Solutions to Chapter 6 Homework: Electromagnetic Radiation and Quantum Mechanics, Assignments of Chemistry

Download Solutions to Chapter 6 Homework: Electromagnetic Radiation and Quantum Mechanics and more Assignments Chemistry in PDF only on Docsity! Chapter 6 Homework: solutions 12. a) incorrect. Light is also a form of electromagnetic radiation, and it can pass through water. Corrected: electromagnetic radiation of at least some frequencies is capable of passing through water. b) correct. c) incorrect. Infrared light has lower energy (and therefore lower frequencies) than visible light. d) incorrect. The glow from fireplace and the energy within the microwave oven are forms of electromagnetic radiation. In contrast, foghorn blast is a form of matter wave – it is a periodic change in air pressure. If there was no matter (air) it would not transmit (while electromagnetic waves transmit readily through vacuum). 14. In order of increasing wavelength (ie decreasing frequency and energy): gamma rays < visible light < radio waves visible light: violet < blue < green < yellow < orange < red radiowaves: GHz waves < MHz waves < kHz waves Therefore: gamma rays < yellow < red < FM < AM 15. a) ! c = " T = " *# $# = c " = 3.00*10 8 m /s 955*10 %6 m = 3.14 *10 11 s %1 & 3.14 *10 11 Hz & 314GHz b) ! c = " *# $ " = c # = 3.00*10 8 m /s 5.50*10 14 s %1 = 5.45*10 %7 m & 545nm c) visible region is roughly between 400 and 760 nm, so the radiation in a) will not be visible to human eye (it will be in the microwave region), but the radiation in b) will be visible by human eye (it’s yellow-green). d) ! c = d t " d = c * t = 3.00*10 8 m /s*50.0*10 #6 s =1.50*10 4 m =15.0km 19. a) When we say that energy is quantized, we mean that it can come only in certain quantities (or assume only certain values) – sort of like coal being delivered by truckload. Examples of quantized energy include energy of electromagnetic radiation E=h.ν and energy levels in hydrogen atom En = -hcRH*(1/n2). b) We live in a macroscopic world, where we are best at detecting macroscopic changes of energy. Even smallest amounts of energy we notice are too large in comparison with the quantization of energy. To our eyes (and other senses) it appears that energy can be transferred in any amount. It’s like looking at a line printed diagonally on a paper. It appears to be smooth, but under magnifying glass (or a microscope) we would see that it looks like a staircase, because it is printed from a dots of certain size (typically, 300 or 600 dots per inch or dpi, depending on the printer). Similarly, the quantization of energy is too fine of an effect to detect with our (macroscopic world-adapted) senses. For example noticing velocity of a tiny aluminum bead with mass of 1mg changing from 1mm/s to 2mm/s is near out naked- eye observation limit. The associated change in kinetic energy Ek = (1/2)mv2 is 0.5*10-6kg*(10-3m/s)2 = 5*10-13J. Similarly, change in temperature of this bead by 1°C would be hardly perceptible. The heat needed to increase the temperature of this bead by 1°C is q = s*m*ΔT = 0.90 J.g-1.K-1*10-6g*1K = 9*10-7J. In comparison, energy of a photon of violet light with wavelength λ = 400 nm is ! E = h" = hc # = 6.626*10 $34 J.s* 3.00*10 8 m /s 400*10 $9 m = 4.97*10 $19 J , or roughly a million times smaller than the amount of energy change we can observe with our eyes. 21. a) ! E = h" = hc # = 6.626*10 $34 J.s* 3.00*10 8 m /s 438*10 $9 m = 4.54 *10 $19 J b) ! E = h" = 6.626*10#34 J.s*6.75*1012s#1 = 4.47*10#21J c) ! E = h" = hc # $ # = hc E = 6.626*10 %34 J.s* 3.00*10 8 m /s 2.87*10 %18 J = 6.93*10 %8 m & 69.3nm This radiation will be found in the ultraviolet part of the spectrum. 30. a) ! E = h" #" = E h = 4.41*10 $19 J 6.626*10 $34 J.s = 6.66*10 14 s $1 % 6.66*10 14 Hz b) ! c = " *# $ " = c # = 3.00*10 8 m /s 6.66*10 14 s %1 = 4.50*10 %7 m & 450nm c) The photons of light with wavelength of 439 nm have energy ! E = h" = hc # = 6.626*10 $34 J.s* 3.00*10 8 m /s 439*10 $9 m = 4.53*10 $19 J . Of this amount, 4.41*10-19J will be used to eject the electron from the metal (aka work function); remaining energy will be the kinetic energy of electron: ! Ek = Ephoton " EWF = 4.53*10 "19 J " 4.41*10 "19 J =1.2*10 "20 J d) If the light was at wavelength of 450 nm (ie all energy is used to eject electrons from metal and none towards the kinetic energy of the electrons), then the amount of electrons freed by a 1µJ burst of light would be ! Etotal = Nelectrons * EWF " Nelectrons = Etotal EWF = 1.00*10 #6 J 4.41*10 #19 J = 2.27*10 12 photons 35. ! E n = "hcR H * 1 n 2 = "2.18*10 "18 J * 1 n 2 ! E2 = "hcRH * 1 2 2 = "2.18*10 "18 J * 1 2 2 = "5.45*10 "19 J E6 = "hcRH * 1 6 2 = "2.18*10 "18 J * 1 6 2 = "6.06*10 "20 J #E6$2 = E2 " E6 = "5.45*10 "19 J " ("6.06*10 "20 J) = "4.84 *10 "19 J The negative sign of energy means the energy is emitted (in the form of radiation). ! E = h" = hc # $ # = hc E = 6.626*10 %34 J.s* 3.00*10 8 m /s 4.84 *10 %19 J = 4.11*10 %7 m & 411nm This line will be in the visible spectrum; it will have blue-violet color.