Physics Flashcards
what factors affect image resolution?
- signal to noise ratio (increases with # of photons– ie ^ mAs)
- kVp (contast slightly decreases with increased kVp–higher energy photons in beam, more penetrating, less photoelectric effect)
- size of lesion (sharper reconstruction kernel will increase contrast in small lesions, BUT will increase noise)
- increasing slice thickness
how does harmonic US work
The transducer sends a lower frequency pulse, but only receives the 1st order harmonic (double the incident frequency) echoes. This leads to:
increased lateral resolution
reduction of side lobe artifacts and reverberation artifacts caused by tissues close to the transducer.
Electron binding energy is what
amount of energy needed to remove electron completely from the atom
(also called ionization potential)
measured in eV (1 eV = 1.6 x 10-19 J)
directly proportional to the atomic number (Z)
indirectly proportional to the distance from nucleus, i.e. inner-shell electrons have greater binding energy than outer-shell electrons
An electron can only be removed from an atom if the applied energy is greater than its electron binding energy. When an inner-shell electron is ejected, the vacancy will be filled by an electron from an outer shell. The excess energy from this shift is emitted as electromagnetic radiation.
Coherent/Raleigh scatter
unmodified, classic, elastic scatter
Upon interacting with the attenuating medium, photon does not have enough energy to liberate the electron from its bound state (i.e. the photon energy is well below the binding energy of the electron) so no energy transfer occurs. No energy deposition, thus no dose from coherent scattering.
Film screen systems
create image by? (exposure)
Dose required?
image exposed by light photons created from x rays
decrease required dose but at expense of spatial resolution (light exposes larger area of the film per photon)
syringe shields
- for radiopharmaceuticals
- Fall under principle of ALARA
- syringe and vial must be labeled with radiopharmaceutical, also shield unless can see labels on the vial/syringe
inverse square law?
I ~ 1/d2
When X rays enter the patient tissue, what interactions are possible?
- Coherent scatter (doesn’t account for many interactions in Dx Rads, happens at lower energy radiation)
- Compton scatter
- Photoelectric absorption (what we want)
kVp?
Kilovoltage peak
voltage applied across x ray tube (cathode–> anode)
determines penetration of x ray beam and affects image contrast by determining maximum energy of x ray photons produced (ie applying 60 kV would result in xray spectrum with max energy 60 keV
if kVp is 100, what is the average energy (keV) of emitted photons in the x ray beam
~33-45 keV
(on average, 1/3-1/2 the energy of the kVp)
What happens at higher kVp/higher keV?
Radiography: Dose ∝ kVp2
CT: Dose ∝ kVp2.6
greater intensity of X ray beam–> more rays penetrate patient –> giving LESS contrast between densities
(this is why mammo uses lower kVp–need better contrast between tissue densities but photons can travel less far before being attenuated by tissue)
doubling kVp QUADruples dose
what is the photoelectric effect(aka photoelectric absorption/PEA)?
one of the 3 forms of photon interaction in tissues
a low energy photon interacts with an electron in the atom and removes it from its shell
Most likely to occur when energy of the incident photon is equal to or just greater than the binding energy of the electron in its shell (k edge) and the electron is tightly bound (as in K shell)
electron that is removed is called a photoelectron and incident photon is completely absorbed. Hence, the photoelectric effect contributes to the attenuation of the x-ray beam as it passes through matter/tissue
To stabilize the atom an outer shell electron fills the vacancy in the inner shell. The energy which is lost by this electron as it drops to the inner shell is emitted as characteristic radiation (an x-ray photon) or as an Auger electron
why is lower kVp better for contrast?
and higher kVp bad for contrast?
more photoelectric effect
higher kVp–> more compton scatter
mA/s?
milli-amperes (/second)
the current applied across the xray tube mA X exposure time s
determines # of electrons produced at the cathode per second
thus high mA increases number of photons(radiation) produced
directly proportional to dose (dose ~ mA)
if you increase kVp, what will x ray machine do automatically?
decrease mA
- so you produce fewer electrons–> producing fewer photons–> BUT they will be higher intensity/more penetrating
- decreases dose to patient
- decreases contrast in image
what is absorbed dose?
the amount of radiation absorbed per unit mass of tissue
(not all rays will be absorbed bc scatter)
was measured in rads, now measured in Gy
1 Gy = 1 J/kg
(~3mGy average annual background dose, ~5Gy acutely pretty much fatal)
Equivalent dose?
not all x rays have same degree of negative effect on our cells
Alpha particles + neutrons more damaging than x rays, gamma rays, electrons
multiply absorbed dose by weighting factor WR
In diagnostic rads, WR is 1, measured in Sv
1 Sv= 1 J/kg = 100mrem
what is effective dose
some tissues receive higher dose than others
accounting for stochastic effects in different organs
weighting factor WT for each tissue, all of these sum to 1 (ie if you radiate the whole body)
measured in Sv
effective dose=equivalent dose if you radiate entire body
**effective dose in CT is DLP x WT for imaged area: eg WT for head is 0.0015, for abdomen = 0.0021
Rad? Rem? Gy? Sv?
relationships?
Rad= absorbed dose, now measured in Gy
100 rad = 1 Gy
Rem= equivalent dose, now measured in Sv
100 rem = 1 Sv
X ray production occurs in the tube, how?
Characteristic and bremstrahhlung
interaction with K-shell electron: causes the production of characteristic radiation
interaction with nucleus: causes bremsstrahlung radiation
interaction with outer shell electrons: causes line spectrum
**less than 1% of the colliding electron energy on the anode creates x rays–>99% dissipates as heat (this is why we use materials with high melting point like tungsten)
what is Compton scatter?
one of the 3 forms of photon interaction
interaction of the photon (x-ray or gamma) with free electrons (unattached to atoms) or loosely bound valence shell (outer shell) electrons. The resultant incident photon is scattered (changes direction) and imparts energy to the electron (recoil electron)
scattered photon has a different wavelength and thus a different energy (E=hc/λ)
Depends on:
- directly proportional to number of outer shell electrons, i.e. the electron density, physical density of the material
- inversely proportional to photon energy
- does not depend on atomic number (unlike photoelectric effect and pair production)
Becomes dominant process in human tissues at 30 keV to 30 meV (doses used in dx and therapeutic rads)
Finite size of x ray focal spot gives rise to?
Penumbra (geographic blurring)
–> this is why we use a smaller focal spot in magnification images
Anode angle and anode heel effect
X rays produced on the focal spot of the anode only (small, rectangular)–
Most have anode angle 12-15*
Smaller anode angle= smaller focal spot (more heat generated)
Heel effect: because of the beveled surface, the rays that must travel farther in and back out of the anode//target material to reach the exit window and contribute to the beam will have lower energy (and their emission angles are more perpendicular to the electron beam). Also there will be fewer of them in general as more are absorbed by target material, which leads to higher intensity of beam at cathode end-
- anode angle: by increasing the angle, the amount of target material perpendicular to the anode is decreased resulting in less resorption of x-rays produced.
- target-to-film distance: increase in distance reduces heel effect by allowing more divergence of the beam which produces a more uniform image.
- field size: the field will be more uniform at the center (i.e. smaller field size) due to the collimator absorbing the peripheral variations.
- positioning: by aligning higher attenuating material towards the cathode and lower attenuating material towards the anode the resulting field is more uniform
what is this and what effect does this have on noise and image quality?
decreased source to object distance (SOD)
also decreases SID (source to image distance) –SID = SOD + ODD
Detector and patient are closer to the source = patient receives higher dose of radiation, image noise is reduced
practical applications?
ie what will increase unsharpness
- larger focal spot
- patient farther from detector (increased ODD-also decreases dose)
- focal spot closer to patient (decreased SOD)
Left- more blur with decreased SOD Middle- less blur with increased SOD Right theoretical, no blur with point source as focal spot
applications?
Patient farther from detector = more magnification
Focal spot closer to patient = more magnification
E.g. If SOD is 1 meter, and ODD is 2 meters, then SID must be 3 meters, so 3/1 = 3x mag
What is fogging artifact?
Addition of unintended charge to the detector from something other than the original x-ray beam, causing increased “blackening” of the exposed portion of the image
ie: patient with implanted radioactive seeds to treat a lung tumor in the right lower lung results in extra radiation exposure to the detector in this area causing fogging
how to differentiate underpenetrated from underexposed x ray?
both appear ‘whiter’
Differentiate from penetration by looking at the relative visibility of different structures (e.g. a chest x-ray will appear whiter if underpenetrated or underexposed, but the vertebral bodies will not be visible in an underpenetrated chest x-ray, and will be visible in an underexposed x-ray)
–underexposed will also have quantum mottle from too little photons in general
overexposed image (on Film-screen system)
Won’t see on Digital system as it is corrected for
to decrease dose to patient during fluoro, do what?
position them close to the II/image receptor
far from the detector
use pulsed fluoroscopy when possible
increase collimation
minimize time
