2 X-ray Interactions

Question 1

The number of electrons each shell

Answer 1

–The number of electrons each shell can contain is 2n2.

Question 2

The electron density of a substance

Answer 2

–The electron density of a substance is ρ N0(Z/A) electrons/cm3, where ρ is the density measured in grams per cubic centimeter (g/cm3) and N0 is Avogadro’s number.

Question 3

Z/A is equal to

Answer 3

–For most atoms making up tissues (e.g., oxygen, carbon, nitrogen, calcium), Z/A is equal to 0.5.

Question 4

The electron binding energy

Answer 4

–The energy required to completely remove an electron from an atom.

Question 5

Characteristic x-ray

Answer 5

–Electrons moving from an outer shell to an inner shell may emit excess energy as electromagnetic radiation.

Question 6

Auger electron

Answer 6

–The excess energy may be transferred to an Auger electron, which then leaves the atom.
–Auger electron energy is the characteristic x-ray energy minus the binding energy of the outer shell electron.

Question 7

Ionization

Answer 7

–Ionization occurs when an electron is ejected from a neutral atom, leaving behind a positive ion.

Question 8

Directly ionizing

Answer 8

–Radiation is directly ionizing when it is in the form of charged particles.
–Electrons and protons are both directly ionizing radiations.

Question 9

Excitations

Answer 9

–Energy lost from energetic particles can eject electrons from atoms or raise atomic electrons to more distant atomic shells.

Question 10

Indirectly ionizing

Answer 10

–Uncharged particles are indirectly ionizing.
–Neutrons are indirectly ionizing radiations that interact with matter by first transferring energy to protons.
–Indirectly ionizing radiations include x-rays, and gamma rays.

Question 11

Three ways that diagnostic energy x-rays interact with matter

Answer 11

–Three ways that diagnostic energy x-rays interact with matter are (i) coherent scatter, (ii) photoelectric (PE) effect, and (iii) Compton scatter.

Question 12

Coherent scatter

Answer 12

–Coherent scatter occurs when a low-energy x-ray photon is scattered from an atom without any energy loss.
–The wavelength of the scattered photon is the same as the wavelength of the incident photon.
–Coherent scatter is referred to as Rayleigh or classical scatter.

Question 13

Photoelectric effect

Answer 13

–This occurs between tightly bound (inner shell) electrons and incident x-ray photons.
–The x-ray photon is totally absorbed by an inner shell electron and that electron is ejected from the atom.
–X-ray photons absorbed in PE interaction therefore “disappear”.
–As a result of the photoelectric interaction, a photoelectron is emitted and a positive atomic ion is left behind.
–The energy of the emitted photoelectron equals the difference between the incident photon energy and the electron binding energy.
–The photoelectron loses energy by ionizing other atoms in the tissue and contributes to patient dose.

Question 14

Auger electron

Answer 14

–Outer shell electrons fill the inner shell electron vacancies, with the excess energy emitted as a characteristic x-ray or Auger electron.
–Auger electron energy is slightly lower than the characteristic x-ray energy.

Question 15

Photoelectric effect probability

Answer 15

–Above the K-edge, photoelectric interactions are proportional to 1/E3.
–The probability of photoelectric absorption increases with atomic number and is proportional to Z3.

Question 16

Compton scatter

Answer 16

–Incident photons interact with outer shell electrons.
–A Compton interaction results in a scattered photon that has less energy than the incident photon and generally travels in a different direction.
–A scattered (ejected or recoil) electron carries the energy lost by the incident photon as kinetic energy.

Question 17

Compton interaction probability

Answer 17

–Electron density.
–Physical density.
–(1/E).

Question 18

The linear attenuation coefficient (μ)

Answer 18

–The fraction of incident photons removed from the beam in traveling unit distance.
–Expressed in inverse centimeters (cm−1).

Question 19

The linear attenuation coefficient (μ) affected by

Answer 19

–Increases with increasing physical density.
–Increase with increasing atomic number.
–Attenuation decreases with increasing photon energy.

Question 20

The mass attenuation coefficient

Answer 20

–The linear attenuation coefficient (μ) divided by the density (ρ).
–Mass attenuation coefficient (μ/ρ) is independent of physical density.

Question 21

Half-Value Layer

Answer 21

–The HVL is the thickness of material that attenuates an x-ray beam by 50%.
–HVL increases with increasing photon energy and decreases with increasing atomic number.

Question 22

Filters

Answer 22

–Filters are added to the x-ray tube window to preferentially absorb low-energy photons.
–Filtration does not affect the maximum energy of the x-ray beam spectrum.
–Filtration always reduces the x-ray tube output.

Question 23

Beam hardening

Answer 23

–The effect of a filter on a polychromatic x-ray beam containing a range of x-ray photon energies.
–Adding filters changes the x-ray spectrum.
–The x-ray beam intensity (i.e., output) is decreased with increased filtration, but the average x-ray energy is increased.
–Hard beams are produced at high voltages using heavy filtration.
–Soft beams are produced at low voltages using less filtration.

Question 24

X-ray beam quality

Answer 24

–The ability of an x-ray beam to penetrate the patient.

Question 25

Heel effect

Answer 25

–This attenuation is greater in the anode direction than in the cathode direction because of differences in the path length within the target.
–results in higher x-ray intensity at the cathode end and lower x-ray intensity at the anode end of the beam.

Question 26

To reduce the heel effect

Answer 26

–The anode angle should be increased, SID increased, and field size decreased.

Question 27

Air gaps and scatter

Answer 27

–Air gaps between the patient and cassette reduce scatter.

Question 28

Antiscatter grids

Answer 28

–Antiscatter grids are made up of many narrow parallel bars of lead or other highly attenuating material.
–Antiscatter grids are used to removing scatter in diagnostic radiology.

Question 29

The grid ratio

Answer 29

–The grid ratio is the ratio of the strip height (h) along the x-ray beam direction to the gap (D) between the lead strips, so the grid ratio is h/D.
–Grid ratios typically range from 4 to 16.

Question 30

The strip line density

Answer 30

–1/(D+d) lines per unit length, where d is the strip thickness.
–Strip line densities range from 25 to 60 lines per centimeter.

Question 31

Primary transmission

Answer 31

–The percentage of incident primary radiation (i.e., not scattered) that passes through the grid.

Question 32

The Bucky factor

Answer 32

– The ratio of radiation incident on the grid to the transmitted radiation.
–The Bucky factor is the increase in patient dose due to the use of a grid.
–Typical values for the Bucky factor range between 2 and 6.

Question 33

The contrast improvement factor

Answer 33

– The ratio of contrast with a grid to contrast without a grid.
–Contrast improvement factors are ∼2.

Question 34

Grid ratio

Answer 34

–Grid ratios may be increased either by increasing the height of the lead strips or reducing the space between the lead strips.
–Increasing the grid ratio increases x-ray tube loading and patient exposure.
–Increasing the grid ratio increases image contrast.

Question 35

Primary transmission

Answer 35

–The percentage of incident primary radiation (i.e., not scattered) that passes through the grid.

Question 36

The Bucky factor

Answer 36

–The ratio of radiation incident on the grid to the transmitted radiation.
–The increase in patient dose due to the use of a grid.
–Typical values for the Bucky factor range between 2 and 6.

Question 37

The contrast improvement factor

Answer 37

– The ratio of contrast with a grid to contrast without a grid.

Question 38

Air kerma

Answer 38

–It is a source-related term used to quantify the x-ray beam intensity.
–Air kerma stands for the Kinetic energy released per unit mass.
–Air kerma is the kinetic energy transferred from uncharged particles (e.g., photons) to charged particles (i.e., electrons).
–The unit of air kerma is joules per kilogram (J/kg), where 1 J/kg is 1 gray (Gy).

Question 39

Intensity

Answer 39

–Intensity is the amount of radiation, and is directly related to the number of x-ray photons.

Question 40

Exposure

Answer 40

–The total charge of electrons liberated per unit mass of air by the x-ray photons.
–Exposure is measured in coulombs per kilogram (C/kg) in the SI system or in roentgens (R) in non-SI units.
–1 R = 2.58 × 10−4 C/kg

Question 41

Absorbed dose (D)

Answer 41

–Absorbed dose (D) measures the amount of radiation energy (E) absorbed per unit mass (M) of a medium: D = E/M.
–Absorbed dose is specified in gray (Gy) in the SI system.
–One gray is equal to 1 J of energy deposited per kilogram.

Question 42

The relation between absorbed dose (D) and air kerma (K)

Answer 42

–D = R × K.

–R depends on the characteristics of the medium irradiated, primarily the atomic number (Z) of the absorbing medium.

Question 43

The outer shell binding energy is

Answer 43

∼5 eV (i.e., 0.005 keV).

Question 44

It takes 30 eV or so to produce

Answer 44

One ionization.

Question 45

Coherent scatter never accounts for

Answer 45

More than 5% of all interactions in diagnostic radiology.

Question 46

In bone, Compton and photoelectric interactions are equally probable at

Answer 46

40 keV.