Chemistry 08 Spring 2010 Assignment 4 Solutions

Chemistry 08 Spring 2010 Assignment 4 Solutions Chapter 7 # 7.14, 7.22, 7.24, 7.28, 7.32, 7.36, 7.38, 7.40. 7.14. Collect and Organize To answer why infrared (IR) light is ―safe‖ and ultraviolet (UV) light may cause cancer, we need to look at their relative energies. Analyze High-energy light may cause cellular damage by disrupting chemical bonds. Energy is inversely related to the wavelength of light: E = hc/. Solve UV light has a shorter wavelength than IR and therefore has a higher energy. This energy is enough to break chemical bonds or ionize molecules. IR light, however, with a longer wavelength and lower energy, does not cause cellular damage—its energy is too low. Think About It Both IR and UV light are outside of the visible (to our eyes) range in the electromagnetic spectrum. 7.22. Collect and Organize Given the frequency of an AM radio station, (680 kHz), we are to calculate its wavelength and compare it to the wavelength of infrared radiation emitted by Earth ( = 15 mm). Analyze Wavelength can be calculated using  = c/ where  is in meters, c = 3.00  10 m/s, and  is in hertz (per second). We have to convert kilohertz to hertz (1 kHz = 1000 Hz). 8 Solve  3.00  108 m/s  441 m 680  103 s –1 The wavelength of the AM radio station (a) is longer than the 15 mm wavelength for IR radiation (b). Think About It Radio waves are measured on the meter scale; light waves (including IR and UV) are measured on the nanometer scale. 7.24. Collect and Organize Chemistry 08 Spring 2010 Red light has a wavelength of about 600 nm. Green light has a wavelength of about 550 nm. We are asked to compare the frequencies of these two colors of lights and state which is higher. Analyze We can calculate the frequency for each color of light using  = c/. Wavelength must be expressed in meters for this calculation (1 nm = 1  10–9 m). Solve 3.00  108 m/s  5.00  1014 s –1 –9 600  10 m 3.00  108 m/s   5.45  1014 s –1 550  10 –9 m  red   green The green light (b) has a higher frequency than the red light (a). Think About It We did not have to calculate the actual frequencies for this problem. If we keep in mind that wavelength decreases along the spectrum of colors in the rainbow (ROYGBIV) and we know that frequency is inversely proportional to wavelength, then frequency increases along the spectrum (ROYGBIV) and green light has a higher frequency (and shorter wavelength) than red light. 7.28. Collect and Organize We are to consider whether Fraunhofer lines are atomic emissions or absorptions. Analyze Fraunhofer lines show up as dark lines in the broad band of color. Solve The colors that are missing from the spectrum have been absorbed, hence they are not observed. 7.32. Collect and Organize / Analyze We are to define photon. Solve A photon is the smallest indivisible packet or particle of light energy. Think About It The energy of a photon can be related to its frequency (E = h) and wavelength (E = hc/). 7.36. Collect and Organize From a list, we are to choose which have quantized values. Chemistry 08 Spring 2010 Analyze Something is quantized if it is present only in discrete amounts and can only have whole-number multiples of the smallest amount. Solve (a) The pitch of a note on a slide trombone can be varied continuously, so this is not a quantized value. (b) The pitch of a note on a flute depends on which holes are covered or open, so this is a quantized value. (c) The wavelengths of light produced by a heating element are continuous, so this is not a quantized value. (d) The wind speed on top of Mt. Everest can vary continuously, so this is not a quantized value. Think About It Any quantity that is quantized has to have changes occurring in discrete steps. 7.38. Collect and Organize The work function equation to answer this problem is KE electron  h –  We are asked to determine the wavelength corresponding to the work function to eject one electron. Analyze The kinetic energy for the electron ejected by the maximum wavelength must be zero because no additional energy is supplied to the electron by the incident radiation other than ejecting it. Therefore, the frequency of the light to eject the electron would be 0  h –  or   /h The work function is the energy required to eject an electron. This is equivalent to the first ionization energy (given as 6.24  10–19 J). The wavelength can then be found from  using  = c/. Solve 6.24  10 –19 J  9.417  1014 s –1 6.626  10 –34 J  s 3.00  108 m/s   3.19  10 –7 m or 319 nm 9.417  1014 s –1  Think About It This wavelength (319 nm) is in the UV region of the electromagnetic spectrum. 7.40. Collect and Organize We can use the equation for the photovoltaic effect to determine if electrons could be ejected from tantalum using light with a wavelength of 500 nm. Chemistry 08 Spring 2010 Analyze As long as the energy of the light that shines on the metal is greater than the work function for tantalum ( = 6.81  10–19 J), the light will eject electrons and tantalum would therefore be useful in a voltaic cell. The energy of the light can be calculated from E = hc/. Solve The energy of light of 500 nm wavelength can be found as E 6.626  10 –34 J  s  3.00  108 m/s  3.98  10 –19 J 5.00  10 –7 m This energy is less than the work function for tantalum ( = 6.81  10–19 J), so tantalum could not be used to convert solar energy at 500 nm to electricity. Think About It The wavelength of light that would be needed to eject electrons from tantalum would be  hc 6.626  10 –34 J  s  3.00  108 m/s   2.92  10 –7 m or 292 nm  6.81  10 –19 J