Attenuation, HVL and Beam Quality Flashcards
attenuation decay rate
exponential decay rate
in an added layer of aluminum, the energy value decreases by half
“Half value layer”
What is a half-value layer?
describes the penetrating ability (quality) of the x-ray beam
defined as the thickness of a material that reduces intensity by 50%
at energy levels below 120kV (diagnostic radiography), HVLs are measured in mm of Al
At energy level of 120-400kV (radiation therapy), HVLs are expressed in mm of Cu
The HVL of a beam of radiation is 3mm of tin. What percentage of the beam will be transmitted through a tin sheet 9mm thick?
12.5%
HVL semilogarithmic plot
one axis is logarithmic the other is linear
the number of x-rays emerging from any thickness will never reach zero
the log of the number of photons transmitted varies linearly with the thickness of the attenuating material, resulting in a straight line
basic logarithmic functions
logb x is the same as b^y = x
log 100 is the same as log10 100 = 2
base of natural log is e instead of 10 - e=2.718
loge x = ln x
lne = loge e = 1
I(x) = I0e^-ux
I(x) = beam intensity transmitted through an absorber of thickness x
I0 = beam intensity at an absorber thickness of zero (no attenuator)
e = 2.718
u = attenuation coefficient
x = absorber thickness
if at a given distance of x through matter, the relative intensity is 50%, what is the value of the attenuation coefficient of matter at 80keV?
I0 = 80 keV
Ix = 40 keV
u = ?
x = 5
e = 2.718
u = -ln(Ix/Io)/5
u = -ln(0.5)/5
u = 0.1386 cm^-1
Tenth value layer
thickness of a material that will reduce the incident intensity by a factor of 10
- 90% attenuation, 10% transmission
Used for shielding calculations for barriers
- determine the amount of attenuating material required to protect individuals working with or near radiation sources or x-ray units
Ix/I0 = 0.1 = e^-uTVL
what happens if the natural logarithm is calculated for each side of TVL at Ix/I0 = 0.1?
ln(0.1) = ln (e^-uTVL)
ln(0.1) = -uTVL
TVL = 2.303/u
homogenous versus heterogeneous beam
as a beam penetrates through matter, its intensity will reduce
as a beam penetrates through matter, it’s u will…..
- remain the same if the beam is HOMOGENOUS
- decreases if the beam is HETEROGENOUS
Polychromatic X-ray spectrum attenuation
beam contains a varied spectrum of photon energies
low energy photons are attenuated more rapidly than higher-energy
- effective energy increases as it penetrates deeper
in practice - our x-ray beams are so heavily filtered that departures from the exponential law of attenuation are relatively insignificant
filtration
primary purpose of added filtration is to remove low-energy photons
effective energy of the beam increases - beam hardening
beam is filtered by a) inherent filtration (glass envelope, insulating oil, exit window) b)added filtration (Al), and c) the patient
- added filtration increases the HVL of an x-ray beam
filtration is expressed in terms of mm of aluminum
edge of absorption edge in a filter
a filter exhibits unusual behaviour at beam energies close to the energies of its k-shell absorption edge
at photon energies just below this level, the u is at a minimum and transmission is at a maximum
all elements will transmit most of their own characteristic radiation if they are both the source and filter
What is the attenuation coefficient of a certain material if we know that 5cm of the material absorbs 80% of gamma rays?
u = ?
x = 5cm
I0/Ix = 0.2
u= -ln(0.2)/5
u = 0.322 cm^-1
If 6x10^7 x-ray photons enter a 2cm thick layer, what will be the number of x-ray photons transmitted through the layer if u=1.42 cm^-1
x = 2cm
u = 1.42
e = 2.317
Ix = ?
I0 = 6x10^7
Ix = I0 e^-ux
Ix = 6 x 10^7 e^(-1.42x2)
Ix = 3.51 x 10^6
