MathConnections
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6.2 The Electromagnetic Spectrum
Visible light is a more accurate term for what we usually refer to simply as light.
The light that our eyes can detect comprises only a small portion of the possible electromagnetic spectrum and accounts for only a small part of the emission of most light bulbs. Other familiar forms of electromagnetic radiation include radio waves, microwaves, and X-rays. The origin of the word electromagnetic lies in the na- ture of light. Historically, light has been described as a wave traveling through space.
One component is an electric fi eld, and another is a magnetic fi eld (Figure 6.3). To understand the wave nature of light, then, we should probably be familiar with the general features of waves.
The Wave Nature of Light
Many of the features of light that we encounter can be explained as properties of waves. The central characteristics of a wave can be defi ned by four variables: wave- length, frequency, velocity, and amplitude. Figure 6.4 illustrates the defi nition of these terms. The wavelength of any wave is the distance between corresponding points on adjacent waves. In the illustration, for example, the wavelength (des- ignated as l and measured in units of length) is defi ned as the distance between peaks. We could also defi ne the wavelength as the distance between valleys, and the value would be the same. The amplitude is the size or height of the wave.
Electric vector
Nodes
Magnetic vector
Direction of propagation Amplitude
Wavelength, l
Figure 6.3 ❚ Light is one form of electromagnetic radiation, which consists of oscillating electric and magnetic fi elds.
The lower wave in Figure 6.4 has a larger amplitude than the upper wave. The frequency is the number of complete cycles of the wave passing a given point per second. Frequency is usually designated as ν and measured in units of 1/s, or hertz (Hz). Frequency and wavelength are not independent of one another. The speed of light in a vacuum (designated by c and measured in units of distance/time) is a con- stant of nature that has been measured with impressive precision and is now defi ned to have an exact value:
c = 2.99792458 × 108 m s−1
Because the speed of light is a constant, the number of waves that will pass a certain point in time will be inversely related to the wavelength. Long wavelengths will have fewer cycles pass by in one second, so they will have lower frequencies (Figure 6.4).
This relationship is expressed mathematically in Equation 6.1:
c = l × n (6.1)
MathConnections
If we want to describe the detailed shape of a wave, we need to write a math- ematical equation for it. Let’s consider a very simple example in which the wave happens to follow the shape of a sine function. (This turns out to be true for many naturally occurring waves.) Imagine that we take a snapshot of the wave at a particular instant and picture it as an x – y graph. The wave originates at x = 0 and travels along the x axis. The y values represent the height of the wave at each point. We could write a very general equation for this wave:
y = A sin (bx)
A is clearly the amplitude of the wave, but what about b? Let’s assume that we want to evaluate the sine function in terms of radians. That means that the quan- tity bx must have no units. Because x is the distance from the origin, it would have units of length. So the b in our equation must have units of 1/length. If we know the wavelength of the wave—which we could read off the graph—then we can also fi x the value of b. The wavelength of the sin (x) function is 2π. So, suppose that the wavelength of our wave is 450 nm, in the range of blue light. For our equation to match the wave, we require that
b × 450 nm = 2p ν is the Greek letter nu.
ν is the Greek letter nu.
Wavelength λ
Amplitude
Low frequency Long wavelength Small amplitude
High frequency Short wavelength Large amplitude Figure 6.4 ❚ Two sine waves are
shown, and the meanings of some important terms are illustrated.
Because light has a fi xed speed, its frequency and wavelength are always inversely related. Amplitude is independent of frequency; the choice of which of these waves would have larger amplitude was made arbitrarily.
or
b = 2p/450 nm
More generally, we can see that the b in our equation must always be 2p/l.
If we wanted to describe a more complicated wave, we might have to use a combination of several sine (or cosine) functions, each of the same general form as we see here but with different wavelengths. By combining enough such functions, we could describe a waveform of arbitrary complexity. If we wanted to describe a wave as a function of time rather than position, we could proceed in a similar fashion.
Speed, wavelength, frequency, and amplitude are characteristics that can describe any wave. But because we are interested in light, it should help to compare these quantities with some familiar properties of light. If someone asks you to compare two different lights, one of the things you would probably consider is how bright each one appears. How does this relate to the wave parameters we’ve defi ned? In the wave model for light, the amplitude determines the brightness of the light: the larger the amplitude of the wave, the brighter the light will appear. Another property that you would surely consider in comparing two lights is color. In the wave model, both wave- length and frequency correspond to the color of the light. How can these two seem- ingly different properties of the wave both correspond to color? The answer can be seen in Equation 6.1. Because the speed of light must remain constant, specifying either the frequency or the wavelength automatically fi xes the value of both of those properties of a light wave. So, we could say that the frequency and the wavelength are simply two different ways of expressing the same information.
The wave nature of light also helps to explain many of the phenomena we experi- ence with light. Refraction is the bending of a wave when it passes from one medium to another of different refractive index. One common manifestation of refraction is the rainbow, in which white light is separated into colors. In passing from air to water or glass, the speed of light changes, and the light bends at an angle that depends on its wave- length. Thus the visible spectrum is separated into its colors, and the colors are dispersed in order of their corresponding wavelengths. Figure 6.5 illustrates this phenomenon.
Because of wave-like phenomena such as refraction, it is common to categorize electromagnetic radiation in terms of its wavelength or frequency. Figure 6.6 shows the entire spectrum, including information about both frequency and wavelength.
Mathematically, the intensity of a light wave is actually proportional to the square of its amplitude.
Mathematically, the intensity of a light wave is actually proportional to the square of its amplitude.
White light Prism
Increasing wavelength Slits to isolate
thin beam
Figure 6.5 ❚ When white light passes through a prism, it is separated into its constituent colors by refraction.
In addition to visible light, the spectrum includes X-ray, ultraviolet (UV), infrared (IR), microwave, and radio wave radiation; the visible falls between UV and IR. This order lists the regions of the spectrum in increasing wavelength. We will often fi nd that we can measure or are provided with the wavelength of light but need to express it as a frequency. Example 6.1 shows this type of conversion.
E X A M P L E P RO B L E M 6 .1
Neon lights emit an orange-red colored glow. This light has a wavelength of 670 nm.
What is the frequency of this light?
Strategy We know that wavelength and frequency are related through Equation 6.1. In using this equation, we will need to be careful of units. Because the wavelength is given in nanometers, we will convert it to meters so that it is consistent with the meters per second in the speed of light.
Solution
670 nm × —————— 1 m
1 × 109 nm = 6.7 × 10−7 m We know that c = lν, so, rearranging, ν = —— c
l . Then we simply substitute the values for the speed of light and the wavelength in the equation:
ν = 2.998 × 108 m s−1
—————————
6.7 × 10−7 m = 4.5 × 1014 s−1
Analyze Your Answer It is diffi cult to have a physical intuition about something like the frequency of light. But with some practice we can develop an idea of the expected order of magnitude. First, notice that to fi nd the frequency, we took a large number (the speed of light) and divided it by a small number (the wavelength). So we should expect a very large numerical answer. It is also handy to become familiar with the typical magnitudes of some quantities. A glance at Figure 6.6, for example, will show us that the frequency of visible light is on the order of 1014 s−1, so our answer
Wavelength increases Energy increases
Energy Increases
400 500 600 700 l (nm)
1024 1022 1020 1018 1016 1014 1012 1010 108 106 104 102 100 n (Hz)
10⫺16 10⫺14 10⫺12 10⫺10 10⫺8 10⫺6 10⫺4 10⫺2 100 102 104 106 108 l (m)
g-rays X-rays UV IR Microwave FM AM Long radio waves
Radiowaves
Visible spectrum
Figure 6.6 ❚ Visible light is just the small portion of the entire electromagnetic spectrum that our eyes can detect. Other regions of the spectrum are used in many technological applications, including remote controls, cell phones, and wireless networks.
seems plausible. These kinds of checks can be very handy because the most common errors in problems like this involve mishandling of units, leading to results that can be off by several orders of magnitude.
Check Your Understanding When trace amounts of xenon gas are added to neon lights, the color becomes blue-green. If the wavelength of this light is 480 nm, what is the frequency?
Generally, we think of a lamp as something emitting visible light. But for some specifi c applications, bulbs are designed to produce radiation in different regions of the electromagnetic spectrum. Heat lamps in fast food restaurants, for example, emit infrared radiation to keep French fries warm. Compared with ordinary incandescent bulbs, heat lamps are designed to emit a range of wavelengths shifted toward the IR region. These bulbs appear red to our eyes because that is the visible color that is adjacent to the infrared. Moving to the opposite end of the visible spectrum, lamps designed to emit ultraviolet radiation are used both as “black lights,” causing various dyes to glow, and in tanning beds. Our eyes do not detect ultraviolet light, but these lamps appear purple because the range of wavelengths they emit overlaps the violet end of the visible spectrum. Microwave ovens, radios, televisions, wireless telephones, and wireless computer networks all function by exploiting other specifi c regions of the electromagnetic spectrum.
The Particulate Nature of Light
The wave model of electromagnetic radiation can explain many observed proper- ties of light. In the early 1900s, however, scientists developed more sophisticated ap- paratus and carried out experiments that challenged the wave model of light. One pivotal group of experiments involved the observation of what came to be known as the photoelectric effect, in which light strikes a piece of metal and causes elec- trons to be ejected. Figure 6.7 is a schematic diagram of a typical photoelectric effect
Meter (current) Light
Electron
Photo cathode (–)
Electron (–)
Anode (+)
3 1 Photons of sufficient energy strike 2
a metal surface and electrons are ejected.
The ejected electrons are collected by the anode.
The movement of the photoelectrons causes a current to flow, and allows the light intensity to be monitored.
Figure 6.7 ❚ In the photoelectric effect, light strikes the surface of a metal and electrons are ejected, as shown in the inset. This is exploited in photoelectric cells: light strikes a surface called the photocathode, and a wire (the anode) collects the emitted electrons. (A positive voltage applied to the anode ensures that the electrons travel in the desired direction.) The resulting current can be measured easily and is related to the intensity of the incident light.
# e–
υ0
υ > υ0
υ > υ0
υ
# e–
I
υ0 υ I
KE KE
(a) (b)
(c) (d)
Figure 6.8 ❚ The graphs sketched here show the most important observations from experiments on the photoelectric effect. Below some threshold frequency (ν0), no photoelectrons are emitted.
For frequencies greater than ν0, the number of electrons emitted is independent of frequency (graph a) but increases with increasing light intensity (graph b). The kinetic energy of the emitted electrons, on the other hand, increases linearly with frequency (graph c) but is independent of intensity (graph d).
experiment. At the heart of the experiment is light shining on a piece of metal. If the metal is enclosed in a container under vacuum so that there are very few surrounding gas molecules to interfere, then we can detect electrons emitted from the metal under certain conditions. (The metal itself is grounded so that it does not develop a large positive charge during this process.) As we learned in Section 2.2 (page 33), electrons carry a negative charge. This makes it fairly easy to measure how many electrons are ejected and even how fast they are going. So in studying the photoelectric effect, the experimenter could vary the color or intensity of the light and see how the number of electrons emitted or their kinetic energy might change. The experiment could also be repeated using different metals.
How can this photoelectric effect be understood? The simplest explanation is that energy from the light is transferred to the electrons in the metal. If an electron picks up enough energy, it can break free from the surface of the metal. The more energy given to the electron, the faster it will travel after it leaves the metal.
Figure 6.8 summarizes the sort of results obtained in these experiments. A close look at these graphs reveals that the detailed observations are not what we would expect based on the wave model for light. In a wave description of a light source, the amplitude of the wave or the intensity of the light determines the energy. So we would expect that as the light gets brighter, the kinetic energy of the outgoing electrons should increase. But Figure 6.8(d) shows that the kinetic energy of the electrons is actually independent of the intensity of the incoming light. Only the number of elec- trons detected increases when the intensity goes up (graph b). But the kinetic energy does depend on the frequency or color of the light used (graph c). This is also at odds with the wave model. So these experiments (and others) caused scientists to reexamine the nature of light.
Despite elaborate efforts to make these observations of the photoelectric effect conform to existing wave concepts, consistent explanations of light based solely on waves ultimately failed. The only way to explain all of the experimental results was to invoke the notion of wave-particle duality, which says that, in some situations, light is best described as a wave, whereas in other cases, a particle description works better.
It’s important to realize that this does not mean that there are two different kinds of light! It simply means that neither the wave model nor the particle model provides an accurate description of all of the properties of light. So in a given situation, we use whichever model works best to describe the properties we are interested in. Someday, advances in our understanding of light may produce a new model that reconciles the wave-particle confl ict. But until that happens, we take a pragmatic approach and use the model that’s best suited to the problem at hand.
The context in which we normally think of light as a particle is when it is impart- ing energy to another object. Albert Einstein proposed that light could be described as a collection of packets of energy, called photons. Bright light has many photons, whereas dim light has few. The energy of a photon of light has been shown to be pro- portional to its frequency, leading to a simple equation:
E = hn (6.2)
Here E represents the photon energy, and n is the frequency of the light. The h term is a constant, called Planck’s constant, after German scientist Max Planck. Its value is
h = 6.626 × 10−34 J s.
Because frequency is related to wavelength we can substitute for ν in this equation, using Equation 6.1:
E = —— hc
l (6.3)
Thus, if we know either the wavelength or the frequency of radiation, we can deter- mine the energy of a photon of that radiation, as shown in Example Problem 6.2.
E X A M P L E P RO B L E M 6 . 2
The laser in a standard laser printer emits light with a wavelength of 780.0 nm. What is the energy of a photon of this light?
Strategy We know the connection between photon energy and wavelength, which is given by Equation 6.3. Again, care with units requires conversion from nanometers to meters.
Solution
780.0 nm × —————— 1 m
1 × 109 nm = 7.800 × 10−7 m E = 6.626 × 10−34 J s × 2.998 × 108 m s−1
—————————————————
7.800 10−7 m = 2.547 × 10−19 J
Analyze Your Answer The result is a very small number. But we should realize that Planck’s constant is incredibly small—many orders of magnitude smaller than the other quantities in the problem. So a very small energy is likely. Beyond that, we can again rely on a general sense of the magnitude of the quantity we are calculating. For visible light, photon energies are typically on the order of 10−19 J, which makes our answer seem plausible.
Recall that in Chapter 3 (p. 68) we used hν to indicate that light is required in a photochemical reaction.
Recall that in Chapter 3 (p. 68) we used hν to indicate that light is required in a photochemical reaction.
Check Your Understanding An infrared laser for use in a fi ber-optic communi- cations network emits at a wavelength of 1.2 mm. What is the energy of one photon of this radiation?
The idea that light is made up of a collection of particles, or “chunks of energy,”
does not seem consistent with our everyday experience. To understand why, look at the magnitude of the photon energy calculated in Example Problem 6.2: 10−19 J.
Compare that with the amount of energy emitted by an ordinary lamp. Even a modest 60-watt bulb emits 60 J of energy each second. So the lights that you are familiar with all consist of vast numbers of photons. As the light gets brighter, the number of pho- tons emitted per second increases. But because each photon is such a small additional amount of energy, it appears to us as if the energy coming from the lamp varies con- tinuously rather than in discrete increments or steps.
Though not readily discernible to us, the energy of a single photon can be very signifi cant from an atomic or molecular vantage point. Atoms and molecules are very small, after all, and their typical energies are small as well. Consider the results pre- sented earlier for the photoelectric effect experiment. How can the concept of photons help us to explain those observations? Imagine that we could do the experiment by shooting a single photon at the metal. That means we would be supplying an amount of energy (equal to hν) to the metal. If this energy is enough to overcome the force that the metal atoms exert to hold onto their electrons, then when the metal absorbs the photon, it will be able to eject one electron. If the energy of the photon is smaller than the binding energy holding the electrons to the metal, then no electron can be emitted. The fact that there is a threshold frequency (ν0 in Figure 6.8), below which no electrons are detected, is evidence that the photoelectric process is carried out by indi- vidual photons. Below the threshold frequency, a photon does not have enough energy to overcome the binding energy and eject an electron. Why does the electron kinetic energy depend on the frequency of the light? This is really a question of conservation of energy. For a particular metal, the electron binding energy is fi xed. For any fre- quency above ν0, a portion of the energy of the absorbed photon will be used to over- come the binding energy. Then, whatever energy remains will be transferred to the electron as kinetic energy. So, as the frequency increases, the photon energy increases.
This leads to an increase in the observed kinetic energy of the ejected electrons.
E X A M P L E P RO B L E M 6 . 3
In a photoelectric effect experiment, ultraviolet light with a wavelength of 337 nm was directed at the surface of a piece of potassium metal. The kinetic energy of the ejected electrons was measured as 2.30 × 10−19 J. What is the electron binding energy for potassium?
Strategy We know that energy must be conserved, and we can use this idea to set up an equation relating the photon energy to the kinetic energy and the binding en- ergy. Because the photon is absorbed in this process, we must be able to account for its energy. Some of that energy does work to overcome the electron binding energy, and the rest appears as the kinetic energy of the ejected electron. We can write this as a simple equation:
E photon= Binding E + Kinetic E
We know how to find the photon energy from the wavelength, and we know the kinetic energy, so we can solve for the binding energy. As always, we must be careful with the units on all quantities involved.
Be careful not to confuse the photon energy with the intensity or brightness of the light. If you remember that photon energy is another way of specifying the color of the light, you will see that it is independent of intensity.
Be careful not to confuse the photon energy with the intensity or brightness of the light. If you remember that photon energy is another way of specifying the color of the light, you will see that it is independent of intensity.