false. The total intensity is proportional to 4 . T (Stefan-Boltzmann Law) (b) true; (c) false. Photons of radio-frequency radiation are lower in energy than photons of ultraviolet radiation. 1.13 (a) Use the de Broglie relationship, 11 () . hp h mv == 28 31 e 1 6 1 (9.109 39 10 g) (1 kg/1000 g) 9.109 39 10 kg (3.6 10 km s ) (1000 m km ) 3.6 10 m s m = × = × 1 34 31 6 1 10 6.626 08 10 J s (9.109 39 10 kg) (3.6 10 m s ) 2.0 10 m hmv = = ×× (b) 34 16 1 17 (6.626 08 10 J s) (2.50 10 s ) 1.66 10 J ν = × 3 23 1 1 19 1 5.00 10 g Na (6.022 10 atoms mol ) 22.99 g mol Na (3.37 10 J atom ) 44.1 J E ⎛⎞ × ⎜⎟ ⎝⎠ =
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SM-48 (c) The photon needs to contain enough energy to eject the electron from the surface as well as to cause it to move at 31 3.6 10 km s . × The energy involved is the kinetic energy of the electron, which equals 1 2 2 . mv 17 2 photon 17 31 6 1 2 17 18 17 1 1.66 10 J 2 1 1.66 10 J (9.109 39 10 kg) (3.6 10 m s ) 2 1.66 10 J 5.9 10 J 2.3 10 J Em v −− + + × × + × But we are asked for the wavelength of the photon, which we can get from Eh v = and cv λ = or 1 . c = nm m s m s m kg s m kg 6 . 8 10 6 . 8 ) 10 99792 . 2 ( ) 10 62608 . 6 ( 10 3 . 2 9 1 1 8 1 2 34 2 2 17 = × = × × × = × (d) 8.6 nm is in the x-ray/gamma ray region. 1.15 To answer this question, we need to convert the quantities to a consistent set of units, in this case, SI units. 1 1 (5.15 ounce) (28.3 g ounce ) (1 kg/1000 g) 0.146 kg 92 mi 1 h 1 km 1000 m 41 m s h 3600 s 0.6214 mi 1 km ⋅= ⎛⎞ =⋅ ⎜⎟ ⎝⎠ Use the de Broglie relationship. 11 1 34 1 34 2 1 1 34 () 6.626 08 10 J s (0.146 kg) (0.041 km s ) 6.626 08 10 kg m s (0.146 kg) (41 m s ) 1.1 10 m hp h mv hmv == = ×⋅ = =
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SM-49 1.17 From the de Broglie relationship, 1 p h λ = or , hm v = we can calculate the velocity of the neutron: 34 2 1 22 27 12 31 (6.626 08 10 kg m s ) (remember that 1 J 1 kg m s ) (1.674 93 10 kg) (100 10 m) 3.96 10 m s h v m −− = ×⋅ == ×× 1.19 Yes there are degenerate levels. The first three cases of degenerate levels are: 1 = 2 = with enerate deg is 2 = 1 = 2 1 2 1 n , n n , n 12 1 2 1, 3 is degenerate with 3, 1 nn 2, 3 is degenerate with 2 1.21 (a) Integrate over the “left half of the box” or from 0 to ½ L: 2 2 2 00 0 2 sin 21 cos sin given is an integer: 2/ 2 1 0 LL L nx dx nx x Ln L L n L L π ππ
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This note was uploaded on 03/31/2008 for the course CHEM 211 taught by Professor Crane, b during the Fall '06 term at Cornell University (Engineering School).

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ch01_odd _final - CHAPTER 1 ATOMS: THE QUANTUM WORLD 1.1...

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