Ionizing Radiation

A photon is a quantum of EM radiation. Its energy is given by \(E=\text{hf}\) and is related to the frequency \(f\) and wavelength \(\lambda \) of the radiation by

\(E=\text{hf}=\cfrac{\text{hc}}{\lambda }\text{(energy of a photon),}\)

where \(E\) is the energy of a single photon and \(c\) is the speed of light. When working with small systems, energy in eV is often useful. Note that Planck’s constant in these units is

\(h=\text{4}\text{.}\text{14}×{\text{10}}^{\text{–15}}\phantom{\rule{0.25em}{0ex}}\text{eV}\cdot \text{s}.\)

Since many wavelengths are stated in nanometers (nm), it is also useful to know that

\(\text{hc}=\text{1240 eV}\cdot \text{nm}.\)

These will make many calculations a little easier.

All EM radiation is composed of photons. This figure shows various divisions of the EM spectrum plotted against wavelength, frequency, and photon energy. Previously in this book, photon characteristics were alluded to in the discussion of some of the characteristics of UV, x rays, and \(\gamma \) rays, the first of which start with frequencies just above violet in the visible spectrum. It was noted that these types of EM radiation have characteristics much different than visible light. We can now see that such properties arise because photon energy is larger at high frequencies.

An electromagnetic spectrum is shown. Different types of radiation are indicated using double-sided arrows based on the ranges of their wavelength, energy, and frequency; the visible spectrum is shown, which is a very narrow band. The radio wave region is further segmented into A M radio, F M radio, and Microwaves bands.

The EM spectrum, showing major categories as a function of photon energy in eV, as well as wavelength and frequency. Certain characteristics of EM radiation are directly attributable to photon energy alone.

Representative Energies for Submicroscopic Effects (Order of Magnitude Only)

Rotational energies of molecules	\({\text{10}}^{-5}\) eV
Vibrational energies of molecules	0.1 eV
Energy between outer electron shells in atoms	1 eV
Binding energy of a weakly bound molecule	1 eV
Energy of red light	2 eV
Binding energy of a tightly bound molecule	10 eV
Energy to ionize atom or molecule	10 to 1000 eV

Photons act as individual quanta and interact with individual electrons, atoms, molecules, and so on. The energy a photon carries is, thus, crucial to the effects it has. This table lists representative submicroscopic energies in eV. When we compare photon energies from the EM spectrum in this figure with energies in the table, we can see how effects vary with the type of EM radiation.

Gamma rays, a form of nuclear and cosmic EM radiation, can have the highest frequencies and, hence, the highest photon energies in the EM spectrum. For example, a \(\gamma \)-ray photon with \(f{\text{= 10}}^{\text{21}}\phantom{\rule{0.25em}{0ex}}\text{Hz}\) has an energy \(E=\text{hf}=6.63×{\text{10}}^{\text{–13}}\phantom{\rule{0.25em}{0ex}}\text{J}=4\text{.}\text{14 MeV.}\) This is sufficient energy to ionize thousands of atoms and molecules, since only 10 to 1000 eV are needed per ionization.

In fact, \(\gamma \) rays are one type of ionizing radiation, as are x rays and UV, because they produce ionization in materials that absorb them. Because so much ionization can be produced, a single \(\gamma \)-ray photon can cause significant damage to biological tissue, killing cells or damaging their ability to properly reproduce. When cell reproduction is disrupted, the result can be cancer, one of the known effects of exposure to ionizing radiation. Since cancer cells are rapidly reproducing, they are exceptionally sensitive to the disruption produced by ionizing radiation. This means that ionizing radiation has positive uses in cancer treatment as well as risks in producing cancer.

An x-ray image of Bertha Röentgen’s hand is shown with a dark circular spot superimposed on the fingers.

One of the first x-ray images, taken by Röentgen himself. The hand belongs to Bertha Röentgen, his wife. (credit: Wilhelm Conrad Röntgen, via Wikimedia Commons)

High photon energy also enables \(\gamma \) rays to penetrate materials, since a collision with a single atom or molecule is unlikely to absorb all the \(\gamma \) ray’s energy. This can make \(\gamma \) rays useful as a probe, and they are sometimes used in medical imaging. x rays, as you can see in this figure, overlap with the low-frequency end of the \(\gamma \) ray range. Since x rays have energies of keV and up, individual x-ray photons also can produce large amounts of ionization.

At lower photon energies, x rays are not as penetrating as \(\gamma \) rays and are slightly less hazardous. X rays are ideal for medical imaging, their most common use, and a fact that was recognized immediately upon their discovery in 1895 by the German physicist W. C. Roentgen (1845–1923). (See this figure.) Within one year of their discovery, x rays (for a time called Roentgen rays) were used for medical diagnostics. Roentgen received the 1901 Nobel Prize for the discovery of x rays.

Connections: Conservation of Energy

Once again, we find that conservation of energy allows us to consider the initial and final forms that energy takes, without having to make detailed calculations of the intermediate steps. This example is solved by considering only the initial and final forms of energy.

A cathode ray tube connected to a high-voltage source is shown in the figure. The image shows electrons coming out of the heated filament at one end of the vacuum tube as tiny balls, and hitting the metal plate at the opposite end of the vacuum tube. X rays are shown coming out from the metal plate in the form of waves.

X rays are produced when energetic electrons strike the copper anode of this cathode ray tube (CRT). Electrons (shown here as separate particles) interact individually with the material they strike, sometimes producing photons of EM radiation.

While \(\gamma \) rays originate in nuclear decay, x rays are produced by the process shown in this figure. Electrons ejected by thermal agitation from a hot filament in a vacuum tube are accelerated through a high voltage, gaining kinetic energy from the electrical potential energy. When they strike the anode, the electrons convert their kinetic energy to a variety of forms, including thermal energy. But since an accelerated charge radiates EM waves, and since the electrons act individually, photons are also produced. Some of these x-ray photons obtain the kinetic energy of the electron. The accelerated electrons originate at the cathode, so such a tube is called a cathode ray tube (CRT), and various versions of them are found in older TV and computer screens as well as in x-ray machines.

Example: X-ray Photon Energy and X-ray Tube Voltage

Find the maximum energy in eV of an x-ray photon produced by electrons accelerated through a potential difference of 50.0 kV in a CRT like the one in this figure.

Strategy

Electrons can give all of their kinetic energy to a single photon when they strike the anode of a CRT. (This is something like the photoelectric effect in reverse.) The kinetic energy of the electron comes from electrical potential energy. Thus we can simply equate the maximum photon energy to the electrical potential energy—that is, \(\text{hf}=\text{qV.}\) (We do not have to calculate each step from beginning to end if we know that all of the starting energy \(\text{qV}\) is converted to the final form \(\text{hf.}\))

Solution

The maximum photon energy is \(\text{hf}=\text{qV}\), where \(\text{q}\) is the charge of the electron and \(\text{V}\) is the accelerating voltage. Thus,

\(\text{hf}=(1\text{.}\text{60}×{\text{10}}^{\text{–19}}\phantom{\rule{0.25em}{0ex}}\text{C})(\text{50.0}×{\text{10}}^{3}\phantom{\rule{0.25em}{0ex}}\text{V})\text{.}\)

From the definition of the electron volt, we know \(\text{1 eV}=\text{1}\text{.}\text{60}×{\text{10}}^{\text{–19}}\phantom{\rule{0.25em}{0ex}}\text{J}\), where \(\text{1 J}=1 C\cdot V.\) Gathering factors and converting energy to eV yields

\(\text{hf}=(\text{50.0}×{\text{10}}^{3})(1.60×{\text{10}}^{\text{–19}}\phantom{\rule{0.25em}{0ex}}\text{C}\cdot \text{V})\left(\cfrac{\text{1 eV}}{1.60×{\text{10}}^{\text{–19}}\phantom{\rule{0.25em}{0ex}}\text{C}\cdot \text{V}}\right)=(\text{50.0}×{\text{10}}^{3})(\text{1 eV})=\text{50.0 keV.}\)

Discussion

This example produces a result that can be applied to many similar situations. If you accelerate a single elementary charge, like that of an electron, through a potential given in volts, then its energy in eV has the same numerical value. Thus a 50.0-kV potential generates 50.0 keV electrons, which in turn can produce photons with a maximum energy of 50 keV. Similarly, a 100-kV potential in an x-ray tube can generate up to 100-keV x-ray photons. Many x-ray tubes have adjustable voltages so that various energy x rays with differing energies, and therefore differing abilities to penetrate, can be generated.

A graph for X-ray intensity versus frequency is shown. Frequency is plotted along x axis and intensity along y axis. The curve has a smooth rise up then at highest point it has two peaks and ends smoothly at f sub max. q V is equal to h f sub max is written in the graph.

X-ray spectrum obtained when energetic electrons strike a material. The smooth part of the spectrum is bremsstrahlung, while the peaks are characteristic of the anode material. Both are atomic processes that produce energetic photons known as x-ray photons.

This figure shows the spectrum of x rays obtained from an x-ray tube. There are two distinct features to the spectrum. First, the smooth distribution results from electrons being decelerated in the anode material. A curve like this is obtained by detecting many photons, and it is apparent that the maximum energy is unlikely. This decelerating process produces radiation that is called bremsstrahlung (German for braking radiation).

The second feature is the existence of sharp peaks in the spectrum; these are called characteristic x rays, since they are characteristic of the anode material. Characteristic x rays come from atomic excitations unique to a given type of anode material. They are akin to lines in atomic spectra, implying the energy levels of atoms are quantized. Phenomena such as discrete atomic spectra and characteristic x rays are explored further in Atomic Physics.

Ultraviolet radiation (approximately 4 eV to 300 eV) overlaps with the low end of the energy range of x rays, but UV is typically lower in energy. UV comes from the de-excitation of atoms that may be part of a hot solid or gas. These atoms can be given energy that they later release as UV by numerous processes, including electric discharge, nuclear explosion, thermal agitation, and exposure to x rays. A UV photon has sufficient energy to ionize atoms and molecules, which makes its effects different from those of visible light. UV thus has some of the same biological effects as \(\gamma \) rays and x rays.

For example, it can cause skin cancer and is used as a sterilizer. The major difference is that several UV photons are required to disrupt cell reproduction or kill a bacterium, whereas single \(\gamma \)-ray and X-ray photons can do the same damage. But since UV does have the energy to alter molecules, it can do what visible light cannot. One of the beneficial aspects of UV is that it triggers the production of vitamin D in the skin, whereas visible light has insufficient energy per photon to alter the molecules that trigger this production. Infantile jaundice is treated by exposing the baby to UV (with eye protection), called phototherapy, the beneficial effects of which are thought to be related to its ability to help prevent the buildup of potentially toxic bilirubin in the blood.

Example: Photon Energy and Effects for UV

Short-wavelength UV is sometimes called vacuum UV, because it is strongly absorbed by air and must be studied in a vacuum. Calculate the photon energy in eV for 100-nm vacuum UV, and estimate the number of molecules it could ionize or break apart.

Strategy

Using the equation \(E=\text{hf}\) and appropriate constants, we can find the photon energy and compare it with energy information in this table.

Solution

The energy of a photon is given by

\(E=\text{hf}=\cfrac{\text{hc}}{\lambda }.\)

Using \(\text{hc}=\text{1240 eV}\cdot \text{nm,}\) we find that

\(E=\cfrac{\text{hc}}{\lambda }=\cfrac{\text{1240 eV}\cdot \text{nm}}{\text{100 nm}}=\text{12}\text{.}\text{4 eV}.\)

Discussion

According to this table, this photon energy might be able to ionize an atom or molecule, and it is about what is needed to break up a tightly bound molecule, since they are bound by approximately 10 eV. This photon energy could destroy about a dozen weakly bound molecules. Because of its high photon energy, UV disrupts atoms and molecules it interacts with. One good consequence is that all but the longest-wavelength UV is strongly absorbed and is easily blocked by sunglasses. In fact, most of the Sun’s UV is absorbed by a thin layer of ozone in the upper atmosphere, protecting sensitive organisms on Earth. Damage to our ozone layer by the addition of such chemicals as CFC’s has reduced this protection for us.

Ionizing Radiation