Ouhsc Exposure & Instrumentation Test 2

Front 1

In diagnostic range of x-ray energies, target interactions produce _____ heat and _____ x-ray photons

Back 1

99%; 1%

Front 2

Characteristic Interactions (inside the tube)

Back 2

- electron from cathode interacts w/ inner shell electron on focal track of anode
- removes electron from atom
- loses the energy necessary to remove the electron

Front 3

What is the energy of a characteristic interaction equal to?

Back 3

The difference in the binding energy of the orbital electrons involved

Front 4

X-rays generated from the K Shell of __________ atoms are the only energies of strength that can be used for characteristic radiation.

Back 4

Tungsten

Front 5

Bremsstrahlung Interactions (inside tube)

Back 5

- incident electron from cathode passes near nucleus of the tungsten atom on anode, completely bypassing all electrons
- electrostatic attraction between incident electron & nucleus occurs

Front 6

Braking Radiation

Back 6

When Bremsstrahlung's interact w/ tungsten, it's Z number makes it slow down, lose some energy, & change direction

Front 7

The closer the electron goes to the nucleus, the _________ (more/less) energy the electron loses, the ___________ (higher/lower) amount of Bremsstrahlung produced.

Back 7

more; higher

Front 8

Polyenergetic

Back 8

X-rays produced at many different energies

Front 9

What's an x-ray emission spectrum?

Back 9

A graph showing the numbers of photons produced at particular ranges of energies

Front 10

Characteristic X-ray Spectrum

Back 10

Discrete energy levels representing the characteristic of each electron shell

Front 11

Bremsstrahlung X-ray Spectrum

Back 11

Bell-shaped curve ranging from 0 to max kV setting

Front 12

Factors Affecting Emission Spectrum - change in mA

Back 12

A change in mA results in a directly proportional change in the amplitude of emission spectrum

- increasing mA will increase the number of incident electrons flowing from cathode to anode

Front 13

Factors Affecting Emission Spectrum - change in kVp

Back 13

- changes will affect amplitude & position of x-ray emission spectrum
- increase in kVp = increase amplitude w/ a curve shift right

Front 14

Factors Affecting Emission Spectrum - Target Material Effect

Back 14

As atomic number increases, efficiency of the production of Bremsstrahlung radiation increases

More high energy x-rays are produced

Front 15

Factors Affecting Emission Spectrum - Changes in Voltage Waveform

Back 15

- operation w/ 3-phase power is equivalent to a 12% increase in kVp over single-phase (almost doubles the mAs)
- more efficient waveform = increased amplitude

Front 16

Factors Affecting Emission Spectrum - Filtration

Back 16

- increases effective energy of x-ray beam by attenuating low-energy photons
- improves overall quality of x-ray beam

Front 17

Filtration

Back 17

Process of eliminating undesirable low-energy x-ray photons by insertion of absorbing materials into primary beam

Reduces patient dose

Front 18

What's the standard filtering material?

Back 18

Aluminum

Front 19

Aluminum Equivalent

Back 19

- any other material other than aluminum that's going to attenuate at the same rate
ex. Copper is slightly more dense, less of it is needed to attenuate the same amount

Front 20

Half Value Layer (HVL)

Back 20

- amount of absorbing material that will reduce the intensity (# of photons) of the primary beam to one-half its original value

1 HVL = 2.5 mm Al/Eq for 100 kVp beam

Front 21

Inherent Filtration

Back 21

- filtration that's a result of composition of tube & housing
- between target & exit port of glass envelope
- can't be altered, remains constant
- typical tube has 0.5-1.0 mm Al/Eq of filtration

Front 22

Added Filtration

Back 22

- any filtration occurring outside the tube & housing and before patient is considered to be added
- ex. filtration occurring in collimator

Front 23

Compound Filtration

Back 23

- composite filters using 2 or more materials to absorb photons
- layered w/ highest atomic number material on top; subsequent layers absorb characteristic photons producing by previous layer

Front 24

Thoreaus Filter

Back 24

- form of compound filter
- highest atomic number = tin
- middle atomic number = copper
- lowest atomic number = aluminum
- see this in radiation therapy

Front 25

Compensation Filtration

Back 25

- external filter used to compensate for unequal absorption due to part density/thickness
- produces more uniform density
- does NOT reduce patient dose
- ex. wedge filters

Front 26

Total Filtration

Back 26

- equal to the sum of inherent & added filtration
- doesn't include compound or compensating filters

Front 27

Recommended Minimum Total Filtration Levels - kVp below 50

Back 27

0.5 mm aluminum

Front 28

Recommended Minimum Total Filtration Levels - 50-70 kVp

Back 28

1.5 mm aluminum

Front 29

Recommended Minimum Total Filtration Levels - kVp above 70

Back 29

2.5 mm aluminum

Front 30

Increasing filtration will _________ (increase/decrease) radiographic density

Back 30

Decrease

Front 31

Significant soft tissue penetration occurs between _________ kVp

Back 31

30-40

Front 32

Prime Factors

Back 32

milliampere (mA), time, kilovolt peak (kVp), distance

factors the radiographer controls that affect x-ray emission

Front 33

Roentgen (R)

Back 33

Unit of measurement of x-ray quantity or intensity

Front 34

X-ray Quality

Back 34

- measurement of penetrating ability of x-ray beam
- represented by HVL
- factors that directly affect x-ray quality are kVp & filtration

Front 35

Millampere (mA)

Back 35

determine number of x-rays produced; radiation QUANTITY

Primary controller radiographic density; doubling the mAs doubles the density produced

Front 36

Law of Reciprocity

Back 36

An increase in mA requires a decrease in exposure time if the same degree of blackening on the film is desired

Blackening on the film remains the same as long as the total energy exposing the film is constant

Front 37

Equation for Law of Reciprocity

Back 37

mA1/mA2 = S2/S1

Front 38

Kilovoltage Peak (kVp)

Back 38

- primary controller of beam QUALITY (penetrability)
- controls the scale of contrast on a finished radiograph (relates to number of shades of gray & difference between them)

Front 39

High kVp / _____ contrast (high/low) / ______ scale of contrast (short/long)

Back 39

low; long

Front 40

Low kVp / _______ contrast (high/low) / ________ scale of contrast (short/long)

Back 40

high; short

Front 41

As we increase kVp, the emission spectrum is shifted ________ (right/left) because of the increase in energy

Back 41

Right

Front 42

The higher the kVp, the ___________ (higher/lower) the energy of the electrons, the _________ (higher/lower) the optical density

Back 42

higher; higher

Front 43

15% Rule

Back 43

An increase in kVp by 15% will approx. double the overall density/exposure on film

Front 44

While trying to maintain density, an increase in kVp by 15% forces you to ________ (double/halve) the mAs

Back 44

Halve

Reverse is also true

Front 45

Increasing kVp increases.....

Back 45

Penetrability

Front 46

Beam intensity decreases with a(n) ___________ (increase/decrease) in distance

Back 46

Increase

Front 47

X-ray beam intensity/exposure is measured in...

Back 47

Roentgens (R) or milliroentgens (mR)

Front 48

Inverse Square Law

Back 48

- deals with x-ray beam intensity
- as distance increases the amount of exposure received by patient or tech decreases

Front 49

There is a(n) _________ (direct/inverse) relationship between distance & radiation exposure

Back 49

Inverse

Front 50

Inverse Square Law Formula

Back 50

I1/I2 = D2^2/D1^2

To determine amount of exposure...

I2 = I1 x (D1^2/D2^2)

Front 51

Density Maintenance

Back 51

- provides a mathematical method to compensate for changes in SID
- determines the amount of mAs necessary to provide enough photons to create an image after a change in distance

Front 52

Density Maintenance Formula

Back 52

Old mAs/New mAs = Old Distance^2/New Distance^2

Use this formula when you're asked to solve for mAs

Front 53

Doubling the distance will require ___ times the original mAs to maintain density

Back 53

4

Front 54

Attenuation

Back 54

Reduction in the number of x-ray photons in the beam

Front 55

5 Basic Interactions between X-rays & Matter

Back 55

- Coherent Scattering
- Photoelectric Absorption
- Compton Scattering
- Pair Production
- Photodisintegration

Front 56

Photoelectric Absorption

Back 56

- incident photon energy is completely absorbed by inner shell electron
- x-ray photon should have a slightly higher energy than a K shell electron
- causes an ion pair

Front 57

Photoelectron Energy Equation

Back 57

Ei = Eb + Eke

Front 58

What does the ionized atom cause?

Back 58

Characteristic Cascade

- atom electrically unstable after losing electron
- when another electron falls down to fill a hole, it emits a characteristic photon when they change energy levels

Front 59

Secondary Radiation

Back 59

- radiation that originates from irradiated material outside x-ray tube
- production similar to characteristic x-ray production within target
- amount is directly proportional to atomic number

Front 60

As photon energy increases, chance of photoelectric interaction ___________ (increases/decreases) dramatically

Back 60

Decreases

Front 61

If the atomic number is doubled, the chance of photoelectric absorption increases by a factor of _____

Back 61

8

Front 62

Coherent Scatter

Back 62

- low energy photons (below 10 keV)
- electrons get excited from incident photon & vibrate @ photon frequency
- no electrons ejected; no ionization

Front 63

Thompson Scatter

Back 63

Single outer-shell electron used

Front 64

Rayleigh Scatter

Back 64

All electrons of atom used

Front 65

Product of Coherent Scatter

Back 65

Photon released with equal energy to incident photon but traveling in a different direction

Front 66

Compton Scatter

Back 66

- incident photon interacts w/ outer shell, loosely bound electron & ejects it
- ion pair is formed
- photon transfer some energy to the recoil (Compton) electron & continues in a different direction
- most relevant to diagnostic imaging

Front 67

Compton Scatter Equation

Back 67

Ei = Es + Eb + Eke

Eke = recoil electron
Es = scattered photon

Front 68

What is the source of the most occupational exposure & radiation fog?

Back 68

Compton Scatter

Front 69

Backscatter

Back 69

Scatter emitted in the exact opposite direction as incident beam

Front 70

Pair Production

Back 70

- incident photon energy must be 1.02 MeV or higher
- photon energy absorbed by nucleus, causing it to be unstable
- it release a positron & negatron
- negatron will join with another atom; positron is unstable antimatter

Front 71

What happens when a positron combines with an electron?

Back 71

Annihilation reaction

Results = 2 photons of .511 MeV traveling at 180 degrees to each other

Front 72

Pair production is more effective in....

Back 72

Radiation Therapy

Front 73

Photodisintegration

Back 73

- extremely high energy photon (10 MeV or greater)
- nucleus absorbs photon; release alpha particle

Front 74

Which interactions have a less likelihood of occurring with an increase in kVp?

Back 74

Photoelectric absorption, Compton interactions

Front 75

What 2 circumstance would photoelectric absorption likely happen?

Back 75

1) Lower energy ranges (40-70 kVp)
2) In elements with higher Z numbers

Image will have short scale contrast (low kVp, high mAs) [opposite for Compton]