mix questions Flashcards
what standard controls patient demography and billing info
Health Level seven (HL7)
Privacy patient controll and anonymization
HIPAA
what is the lower limit for fringe field for MRI
0.5mT
The peripheral magnetic field around a main MR scanner is called fringe field, and it can affect magnetically sensitive devices. (5G or 0.5 mT) & below is considered as safe level exposure to general public. [Huda, p198-199] (Ken) 1T = 104 G
2T field, 10 ppm what’s noise in Hz
42.58 MHz/T
(2T) => 85.16 MHz (10 p/M) =>851.6 Hz
(Ken)Additional info from Bushberg p.453 - 454, chemical shift refers to the slightly different precessional freq. of protons in different materials or tissues. The shifts (in parts per million, ppm) are referenced to water.
1.5 T field, 3 ppm chemical shift between fat and tissue, what’s noise in Hz;
1.5 * 42.58MHz * 3 ppm = 191.61 Hz
More info taken from the web: Although the majority of the MR signal comes from the hydrogen nuclei of water molecules, lipid protons in fatty tissues can also contribute. Fat and water protons do not resonate at exactly the same frequency. Fat protons resonate at a slightly higher frequency, the difference of which is usually described as a fraction of the resonant frequency. Since the difference in frequencies (or the “chemical shift”) is very small, it is expressed as “parts per million” or “ppm”. Fat protons resonate at 3.5 ppm, a higher frequency than water protons (figure 8). The absolute frequency difference depends on the strength of the main magnetic field (which determines the Larmor frequency). Thus at 0.35 Tesla (15 MHz resonant frequency), a 3.5 ppm chemical shift will result in fat resonating at a frequency of 52.5 Hz (15 MHz x 3.5 ppm) higher than water. At 1.5 Tesla (64 MHz) fat resonates at a frequency of 224 Hz higher than water.
given E(keV)=1.24/lamda(nm), given gyromagnetic ratio and B in Tesla, given Lamor equation, calculate E for proton state inversion
Bushberg, p 382
Assume gyromagnetic ratio (r) in units of MHz/T, ex: Proton r = 42.58 MHz/T
Lamor fequency w0(MHz) = rB, and c = lambda*w0
So, E(keV) = 1.24/(c/rB in nm)
What’s the Larmor frequency under 1T.
For proton w0 = 42.58 MHz/T x 1T = 42.58 MHz
Bright on T1
fat subacute hemorrhage melanin protein-rich fluid slowly flowing blood paramagnetic substances: gadolinium, manganese, copper calcification (rarely) laminar necrosis of cerebral infarction
Dark on T1
increased water, as in edema, tumor, infarction, inflammation, infection, hemorrhage (hyperacute or chronic)
low proton density, calcification
flow void
Bright on T2
increased water, as in edema, tumor, infarction, inflammation, infection, subdural collection
methemoglobin (extracellular) in subacute hemorrhage
Dark on T2
low proton density, calcification, fibrous tissue
paramagnetic substances: deoxyhemoglobin, methemoglobin (intracellular), iron, ferritin, hemosiderin, melanin
protein-rich fluid
flow void
T1 represents
longitudinal relaxation time.
The time constant which determines the rate at which excited protons return to equilibrium. It is a measure of the time taken for spinning protons to realign with the external magnetic field
T2 represents
transverse relaxation time.
The time constant which determines the rate at which excited protons reach equilibrium or go out of phase with each other. It is a measure of the time taken for spinning protons to lose phase coherence among the nuclei spinning perpendicular to the main field.
Why is copper added to beam?
For electron beam, the high z material (copper or lead) is used as a scattering foil to induce scattered electron to achieve large treatment field size, while the pencil beam electron hits on the scattering foil. (Podgorsak p124)
what’s specificity given TN, TP, FP, FN;
TN/(all negative)=TN/(TN+FP)
what’s SENSITIVITY given TN, TP, FP, FN;
TP/(all positive)=TP/(TP+FN)
50 cm SID, 20 cm thick patient, entrance exposure is 1000 mR, if SID changes to 100 cm and keep the radiography exit exposure the same, what’s the entrance exposure?
If patient is close to imager, we can write down
1000 *(30/50)^2 = y *(80/100)^2–> Entrance exposure (y) = 562.5 mR
focal spot is 2.4 mm, 100 cm source to film distance and 20 cm source to object distance, what’s the image unsharpness
image unsharpness: f.spot(M-1)= 2.4mm((100/20)-1)=9.6 (Ken) additional info: Bushberg p 147-148
Increasing window width would
decrease contrast
Calculate the magnification given source to image distance (SID) and object to image distance (OID).
M = SID/(SID – OID)
SID-OID =SOD (Source object distance)
Bushberg 147
A person’s heart is imaged in a diagnostic procedure. The source to film distance is 100cm and the object to film distance is 80cm. How much will the heart be magnified on the film.
M = SID/SOD SID = 100cm SOD = SID-OID=100-80= 20cm M = 100/20 = 5 The heart will be magnified 5x
If a system has x lp/mm resolution, what is the max pixel size
x = 1/(2d), d = width of each line, d = 1/(2x) Bushberg p270, which can be clearly to see our resolution is x lp/mm, it means in 1 mm there is 1/x line pair, and each line is 1/x/2 = 1/2x (mm) as one pixel.
one line pair per mm, what is the resolution?
resolution = 1 lp/mm? for spatial resolution = 0.5 mm
FOV is 40, X line pairs/mm, now FOV is 30,
What is the line pairs per mm? 40X/30 (Bushberg 369)
MRI problem involving Nyquist theory (had to know that the sampling rate has to be at least 2x the highest frequency expected to prevent artifacts).
F_N = 1 / (2d), where d = pixel pitch or sampling pitch. Here 1/2d means 1 cycle we need at least sampling twice, so FN unit is cycle/length. The input signal frequency must be equal or less than Nyquist Freq. FN to prevent aliasing. (Bushberg p 284)
Quantum mottle has its greatest effect on:
Low contrast imaging, Huda p73. Quantum mottle is caused by the discrete nature of x-ray photons and is the most important source of noise in radiotherapy.
To reduce image noise by a facor of 10 the number of information carriers must be increased by a factor of
we know doubling the radiation exposure will reduce the relative fluctuations about the mean value (quantum noise) by sqrt(2). And, the percentage noise can be calculate as sqrt(N)/N (assuming 1 std). Therefore sqrt(SN)/SN = 0.1 sqrt(N)/N -> S = 100
Emission has only neutrino and characteristic x-ray, what’s the decay scheme?
EC
Gamma emission from a metastable state to stable state is forbidden, what’s the alternative decay?
IC
Parent has a very long half life, daughter has 1 hr, after 5 hr,what’s the activity ratio of daughter to parent?
A2/A1 ~ 1 if t»T2
transient equilibrium activityratio
A2/A1~T1/(T1-T2)(1-e^-(L2-L1)*t)
mean energy of a beta emission
~1/3 of max energy
C to F
F=C*1.8+32
C to K
K=C+273.15
energy/ion pair
33.97 eV
why Mo in mammography
characteristic xray is at low energy~20keV. which is good for breast tissue contrast
ICRP risk of cancer
0.1/Sv
probability of photon interaction between 1 and 2 cm
mu
Given attenuation coefficient. What percentage of the incident photons is absorbed between a depth of 1 cm and 2 cm?
it is
(e(-mu1cm) - e(-mu2cm))/e(-mu1cm) =
1 - e(-mu1cm)
Kerma assume uTR
K=(photon energy fluence)*(uTR/roh)
An x-ray beam is composed of two energies. Given the attenuation coefficients of these
energies, what is the effective energy of the beam?
(mu1E1 + mu2E2)/(mu1 + mu2)
Optical density of a film given transmittance
T=10^-OD
Heel effect – what is it and what causes it
The Heel effect is caused by the difference of the photon beam attenuation when the photon traveling through different path length of anode target. Therefore, the x-ray beam is with higher intensity at cathode node than anode node.
The heel effect in a diagnostic beam depends on
Anode angle, source to image detector distance, and field size
The output of the diagnostic x-ray (or CT scanner) is related to: square of the voltage, square of the current, etc
The output x-ray intensity is proportional to mAs x kVp2
1R=
2.58 *10^-4 C/Kg
occupational body/eye/extremities dose limit
20mSv/150mSV/500mSv
public body/eye/extremities dose limit
1/15/50mSv
Lower Limit of Detection
LLD = 4.66σ + 3
Sample and background counting times
are equal
Minimum Detectability Activity
MDA = LLD/(K + t)
K = counter efficiency
