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UW-Madison PHYSICS 107 - Homework Answer Key

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107 Answer Key HW #8- Chapter 13assignmentConceptual: 4, 8, 12, 14Problems: 2, 8November 3, 20051 Conceptual Exercises4.) “Suppose that the photoelectric threshold frequency in a certain metallies in the red region of the spectrum (Figure 9.7). Yellow, green, and violetlight are all directed at the surface of this metal. Which, if any, of thesecolors will cause electrons to b e ejected from the surface?”If the frequency of incident light is larger than the threshold frequency,then electrons will be ejected (since the photons will have more than the re-quired energy), otherwise no electrons will be ejected. We need to determine,then, whether yellow, green, and violet light have frequencies higher than thethreshold: red light.Looking at Figure 9.7, red light is listed as having the smallest fre-quency (largest wavelength) of all the colors. Since the frequencies of yel-low, green, and violet light are all larger than the frequency of red light,all of these colors will eject electrons.8.) “A red light beam has a variable frequency. As you increase thefrequency, does the color change? Do the energies of the individual photonsincrease, decrease, or remain the same? Do the photons’ speeds change?What does it mean for a red light beam to have a variable frequency?Let’s assume they mean that as we change the frequency a little bit we seedifferent ”shades” of red. In this case, the color should not change much,since we’re still saying it’s ”red”. However, as you increase the frequencies,you increase the energies of the photons, since Energy = h ∗ (freq.), whereh = 6.67∗10−34J∗s, and a higher frequency means a higher energy. Since elec-tromagnetic radiation always travels at the same speed regardless of its fre-1quency (namely the speed of light, c = 3 ∗ 108m/s), all photons will travel atthis speed regardless of their frequency, and the photons’ speed will not change.12.) “What is meant by a red photon? A yellow photon? Which one hasgreater energy? Longer wavelength?A red photon is a particle of light that behaves as if it had a wavelengthand frequency corresponding to an electromagnetic wave that stimulates ourcones to produce a signal in our brain that we interpret as red. Thesefrequencies are f ∼ 1013Hz. Similarly, a yellow photon is a particle oflight that behaves as if it had a wavelength and frequency corresponding tothat of a yellow electromagnetic wave (namely f ∼ 1014Hz). Since thefrequency for yellow light is larger than the frequency for red light, theyellow photon has a greater energy, using Energy = h ∗ (freq). Since redlight has a smaller frequency, the red photon has a longer wavelength, using(wavelength) =c(freq).14.) “Which has greater energy, a microwave photon of a visible photon?”Looking at Figure 9.7, we see that the frequency of a microwave photon(f ∼ 1011Hz) is smaller than the frequency of any visible light (for red light,f ∼ 1013Hz). Since Energy = h ∗ (freq), the photon with the larger fre-quency has the larger energy; thus, a visible photon has more energy thana microwave photon.2 Problems2.) “In the preceding problem, find the amount of energy a photon needsto knock electrons out of this surface. Do either the orange photons or theviolet photons have this much energy?”From the preceding problem, the threshold frequency is 6.2∗ 1014Hz. Thethreshold energy (the energy needed to knock an electron out of the surface) isthe energy of a photon at the threshold frequency:Ethres= h ∗ 6.2 ∗ 1014Hz = 4.09 ∗ 10−19J.From the preceding problem, the energy of an orange photon is: h ∗forange= 3.3∗10−19J. The energy of a violet photon is h∗fviolet= 4.6∗10−19J.We see that only the violet photon has an energy larger than the threshold energy.8.) “Making estimates. The human eye can detect as few as 10,000photons per second entering the pupil. About how much energy is this, persecond?”We want to find the amount of e nergy entering a human eye per second;2using units, the equation for this will be something like:Energysec= 10, 000photonssec∗Energyphoton. (1)Notice that the units of photons cancel, and we are left with units of en-ergy/sec, which is what we want. But what is the energy per photon?From the photoelectric effect, we have learned that the energy of one pho-ton is Energy = h ∗ (freq), where h = 6.6 ∗ 10−34J ∗ s. Let’s assumewe have green light (in the middle of the visible part of the spectrum)with a frequency of fgr een= 4 ∗ 1014Hz. Then the energy per photon isEnergy/photon = h ∗ fgr een= 2.64 ∗ 10−19J/photon. Putting this in theequation (1), we have:Energysec= 10, 000photonssec∗ 2.64∗ 10−19J/photon = 2.64 ∗ 10−15J/sec.


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UW-Madison PHYSICS 107 - Homework Answer Key

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