98 Photon Energies and the Electromagnetic Spectrum

Ionizing Radiation

In the previous section, we introduced the idea that light is composed of photons—individual packets, or quanta, of electromagnetic (EM) radiation. Each photon carries a specific amount of energy that depends only on the frequency (or equivalently, the wavelength) of the radiation. This relationship is fundamental to understanding why different types of electromagnetic radiation interact with matter in dramatically different ways.

The energy of a single photon is given by

where E is the energy of one photon, h is Planck's constant, f is the frequency, c is the speed of light, and λ is the wavelength.

When working with atoms and molecules, it is often more convenient to express energy in electron volts (eV) rather than joules. In these units, Planck's constant becomes

Likewise, because wavelengths are frequently expressed in nanometers, another useful relationship is

These convenient forms allow photon energies to be calculated quickly without repeatedly converting between SI units.

Every type of electromagnetic radiation—from radio waves to gamma rays—is made of photons. The primary difference between these forms of radiation is the energy carried by each photon. As photon energy increases, the interaction between radiation and matter changes dramatically. Low-energy photons may simply cause molecules to rotate or vibrate, whereas high-energy photons can remove electrons from atoms or even damage biological tissue.

Earlier chapters introduced ultraviolet (UV) radiation, X-rays, and gamma rays as forms of electromagnetic radiation with properties very different from those of visible light. The photon model explains these differences naturally: photons at higher frequencies simply carry much more energy than photons at lower frequencies.

Because photons interact one at a time with electrons, atoms, and molecules, the energy carried by an individual photon determines the type of interaction that can occur. Comparing the photon energies shown in Figure 98.1 with the characteristic energies listed in Table 98.1 helps explain why different regions of the electromagnetic spectrum produce different physical effects.

Gamma rays have the highest frequencies and therefore the highest photon energies of any electromagnetic radiation. For example, a gamma-ray photon with a frequency of

has an energy of

This enormous amount of energy is sufficient to ionize thousands of atoms or molecules, since most ionization processes require only about 10–1000 eV. Radiation capable of removing electrons from atoms is called ionizing radiation. Gamma rays, X-rays, and the highest-energy ultraviolet photons all belong to this category.

Ionization can damage biological molecules, including DNA. A single high-energy photon may kill a cell or interfere with its ability to divide correctly. Because uncontrolled cell division is one of the hallmarks of cancer, exposure to ionizing radiation increases cancer risk. At the same time, this same property makes ionizing radiation an effective treatment for many cancers because rapidly dividing cancer cells are particularly sensitive to DNA damage.

Because gamma-ray photons carry such large amounts of energy, they can penetrate many materials before being completely absorbed. This penetrating ability makes them valuable not only in medicine but also in scientific research and industrial inspection.

X-rays occupy the region of the electromagnetic spectrum immediately below gamma rays in photon energy. Their energies typically range from several thousand electron volts (keV) upward, making them energetic enough to ionize atoms and molecules as well. Although somewhat less penetrating than gamma rays, X-rays remain highly effective for imaging internal structures of the body because they pass through soft tissues more readily than through bone.

The discovery of X-rays by Wilhelm Conrad Röntgen in 1895 transformed medicine almost immediately. Within a year, physicians were using X-rays to diagnose broken bones and locate foreign objects inside the body. For this groundbreaking discovery, Röntgen received the first Nobel Prize in Physics in 1901.

Unlike gamma rays, which originate from processes inside atomic nuclei, X-rays are produced when energetic electrons are rapidly decelerated. In a typical X-ray tube, electrons are emitted from a heated metal filament (the cathode) by thermal emission. A large electric potential difference accelerates these electrons across a vacuum toward a metal target called the anode.

As the electrons strike the anode, their kinetic energy is converted into several forms. Most of the energy becomes thermal energy, heating the target, while a small fraction is emitted as X-ray photons. Because accelerating electric charges emit electromagnetic radiation, some electrons lose nearly all of their kinetic energy in a single interaction, producing very energetic X-ray photons.

This same basic design has been used in medical X-ray systems for more than a century. Although modern X-ray tubes incorporate improved cooling systems, rotating anodes, and sophisticated beam control, the underlying physics remains unchanged.

The spectrum of X-rays produced in an X-ray tube contains two distinct components. The broad, continuous portion results from electrons slowing down as they pass through the target material. This process, known by the German term bremsstrahlung ("braking radiation"), produces photons with a wide range of energies, extending from nearly zero up to the maximum energy determined by the tube voltage.

Superimposed on this continuous spectrum are narrow, intense peaks called characteristic X-rays. These arise when incoming electrons eject electrons from the inner shells of atoms in the anode. Electrons from higher energy levels then fill the vacancies, emitting photons with precisely defined energies that are characteristic of the target material. These discrete photon energies provide further evidence that atomic energy levels are quantized.

Ultraviolet (UV) radiation occupies the region of the electromagnetic spectrum between visible light and X-rays, with typical photon energies ranging from approximately 4 eV to 300 eV. UV photons are energetic enough to ionize atoms and molecules, giving them biological effects similar to—but generally less severe than—those of X-rays and gamma rays.

Ultraviolet radiation can damage DNA, increase the risk of skin cancer, and destroy microorganisms, making it an effective sterilization tool. At the same time, moderate exposure to UV radiation plays beneficial roles in human health, including stimulating vitamin D production in the skin. UV phototherapy is also used to treat certain medical conditions, such as neonatal jaundice and several inflammatory skin diseases.

Visible Light

Visible light occupies only a small portion of the electromagnetic spectrum, yet it is the part to which the human eye is sensitive. The energies of visible-light photons range from approximately 1.63 eV for red light to 3.26 eV for violet light. (You will verify these values in the Problems and Exercises at the end of the chapter.)

These photon energies are comparable to the energy differences between the outer electron shells of atoms and molecules. As a result, visible photons can be absorbed or emitted when electrons move between these quantized energy levels. Because atomic and molecular energy levels are discrete, a photon can only be absorbed if its energy exactly matches the energy difference between two allowed states.

One remarkable consequence of this quantization is that a single photon can trigger a biological response. In the retina of the eye, for example, a single visible photon can be absorbed by a receptor molecule, causing a chemical change that ultimately generates a nerve impulse. This extraordinary sensitivity allows humans to detect extremely faint light under dark conditions.

Suppose a red photon with frequency f encounters a molecule whose energy levels differ by

The photon can then be absorbed, promoting the molecule to the higher-energy state. If no available energy level matches the photon's energy, the photon is not absorbed and instead may be reflected or transmitted. For example, violet flowers appear violet because they reflect violet photons while absorbing photons of other colors, particularly red light.

Although all visible light behaves according to the same physical laws, differences in photon energy produce noticeable differences in everyday life. Red photons have relatively low energies, while violet photons carry about twice as much energy. This additional energy allows violet light to produce chemical changes that red light cannot.

For example, traditional black-and-white photographic film is relatively insensitive to red light, allowing photographers to use red illumination in darkrooms without exposing the film. In contrast, violet and blue photons carry enough energy to break weak chemical bonds more readily. As a result, dyes that absorb blue or violet light often fade more quickly after prolonged exposure to sunlight.

This same principle explains why many colored posters lose their reds, yellows, and greens before their blues and violets. Pigments that appear red or green absorb blue and violet photons, whose greater energies gradually break apart the weak molecular bonds that give the pigments their color. Blue and violet pigments, however, primarily reflect these high-energy photons rather than absorbing them, making them more resistant to fading.

Transparent materials, such as ordinary window glass, transmit visible light because their atoms and molecules do not possess energy-level differences that match the energies of visible photons. Since photons interact individually with atoms and molecules, it is extremely unlikely that two visible photons will be absorbed simultaneously to produce a larger energy transition.

Because visible photons have relatively low energies, visible light can travel through many transparent materials over long distances with little absorption. In contrast, higher-energy ultraviolet radiation, X-rays, and gamma rays possess enough energy to ionize atoms and molecules, making them much more readily absorbed by matter.

Lower-Energy Photons

Below the visible portion of the electromagnetic spectrum lie infrared (IR) radiation, microwaves, radio waves, and other forms of low-frequency electromagnetic radiation. These photons carry much less energy than visible-light photons and are generally incapable of producing the ionization or chemical changes associated with ultraviolet radiation, X-rays, and gamma rays.

Because their photon energies are so low, infrared photons usually interact with matter by causing molecules to vibrate or rotate rather than by altering their electronic structure. Infrared radiation is therefore closely associated with heat and thermal energy transfer.

Water molecules, in particular, absorb infrared radiation very efficiently because they possess numerous vibrational and rotational energy levels separated by energies ranging from approximately

which fall within the infrared and microwave regions of the electromagnetic spectrum. Since the human body is composed largely of water, skin is an excellent absorber and emitter of infrared radiation, giving it an emissivity close to 1. This property is the basis of thermal imaging cameras, which detect the infrared radiation naturally emitted by warm objects.

Not all substances absorb infrared radiation equally well. Earth's atmosphere, for example, is relatively transparent over several infrared wavelength ranges, allowing infrared radiation to travel long distances through air. This transparency is exploited in remote sensing, astronomy, and certain medical imaging applications.

Microwaves have even lower photon energies than infrared radiation and are commonly produced by electronic circuits, although they also occur naturally. Like infrared radiation, microwave photons interact strongly with water molecules because their energies closely match molecular rotational transitions, typically on the order of

This resonance allows microwave energy to be transferred efficiently into water-rich materials, making microwave ovens an effective method of heating food from the inside. The photons themselves are not especially energetic; rather, an enormous number of microwave photons collectively transfer thermal energy to the food.

Because the energy carried by an individual infrared or microwave photon is extremely small, an enormous number of photons must act together to transfer a noticeable amount of energy. Warming yourself under a heat lamp or heating food in a microwave oven involves countless low-energy photons interacting simultaneously with matter.

Individual photons of visible light, infrared radiation, microwaves, and lower-frequency electromagnetic waves do not possess enough energy to ionize atoms or molecules. Consequently, these forms of radiation generally do not present the same biological hazards as ultraviolet radiation, X-rays, or gamma rays.

When visible light, infrared radiation, or microwaves do cause injury, the damage is usually the result of thermal effects rather than ionization. For example, excessive exposure can heat tissues enough to cause burns or, in some circumstances, contribute to cataract formation by warming the eye's lens. These injuries arise because many photons deposit energy together—not because individual photons are highly energetic.

At extremely high intensities, however, large numbers of photons can generate very strong electric and magnetic fields. Under such unusual conditions, the resulting electromagnetic fields may become strong enough to ionize matter even though the individual photon energies remain low.

Detecting individual photons becomes increasingly difficult as photon energy decreases. For radiation below the microwave region, single photons are so low in energy that they are essentially impossible to detect directly with ordinary instruments. Nevertheless, these photons still exist. A continuous electromagnetic wave can always be viewed as an enormous collection of photons.

For low-frequency electromagnetic radiation, the number of photons involved is so large that treating the radiation as a continuous electric and magnetic wave provides an excellent approximation. This is another example of the correspondence principle, in which quantum behavior naturally approaches the predictions of classical physics when very large numbers of particles—or photons—are involved.

Interactive Exploration: Color Vision

Our perception of color depends on two factors: the wavelength of visible light and the way the photoreceptor cells in our eyes respond to that light. In this interactive simulation, you will investigate how the three primary colors of light—red, green, and blue (RGB)—combine to produce the wide range of colors that humans perceive. You will also explore how monochromatic light and color filters affect the light that ultimately reaches the eye.

As you work through the simulation, experiment with changing the intensity of the red, green, and blue light sources, adjusting the wavelength of monochromatic light, and placing different color filters in front of a white light source. Relate your observations to the photon model of light presented in this chapter and consider how these same physical principles are used in human vision, digital cameras, computer monitors, televisions, and smartphone displays.

Section Summary

• Every photon carries an energy given by [latex]E=hf=\dfrac{hc}{\lambda}[/latex]. As photon frequency increases (and wavelength decreases), photon energy increases.
• Photon energy determines how electromagnetic radiation interacts with matter. Low-energy photons generally produce molecular rotations, vibrations, or heating, whereas high-energy photons can excite electrons, break chemical bonds, or ionize atoms and molecules.
• Ultraviolet radiation, X-rays, and gamma rays are forms of ionizing radiation because individual photons have sufficient energy to remove electrons from atoms and molecules.
• X-rays are commonly produced when high-speed electrons are rapidly decelerated in a metal target. Their maximum photon energy is determined by the accelerating voltage of the X-ray tube.
• Visible-light photons have energies that match electronic transitions in atoms and molecules, making them responsible for color vision and many photochemical processes.
• Infrared radiation and microwaves have photon energies too small to ionize matter. Their biological effects are primarily due to heating produced by the combined action of enormous numbers of photons.
• Because ordinary light sources emit extraordinarily large numbers of photons each second, light usually appears continuous even though it is fundamentally quantized.
• Light exhibits both wave-like and particle-like behavior. Many optical phenomena require the wave description, while interactions between light and matter often require the photon model.

Conceptual Questions

11. Why are ultraviolet radiation, X-rays, and gamma rays collectively called ionizing radiation?
12. How can exposing food to ionizing radiation help prevent spoilage? Since ultraviolet radiation has limited penetrating ability, what other types of radiation could be used for thicker foods?
13. Some older televisions use cathode ray tubes (CRTs), in which electrons are accelerated through a potential difference of approximately 30 kV before striking the screen. Would you expect X-rays to be produced? Explain.
14. Tanning salons often advertise the use of "safe" ultraviolet radiation with longer wavelengths than some of the UV found in sunlight. If this radiation has enough photon energy to stimulate tanning, is it also likely to damage cells and increase cancer risk after prolonged exposure? Explain.
15. Your pupils dilate when the intensity of visible light decreases. If you wear sunglasses that block visible light but not ultraviolet radiation, does this increase or decrease the risk of UV damage to your eyes? Explain.
16. A person standing 75 km from a large atmospheric nuclear explosion could feel the infrared heat but would not be exposed to most of the emitted X-rays or gamma rays. Explain why.
17. Can a single microwave photon damage a cell? Explain your reasoning based on photon energy.
18. For an X-ray tube, the maximum photon energy is often written as [latex]hf=qV[/latex]. Would it be more technically correct to write [latex]hf=qV+\text{BE}[/latex], where BE is the binding energy of electrons in the target material? Why is the simpler expression normally used?

Problems & Exercises

20. What is the energy, in both joules and electron volts, of a photon from an AM radio station broadcasting at 1530 kHz?
21. (a) Find the energy, in joules and electron volts, of photons from an FM radio station broadcasting at 90.0 MHz. (b) Based on your answer, what does this imply about the number of photons emitted each second by the station?
22. Calculate the frequency, in hertz, of a 1.00-MeV gamma-ray photon.
23. (a) What is the wavelength of a 1.00-eV photon? (b) Determine its frequency. (c) Identify the region of the electromagnetic spectrum to which it belongs.
24. Perform the necessary unit conversions to show that [latex]hc=1240\ \text{eV}\cdot\text{nm}[/latex].
25. Verify that visible light, with wavelengths between 380 nm and 760 nm, corresponds to photon energies ranging from approximately 3.26 eV to 1.63 eV.
26. (a) Calculate the energy of an infrared photon having a frequency of [latex]2.00\times10^{13}\ \text{Hz}[/latex]. (b) How many such photons would have to be absorbed simultaneously to break apart a tightly bound molecule? (c) Calculate the energy of a gamma-ray photon with frequency [latex]3.00\times10^{20}\ \text{Hz}[/latex]. (d) Approximately how many tightly bound molecules could one such gamma-ray photon potentially break apart?
27. Show, to three significant figures, that Planck's constant can be written as [latex]h=4.14\times10^{-15}\ \text{eV}\cdot\text{s}[/latex].
28. (a) What is the maximum photon energy produced by a cathode ray tube operating at an accelerating potential of 25.0 kV? (b) Determine the corresponding photon frequency.
29. What accelerating voltage is required for an X-ray tube that produces X-rays with a minimum wavelength of 0.0103 nm?
30. (a) Two microwave ovens operate at frequencies of 950 MHz and 2560 MHz while emitting the same number of photons each second. What is the ratio of their power outputs? (b) If instead they have the same power output, what is the ratio of the number of photons they emit each second?
31. A microwave oven has a power output of 1.00 kW and operates at a frequency of 2560 MHz. How many microwave photons are emitted each second?
32. Some satellites are powered by radioactive sources. (a) If a satellite emits 1.00 W of gamma rays having an average energy of 0.500 MeV, how many gamma-ray photons are emitted each second? (b) How far away must another satellite be to receive, on average, only one gamma-ray photon per second per square meter?
33. (a) A 650-kHz radio station broadcasts with a power of 50.0 kW. How many photons are emitted each second? (b) Assuming the waves spread uniformly in all directions and neglecting atmospheric absorption, how many photons per second per square meter reach a point 100 km away?
34. An X-ray tube produces an X-ray beam with a total power of 1.00 W. Assuming the average photon energy is 75.0 keV, how many X-ray photons are emitted each second?
35. (a) How far from a 650-kHz radio station transmitting 50.0 kW would you have to be so that only one photon per second passes through each square meter? Assume isotropic radiation with no absorption. (b) Discuss what this result suggests about the possibility of detecting radio broadcasts from intelligent civilizations in distant planetary systems.
36. Assume that 10.0% of the output of a 100-W incandescent light bulb is emitted as visible light with an average wavelength of 580 nm. If the light spreads uniformly in all directions and the atmosphere absorbs none of it, how far away would the bulb be if 500 photons per second entered the 3.00-mm-diameter pupil of your eye? (This photon rate is sufficient to stimulate the retina.)
37. Construct Your Own Problem. Consider a laser pointer. Develop and solve a problem in which you calculate the number of photons emitted each second. Include realistic values for the laser's wavelength and power. As an extension, estimate the minimum beam spreading due to diffraction and calculate the photon flux reaching a distant target, taking into account the beam diameter and any absorption or scattering if appropriate.