# The Nature of Electromagnetic Radiation

Submitted by ChemPRIME Staff on Thu, 01/13/2011 - 13:57

Figure 1 Fluctuating electric and magnetic fields associated with electromagnetic waves. A0 is the maximum amplitude of the wave; λ is the wavelength.
Visible light, gamma rays, x-rays, ultraviolet (“black”) light, infrared radiation, microwaves, and radio waves are all related in that many of their properties can be explained by a wave theory. They may be thought of as periodically varying electric and magnetic fields (electromagnetic waves). Figure 1 indicates the relationship of these fluctuating electric and magnetic fields. It also illustrates the maximum amplitude A0 and the wavelength λ associated with the wave. The intensity of a wave is associated with the square of its amplitude.

Electromagnetic waves travel through a vacuum at the speed of light, c = 2.9979 × 108 m s–1. The entire wave shown in Fig. 1 may be thought of as moving from left to right. Thus at position P, where the electric field had maximum amplitude at the instant the figure was drawn there is a progressive decrease in amplitude with time. The amplitude reaches its smallest (most negative) value when the wave has moved a distance equal to that separating points P′ and P. Eventually the amplitude increases to its maximum value again, corresponding to movement of a distance P″ to P, or one wavelength λ. A moving wave can be characterized by the frequencyThe rate at which a periodic event occurs; specifically, the rate at which the waves of electromagnetic radiation pass a point. v at which points of maximum amplitude pass a fixed position. The speed of the wave (distance traveled per unitA particular measure of a physical quantity that is used to express the magnitude of the physical quantity; for example, the meter is the unit of the physical quantity, length. time) must be the productA substance produced by a chemical reaction. of the wavelength (distance between maxima) and the frequency (number of maxima passing per unit time):

c = λv      (1)

Since the speed of electromagnetic radiationEnergy in the form of oscillating, mutually perpendicular electric and magnetic fields. The energy is quantized in units called photons. in a vacuum is always the same, radiation may be characterized by specifying either λ or v. The other quantity can always be calculated from Eq. (1).

EXAMPLE 1 Specify the frequency, wavelength, and speed in a vacuum of each of the types of electromagnetic radiation listed below:

a) Blue-green light; λ = 500 nm.

b) HeatEnergy transferred as a result of a temperature difference; a form of energy stored in the movement of atomic-sized particles. rays emitted from hot asphalt pavement; v = 1.5 × 1014 s–1

c) A gamma ray emitted from ${}_{\text{53}}^{\text{131}}\text{I}$; λ = 3.402 pm.

d) An FM radio transmission; v = 91.5 MHz.

SolutionA mixture of one or more substances dissolved in a solvent to give a homogeneous mixture. In each case we make use of the relationship c = λv = 2.998 × 108 m s–1.

a)      $v=\frac{c}{\lambda }=\frac{\text{2}\text{.998 }\times \text{ 10}^{\text{8}}\text{ m s}^{-\text{1}}}{\text{500 }\times \text{ 10}^{-\text{9}}\text{ m}}=\text{6}\text{.00 }\times \text{ 10}^{\text{14}}\text{ s}^{-\text{1}}$

b)      $\lambda =\frac{c}{v}=\frac{\text{2}\text{.998 }\times \text{ 10}^{\text{8}}\text{ m s}^{-\text{1}}}{\text{1}\text{.5 }\times \text{ 10}^{\text{14}}\text{ s}^{-\text{1}}}=\text{2}\text{.00 }\times \text{ 10}^{-\text{6}}\text{ m}=\text{2}\text{.0 }\mu \text{m}$

c)      $v=\frac{c}{\lambda }=\frac{\text{2}\text{.998 }\times \text{ 10}^{\text{8}}\text{ m s}^{-\text{1}}}{\text{3}\text{.402 }\times \text{ 10}^{-\text{12}}\text{ m}}=\text{8}\text{.812 }\times \text{ 10}^{\text{19}}\text{ s}^{-\text{1}}$

d) The unit hertz Hz is 1 s–1, therefore      $\lambda =\frac{\text{2}\text{.998 }\times \text{ 10}^{\text{8}}\text{ m s}^{-\text{1}}}{\text{91}\text{.5 }\times \text{ 10}^{\text{6}}\text{ Hz}}\text{ }\times \text{ }\frac{\text{1 Hz}}{\text{1 s}^{-\text{1}}}=\text{ 3}\text{.28 m}$

The results obtained in Example 1 indicate that the frequency and wavelength of electromagnetic radiation can vary over a wide range.

The experiments which did most to convince scientists that light could be described by a wave model are concerned with interference. In 1802 Thomas Young (1773 to 1829), an English physicist, allowed light of a single wavelength to pass through a pair of parallel slits very close to each other and then onto a screen. Young observed the interference pattern of alternating dark and bright strips, shown in Fig. 2. Instead of two strips of light, three appeared on the screen, the most prominent being in the center.

Figure 2 Young’s double-slit experiment. The circular lines represent maxima in light waves emitted by the point source. Each slit serves as a secondary source. When maxima from both slits reach the screen at the same time, a bright line is produced. When one maximum and one minimum reach the same point, a dark line appears.

The appearance of these bright and dark strips on the screen is easy to explain if light is regarded as a wave. The bright areas are the result of constructive interference, while the dark ones result from destructive interference. Constructive interference occurs when the crests of two waves reach the same point at the same time. The amplitudes of the two waves add together, giving a resultant larger than either. In the case of destructive interference a maximum in one wave and a minimum in the other reach the same point at the same time. Thus one cancels the effect of the other, and the resultant wave is smaller. This is illustrated in Fig. 3. When destructive interference occurs between two waves which have the same amplitude, the resultant wave has zero amplitude (and zero intensity). Hence the dark strips observed in the double-slit experiment.

Although the behavior of light and other forms of electromagnetic radiation can usually be interpreted in terms of wave motion, this is not always so. When radiation is absorbed or emitted by matterAnything that occupies space and has mass; contrasted with energy., it is usually more convenient to regard it as a stream of particles called photons.

Figure 3 Interference of waves: (a) constructive interference; (b) destructive interference.

Thus electromagnetic radiation has the same kind of wave-particle duality we encountered in the case of the electron. However, photons have some properties which are very different from those of electrons and other particles. Although photons have massA measure of the force required to impart unit acceleration to an object; mass is proportional to chemical amount, which represents the quantity of matter in an object. and energyA system's capacity to do work., and although we can count them, they can travel only at the speed of light. We cannot slow down a photon or stop it without changing it into something else.

The wave-particle duality of photons and electromagnetic radiation is enshrined in an equation first proposed by the German physicist Max Planck (1858 to 1947). The energy of a photon E and the frequency of the electromagnetic radiation associated with it are related in the following way:

E = hv      (2)

where h is a universal constant of nature called Planck’s constant with the value 6.6262 × 10–34 J s. The application of Eq. (2) is best shown by an example.

EXAMPLE 2 Calculate the energy of photons associated with each kind of electromagnetic radiation mentioned in Example 1. Compare each result with the mean bond enthalpyThe change in enthalpy when a mole of chemical bonds of a particular type are broken; the molecules whose bonds are broken must be in the gas phase. Closely related to bond energy. for a C—C single bondAttraction between two atoms (nuclei and core electrons) that results from sharing a single pair of electrons between the atoms; a bond with bond order = 1. (348 kJ mol–1).

Solution In each case use the formula E = hv. If wavelengths are given instead of frequencies, the formula E = hc/λ may be obtained by combining Eqs. (1) and (2).

a)      $E=\frac{hc}{\lambda }=\frac{\text{6}\text{.626 }\times \text{ 10}^{-\text{34}}\text{ J s }\times \text{ 2}\text{.998 }\times \text{ 10}^{\text{8}}\text{ m s}^{-\text{1}}}{\text{500 }\times \text{ 10}^{-\text{9}}\text{ m}}=\text{3}\text{.97 }\times \text{ 10}^{-1\text{9}}\text{ J}=\text{0}\text{.397 aJ}$

b)      $E=hv=\text{6}\text{.626 }\times \text{ 10}^{-\text{34}}\text{ J s }\times \text{ 1}\text{.5 }\times \text{ 10}^{\text{14}}\text{ s}^{-\text{1}}=\text{9}\text{.9 }\times \text{ 10}^{-\text{20}}\text{ J}=\text{0}\text{.099 aJ}$

c)      $E=\frac{\text{6}\text{.626 }\times \text{ 10}^{-\text{34}}\text{ J s }\times \text{ 2}\text{.998 }\times \text{ 10}^{\text{8}}\text{ m s}^{-\text{1}}}{\text{3}\text{.402 }\times \text{ 10}^{-\text{12}}\text{ m}}=\text{5}\text{.839 }\times \text{ 10}^{-1\text{4}}\text{ J}=\text{58 390 aJ}$

d)      $E=\text{6}\text{.626 }\times \text{ 10}^{-\text{34}}\text{ J s }\times \text{ 91}\text{.5 }\times \text{ 10}^{\text{6}}\text{ s}^{-\text{1}}=\text{6}\text{.06 }\times \text{ 10}^{-\text{26}}\text{ J}=\text{6}\text{.06 }\times \text{ 10}^{-\text{8}}\text{ aJ}$

Since the bond enthalpy quoted refers to 1 mol C—C bonds, we must divide by the Avogadro constant to obtain a quantity which is appropriate to compare with the energy of a single quantum of radiation:

Enthalpy to dissociate one C—C bond = $\frac{\text{348 kJ mol}^{-\text{1}}}{\text{6}\text{.022 }\times \text{ 10}^{\text{23}}\text{ mol}^{-\text{1}}\text{ }}$ = 5.78 × 10–19 J = 0.578 aJ

Clearly the energies of visible photons are comparable with the energies of chemical bonds. Infrared and radio waves have far less energy per photon. Gamma-ray photons have enough energy to break open about 100 000 chemical bonds. As a consequence chemical changes often occur when gamma rays or other high-energy photons are absorbed by matter. Such changes are usually detrimental to living systems, and materials such as lead are used to shield humans from sources of high-energy radiation.

The entire spectrum of electromagnetic radiation may be characterized in terms of wavelength, frequency, or energy per photon, as shown in Fig. 4. The example calculations we have done so far and the figure both indicate the broad range covered by λ, v, and E. Electromagnetic radiation which can be detected by the human retina is but a small slice out of the total available spectrum.

Figure 4 The electromagnetic spectrum. The visible region has been expanded in the lower part of the figure so that the wavelengths corresponding to different colors can be distinguished.