Physics Flashcards
Terminology to characterise an atom?
A/Z(X) where A is number of nucleons, Z protons, X chemical symbol
The atomic number and mass numbers are denoted by which characters?
An element is defined by?
Atomic number = Z the number of protons
Mass number = A proton + neutrons
Element defined by Z, A defined the isotopes of X.
The nuclear binding energy means the mass of A is less than the sum of its parts
In the Bohr model each shell can hold how many electrons?
2(n^2) where n is the shell number
1st shell = k = 2
2nd = l = 8
Electron binding energy is the energy that be supplied to remove an electron from its shell.
Elevating an electron from K to l requires?
An amount of energy equal to the difference in binding energies
E>Ek
Define the electron volt
eV is the energy gained by an electron as it is accelerated by a potential difference of 1 volt
Characteristic radiation occurs when? Under what condition can this process be a cascade?
As a consequence is this emission continuous or discrete?
An outer shell electron falling to a more proximal shell - requires a vacancy.
High atomic number atoms have many shells and so the loss of a proximal election will trigger a cascade of electrons falling from higher shells. Each releasing a discrete amount of energy proportional to the binding energy of their shell.
Define radioactive decay and radioactivity?
Instability in ratio of neutrons to proton leads to a transition to a more stable configuration, this change emits particles and EMR with the energy corresponding to the increase in binding energy at final configuration. The change is referred to as decay, the emission as radioactivity.
In a sample of radionuclide the rate of decay is directly proportional to:
the number of atoms of the radionuclide present. A these change in the their progeny nuclide the rate decreases - is a decaying exponential function of initial number of atoms and a decay constant (ln2/T half)
Sometime used to determine dose delivery from permanent brachy implant Average life or mean life is defined as
1/the decay constant, where the decay constant is ln2/over half life = lamda). ie. 1/lamda = 1.44(T half)
Planck’s constant
h=6.62 x 10^-34 J.sec
Relate the energy of a photon to its wavelength
E=hv, where h is Planck’s and v frequency, where v=speed of light (c)/wavelength (i.e wavelength x v = c)
Production of gamma rays
By decay or capture. Decay can be either (or a mix of both - see decay schemes) betaminus (electron emitted by a neuron turning into a positron), or betaplus (when proton and electron combine to make a neutron).
In capture - inner orbital electron captured to make neutron from positron, also releases characteristic radiation (with possible cascade).
Principle elements of an Xray tube
1) Filament (cathode) - releases electron by thermionic emission
2) Target - Large atomic nuclei such as tungsten
3) Anode - At the site of the target,
4) High Voltage between cathode and anode accelerates electron onto focal spot (determined by size of filament) on target
In Kv imaging,what is the focal spot? How is it changed? What is the effect of shrinking the focal spot?
The focal spot is the site where accelerated electrons strike the target. The size of the filament/cathode determines the size of the focal spot (modern systems often have 2 or more filaments).
Smaller focal spots result in a smaller region from which Xrs are emitted (the apparent source) producing a more detailed picture. To produce sharp images, focal spots need to be small but able to withstand heat loading without melting the anode target. A small focal spot is used when spatial resolution is important, while a large focal spot is employed when a short exposure time is the priority.
The possible outcomes of XR/Gamma ray interaction w/matter?
Photon may be 1) Scattered 2) Absorbed 3) Transmitted without interaction Photons that traverse the medium uninterrupted are called primary radiation. Those that are scattered or absorbed are termed attenuated.
The amount by which a photon beam is attenuated is a characteristic of:
1) Attenuating material
2) Photon beam spectrum
What is HVL
Half value layer, in units of cm or mm is the thickness of the attenuating material that reduces the intensity of a radiation beam to half its original value. Sometimes measured in air kerma rate.
Under conditions of good geometry a monoenergetic beam will be attenuated ______ with increasing thickness of the absorber
Exponentially
A polyenergetic beam is not attenuated _____ as the absorber preferentially absorbs _______
Exponentially
Low energy photons
True or False: In the case of a polyenergetic beam the second HVL is equal to or less than the 1st HVL
False. to reduce from 1/2 to 1/4 A much larger HVL may be needed to attenuate high energy photons.
Define the linear attenuation coefficient u
u = 0.693/HVL
The transmitted beam intensity I for any thickness of absorbing material:
I transmitted = I0^-ux where I0 = initial beam intensity u = linear attenuation coeef 0.693/HVL x is thickness u is in units cm^-1
Define Coulomb
What is the charge of the following particles: Proton Neutron Election Positron
1 Columb = it is the charge carried by a current of 1ampres in 1 second. Proton = 1.602 x10^-19C neutron = 0 Electron = -1.602 x10^-19 C Positron = +1.602 x10^-19
Rest mass of an electron
9.109 x10^-31 hbpKg = 0.511 MeV
If atoms are described in terms of A, Z, X What is Z What is the Neutron number? what is the notation for: Proton Neutron Positron Electron
Z= number of protons (and electrons)
N = A-Z
Proton: a=1 (i.e. n+p), Z=1 (i.e 1 proton/electron) symbol lower case p (A/P p)
Neuton: A = 1, Z=0, symbol n
Electron, A=0, Z=-1, symbol = e
Positron, A=0, Z=-1, symbol = e (same as electron)
Contrast Atomic mass and Atomic weight
Atomic mass (ma) is the mass of an atom. A single atom has a set number of protons and neutrons, so the mass is unequivocal (won’t change) and is the sum of the number of protons and neutrons in the atom. Electrons contribute so little mass that they aren’t counted.
Atomic weight is a weighted average of the mass of all the atoms of an element, based on the abundance of isotopes.
Both rely on the atomic mass unit (amu), which is 1/12th the mass of an atom of carbon-12 in its ground state
Define isotope
An atomic species with the same number of proton (atomic number Z) but different number of neutrons (mass number A)
Define energy
The ability to do work - in units of J (Kg.m^2/(s^2))
Work = Force x distance (Nm = J)
Force = any of the known forces (e.g strong, weak, electromagnetic) that causes an object to undergo change. (units of N = Kg.m/(s^2))
Define Force
Force = any of the known forces (e.g strong, weak, electromagnetic) that causes an object to undergo change. (units of N = Kg.m/(s^2))
Define kinetic energy
Kinetic energy = 1/2(mv^2)
The work needed to accelerate a body of a given mass from rest to its stated velocity. Having gained this energy during its acceleration, the body maintains this kinetic energy unless its speed changes.
Electromagnetic radiation (EMR) is made up of
AN electric field and magnetic field propagating through space
All EMR travels at the speed of light c in a vacuum. How does this relate to the frequency and wavelength of EMR
Where c is constant, lama = wavelength
C= f.lamda,
f=C/lamda
ect
A photon has energy E =
E photon = Planck’s (h) x frequency
(ie. f= c/wavelength, so E = h x (c/wavelength
A 25MeV Photon which undergoes Pair Production is an example of an ……… collision. The outcome of which produces
Inelastic collision - where some kinetic force is transferred into a different of energy.
In this case a positron and electron pair - each with mass 0.511 Mev and kinetic energy of (25 - 2(0.511) ) /2= (23.78)/2= 11.9Mev
A photon may interact with an atom either at the orbital electrons or nucleus. At each site list the type of interaction that may occur:
Photon-Orbital (in order of low to high energy): 1) Coherent Scattering 2) Photoelectric effect 3) Compton (Incoherent) Scattering Photon-Nucleus: 1) Pair Production 2) Photdisintegration
Coherent scattering is proportional to what?
When would scatter be highest
Energy of photon and atomic number (squared)
Scatter proportional to (Z^2)/E
i.e highest scatter for high atomic number (high electron density) material hit with lower energy beams
All electrons are not equally attractive to a photon. What makes an electron more or less attractive is its binding energy. The two general rules are:
- Photoelectric interactions occur most frequently when the electron binding energy is slightly less than the photon energy.
- Compton interactions occur most frequently with electrons with relatively low binding energies.
The total attenuation rate depends on the individual rates associated with
photoelectric and Compton interactions.
Utotal = Ucompton + Uphoto
kV beams produced with ….. target
MV beams produced with ….. target
kV beams produced with reflective target
MV beams produced with transmission target
In XR production what is the heel effect, what type of target creates them/beam strength
Seen in kV beams (i.e produced with reflective target):
Heel effect- beam intensity lower on the anode side due to increased anode material needed to pass through. ‘self attenuation’
In Mv photon production:
Mean energy production is 1/? less than nominal
Mean energy ⅓ of nominal energy
In XR production, for a tungsten target X% is characteristic, Y% bremsstrahlung
For tungsten target 20% characteristic, 80% bremsstrahlung
Inverse Square Law
Inverse Square Law
● beam diverges as it moves from its source
● intensity is inversely proportional to the square of the distance
The vast majority of Mv photons from a linear accelerator are produced from
Mostly bremsstrahlung
Beta minus decay:
Electron emitted by a proton turning into a neutron
Boom there’s your minus
Beta plus decay:
A proton turning into a neutron by combing with an electron
Electron capture
When a positron captures an inner shell electron to make a neutron, triggering characteristic radiation
The photoelectric effect requires a photon to interact with
A tightly bound inner orbital electron
The photoelectric effect dominates at?
low energies = less than 0.2 MeV
In the photoelectric effect, the maximum energy an electron can receive in any one interaction is
hv
Letters denoting the 1st 4 orbital shells from inner to outer
K, L, M, N
Photoelectric interactions usually occur with electrons that are ………. bound to the atom, that is, those with a relatively …….. binding energy.
Photoelectric interactions are most probable when?
Photoelectric interactions usually occur with electrons that are firmly bound to the atom, that is, those with a relatively high binding energy.
Photoelectric interactions are most probable when the electron binding energy is only slightly less than the energy of the photon.
Think K edge!!
Describe the photoelectric effect or draw a diagram of it
Photon (typically with energy only slightly more than the binding energy of inner shell electron is) is absorbed by inner shell electron removing from the atom with energy = hv-Ebinding.
Subsequently outer shell electrons (if present) will fill inner shell vacancies, emitting photons (characteristic XRs) with energies = to the difference between new inner orbital shell and previous orbital binding energies.
A further electron may be ejected from the atom if a characteristic XR with sufficient hv generated from the above process interacts with a loosely bound outer shell electron (Ebinding < hv) removing from the atom.
This electron is the Auger electron
How do Auger electrons occur?
An electron may be ejected from the atom if a characteristic XR with sufficient hv generated from the photoelectric effect interacts with a loosely bound outer shell electron (Ebinding < hv) removing from the atom.
This electron is the Auger electron
The probability of the photoelectric effect is dependent on?
Interaction most often in the …. shell? Because?
Z and E - with Z being much more important
Probability proportional to (Z^3) and inversely proportional to E.
Most Photoelectric interactions occur in the K shell because the density of the electron cloud is greater in this region and a higher probability of interaction exists. About 30% of photons absorbed from a clinical x-ray beam are absorbed by the photoelectric process.
Contrast the site of interaction between the photoelectric and compton effects
PE = inner shell (usually k) Compton = loosely bound outer shell electron (no Auger's produced)
The probability of compton interactions is largely dependent on ……
Since most materials are similar in this regard, attenuation in units of (….) is the same, but differs in
Electron density - which is approximately the same for most atoms.
Therefore in terms of grams/cm^3 compton attenuation is the same for most materials.
BUT attenuation per cm is different.
1cm of bone has more electrons/cm^3 than 1cm of soft tissue
The K-edge is a sudden increase in
The K-edge is a sudden increase in XR absorption/attenuation as the photon energy just exceeds the inner shell binding energy such that photoelectric absorption of the photon is more likely to occur.
Note: there are also L3,2,1 edges at very low energies (around 10KeV)
The threshold for pair production is:
Why?
What happens to the excess energy?
It is due to a photon interaction with
1.022Mev (remember this number!)
Because an electron has a rest mass equivalent to 0.511 MeV of energy, a minimum gamma-energy of 1.02 MeV is required to produce 2. Any excess energy of the pair-producing gamma-ray is given to the electron–positron pair as kinetic energy. i.e. hv-1.022 MeV
It is due to a photon interaction with the coulomb field of the nucleus
Mass attenuation from photo electric effect is dependent on:
Z - i.e. the bigger the nucleus column field the more likely
What is fate of the products of pair production?
Electron - undergoes electron interactions/energy transfer
Positron - ANNIHILATION reaction with an electron - creating 2 photons each with hv = 0.511 = (the rest mass of the positron + the electron it collided with)/2.
The 2 types of photon interactions with nuclei in order from low to high of energies involved
1) Pair production - threshold is 1.022 Mev (2x the rest mass of an electron)
2) Photodisintergration - above 10MeV
Photodisintegration occurs above?
10MeV
Photo disintegration is an interaction between a high energy photon and …………… It results in:
Photo disintegration is an interaction between a high energy photon and an atomic nucleus.
It results in: the emission of 1 or more nucleons (typically a neutron), i.e leads to unstable rations and radioactivity.
For photodisintegration to occur threshold energy is:
The rest energy of nucleus - (rest energy residual nucleus + emitted nucleon)
If you were to graph attenuation in terms of which photon interaction with matter contributed at a given energy how would you plot it?
x-axis: Energy (KeV) on log10 10, 100, 1000, 10,000
y axis: Linear attenuation (cm^-1) 100, 1, 0.01, 0.0001
P.electric start at 100cm^-1 finish at 100KeV, K-edge at 10KeV
Compton: Tubby curve crossing y axis at 1, slow decrease to 1MeV then nonlinear drop to 10MeV
PP: starts at 1.022Mev nonlinear approach to flatten just over 0.01/cm at a bit past 10MeV
Show total attenuation curve whichh overlies the above curves
What would be the key components if you were to graph the relationship between atomic number and which photon interactions dominate at a particular energy.
Y-Axis: Zeff (effective proton number), from 0 to 100
X: Energy Log(MeV), 0.1, 1, 10, 100
PE: line where SIGMApe = SIGMAcompton = right half of Paraboler asymtope at Zef =100, MeV=1
Photon Energy is imparted to matter in a two stage process:
What energy is not counted as absorbed energy?
Thus the equation for average energy absorbed is:
Energy is imparted to matter in a two stage process:
● when a photon interacts with material, all or part of its energy is transferred into kinetic energy of electrons.
● most of these electrons lose their energy by inelastic collisions (ionization and excitation) with atomic electrons.
● A few will lose energy by bremsstrahlung interactions with the nuclei.
● Bremsstrahlung energy leaves the local volume as x-rays and is not included in the calculation of absorbed energy.
▁Eab= ▁Etr - ▁Erad (note: line should be under letters)
▁Etr is the average energy transferred from the primary photon to kinetic energy of charged particles
▁Erad is the average energy which the charged particles lose to bremsstrahlung and is not absorbed in the volume.
Define the linear attenuation coefficient
mu = fraction of absorbed photons per unit path length (cm)
Therefore in units of cm^-1
It is constant for monoenergetic beams (ie. takes the same form as half life).
For a beam of intensity Io, the exponentially decaying intensity observed at I(x cm) is described by:
I(x)= Io.e^(-mu.x)
The linear attenuation coefficient is dependent on?
To remove this dependency, such that it only reflects Z and beam intensity, what can be done?
Dependent on beam intensity, and density.
Divide by density, leaving the atomic number and initial beam intensity (Izero) as independent variables.
symbol mu/p
I.e cm^-1/(g.cm^-3) = (cm^2).g^-1
Units for mass attenuation
(cm^2).g^-1
i.e linear attenuation/density, cm^-1/(g.cm^-3)
mass attenuation is dependent on
Z and beam intensity
Electronic and atomic attenuation are derived from?
What are their units
Both atomic and electronic attenuation are derived from mass attenuation (u/d). With units of electrons/cm2 and atoms/cm2
Where Nzero is the number of electrons/gram: eU = (u/d)*1/Nzero
units: electrons/cm2
aU = (u/d)/(Z/No) (units atoms/cm2)
How would you derive the linear attenuation from mass attenuation
If mass attenuation is linear attenuation divided by density, then you can calculate linear attenuation by multiplication if you know the density of the absorber in question.
mu = (mu/p)*p
Describe the Linear Energy Transfer Coefficient:
The LETC is the fraction of energy that is transferred to kinetic energy per unit thickness of the absorber. Using the ratio of the average amount of energy transferred per interaction to the amount of energy (_Etr/hv) which is then scaled to the amount of attenuation:
Utr = U*(_Etr/hv)
Describe the Linear Energy Absorption Coefficient
Fraction of PHOTON energy that is absorbed (converted to inelastic interactions of secondary particles) per unit thickness of the absorber.
If Utr is the attenuation of kinetic energy, then the energy absorption coeffiecient differs in that it captures what happens to this kinetic energy by removing the energy lost by secondary charged particles as bremsstrahlung (B).
If g is the fraction of energy lost as bremsstrahlung 1-g = the remaining energy.
Therefore Uenergy absorbed = (Fraction of energy converted to kinetic energy)*Fraction not lost to B by secondary particles.
Uen = Utr(1-g)
Derive the mass energy transfer coefficient:
Linear energy transfer coefficient/density.
Under what conditions does Utr = Uen?
When Z is low, electrons lose most of their energy by ionisation collisions (very little is lost due to bremsstrahlung).
Therefore the fraction energy transferred to kinetic energy per unit thickness = fraction of energy lost to inelastic collisions with no secondary electons loosing energy to Brem:
U*(_Etr/hv)=Utr(1-g) IF g =0,
Hence,
Utr=Utr x ~1
Attenuation of a photon beam by an absorbing material is depenent on the 5 major types of interactions. Each associated with an attenuation coefficient which vary with?
Beam strength and the atomic number of the absorber.
Electrons undergo large number of ………. interactions before losing all their energy. They are also easily scattered due to their low …….
Electrons undergo large number of Coulomb interactions before losing all their energy. They are also easily scattered due to their low mass.
Electron Inelastic collisions:
○ with atomic electrons resulting in?
○ with atomic nuclei resulting in?
Typical energy loss is about ? in water
● Inelastic collision:
○ with atomic electrons resulting in ionization and excitation (collisional loss)
○ with atomic nuclei resulting in bremsstrahlung (radiative loss)
● Typical energy loss is about 2 MeV/cm in water
Electron Elastic collisions result in:
Elastic collisions resulting in scattering without energy loss
The two forms of electron interactions with matter are described by:
Scatter power
Stopping power
What is stopping power?
What does it depend on?
The retarding FORCE acting on CHARGED particles (typically alpha or beta)
The stopping power depends on the type and energy of the radiation and on the properties of the material it passes.
The stopping power of a material is numerically equal to:
The stopping power of the material is numerically equal to the loss of energy E per unit path length, x
S(E)=dE/dx
This eqn is the linear stopping power which in the international system is in N but is usually indicated in other units like MeV/cm
The range of a charged particle:
In passing through matter, charged particles ionize and thus lose energy in many steps, until their energy is (almost) zero. The distance to this point is called the range of the particle.
The range depends on: type of particle, its initial energy and the material through which it passes.
How is Stopping power graphed?
Bragg curve:
X axis: Path length (cm) e.g 0 - 4cm
Y axis: Force (Stopping power)
From a clinical point of view, what is the key feature of a Bragg curve?
What is the explanation for this feature
The Bragg peak.
Force significantly increases towards the end of the particle’s range.
This is due to the slowing particle interacting more with the medium through which it passes (increased cross-sectional area). Energy lost by charged particles is inversely proportional to the square of their velocity, which explains the peak occurring just before the particle comes to a complete stop.
In terms of an absorber, electrons/gram differ between low an high Z material how?
What are the implications for the type of Energy lost by an electron moving through a low, or high Z medium?
High Z materials have less electrons/gram and more tightly bound electrons
Low Z materials have more electrons/gram, which are less tightly bound leading to excitation and ionisation/
Electrons, moving through a material are more likely to have collisions with other electrons in Low Z materials. Whereas heavy atoms are more likely to cause braking radiation.
What is Mass Stopping Power?
What are its units?
Linear Stopping Power/Density (s/d)
S(E)=dE/dX in units of MeV/cm
Density in Gram/cm3
Therefore units of MeV.cm2/gram
For a solid medium, total mass stopping power is the sum of?
(s/d)tot=(s/d)collisional + (s/d)radiation
Where collisions leads to ionisation and excitation
Collisional stopping power is dependent on?
Energy bellow ~ 1.02MeV and Electrons per/gram, that are in turn weakly bound. Therefore dependent on Z:
(S)col proportional to 1/Z
(S) col proportional to 1/Esqr after 1.02Mev Becomes constant at around 2 MeV/gram.cm^-2
Radiation Stopping Power is dependent on?
What does this imply for XR production?
Z and E
(S)rad proportional to Z^2
(S)rad proportional to E
This implies the generation of bremsstahlung radiation is more efficient at high Z and E.
What is Delta radiation? What is another name for it?
Aka “Knock on electrons” are elections knocked out of their orbits by primary particles with sufficient energy to ionise other atoms.
Appears as branches off the main particle track in a cloud chamber
Total collisional stopping power is equal to?
Hard collisional and soft collisional stopping power
Therefore, if
(S)total = (S)collisional + (S)rad
(S)total = ((S)hard +(S)soft) + (S)rad
In terms of collisional stopping power. What are soft and hard collisions?
Soft (aka distant collisions): when the particle “collision” is far from the atom, the particle interacts weakly with all the orbital electrons - transferring relatively little E to each. But the large number of interaction entail the particle loses a lot of E (~50%)
Hard (Aka Close collisions). Particle interacts with an orbital electron transferring lots of energy to it - the electron turning into delta radiation. The chances of a hard collision are relatively small, but energy transferred is high - particles lose approx 50% of their Ek.
Restricted mass collision stopping power is introduced to calculate the energy transferred to?
By limiting the energy transferred to a ……… ……….. to a threshold (often denoted as D), highly energetic …… ……….. are allowed to escape the region of interest.
The restricted stopping power is lower than the unrestricted stopping power. The choice of the energy threshold depends on the problem at hand. For problems involving ionization chambers a frequently used threshold value is 10 keV (the range of a 10 keV electron in air is of the order of 2 mm).
Restricted mass collision stopping power is introduced to calculate the energy transferred to a localized region of interest.
By limiting the energy transfer to delta particles to a threshold (often denoted as D), highly energetic delta particles are allowed to escape the region of interest (and the calculation).
(S)RLMASS = (S)collisional - Edelta>threshold
The restricted stopping power is lower than the unrestricted stopping power. The choice of the energy threshold depends on the problem at hand. For problems involving ionization chambers a frequently used threshold value is 10 keV (the range of a 10 keV electron in air is of the order of 2 mm).
Define scatter power:
The scattering angle of a charged particle per unit path length within the absorber.
T= change in angle^2 per path length
Units rad^2/cm
Electron scattering depends on?
How?
E and Z
Z: Larger nuclei = large coulomb force = more scatter
Scatter proportional to Z^2
E: Higher the energy the less time near the nucleus therefore the less scatter.
Scatter proportional to 1/E^2
Define mass scattering power
Scatter power/density
(rad^2/cm)/(gram/cm^3) = rad^2.cm^2/gram
There are two major sources of gamma rays.
in one of these a very specific amount of gamma ray energy is produced…
One way in which gamma rays are emitted occurs when a nucleus moves from an excited energy level to a lower energy level. Energy from the transition is given off in the form of gamma rays.
The other way gamma rays are produced results from the annihilation of positron electron pairs. When this event occurs, two gamma rays, both of energy 511 keV, are emitted (i.e the energy equivalent of the rest mass of the colliding particles)
Define radiant energy
for monoenergetic beams
Is the number of particle emitted/transferred/received ect multiplied by their energies E.
R= NxE (units of joules)
E.g Energy fluence = R/dA = particle fluence x Energy of those particles (dN x Energy/dA)
Define flux:
for monoenergetic beams
Particle flux = Number of particles counted (dN)/time interval (dT) (units num/time)
Energy flux = Energy of parties (i.e N particles x E)/dT (units Joules/sec)
(for monoenergetic beams)
Define particle fluence:
This is a measure of the particle………
dN number of particles incident of sphere with cross-sectional area dA:
phi=dN/dA (units M^2)
This is really a measure of the particle intensity, the intensity of the particle beam.
If we have 106 particles incident on a sphere of radius 1 cm, what is the particle fluence?
The cross-sectional area subtended by the sphere is π times the square of the radius of the sphere, or π cm2, so the particle fluence is 106 divided by π particles per square cm.
Define energy fluence:
relax psy to phi
for monoenergetic beams
Radiant energy (RE) (=dN.E)/cross sectional area of sphere (dA)
Psy = dE/dA = phi*E (units joules/M^2)
phi is particle fluence
(i.e phi=dN/dA (units M^2) where dN is number of particles with energy E)
To relate fluence (particle or energy) to a polyenergetic beam it is necessary to:
Use an energy spectrum. Ie. a energies within a range [E,E+dE]
Therefore count the number of incident particles with energy within the range [E,E+dE] incident of sphere with cross-sectional area dA (i.e. the combined mono energetic particle fluences):
combined mono energetic particle fluencies =dPsi
phiE = dPsi/dE
psyE = E*(psiE)
Fluence rate can be described in terms of?
Units for each?
Particle Rate (just called fluency rate psi w/dot on top) or Intensity
Fluence (particle) Rate = dPsi/dt, Fluence rate is the fluence per unit time, or the number per unit area per unit time. So more particles going through this area per unit time, the more intense the beam than if fewer particles per unit time.
(units m^2/second)
Intensity (Energy Fluence Rate) = Energy Fluence/second = (dPsi*E)/dt = dPsy/dt
(units J/m per second)
The energy of photons is imparted in a 2-step process:
How does this relate to the concepts of Kerma and dose?
The energy of photons is imparted in 2-step process:
○ First step is energy TRANSFERRED to charged particles
○ Second step is energy ABSORBED by secondary charged particles
● Kerma refers to energy transferred to a volume: Taking into account the energy transferred from (photons) to charge particles within a volume:
K = (Rin)un - R(out)un
● Dose refers to energy imparted to a volume: Takes into acount energy transferred from photons and charged particles to charged particles within a volume.
D = (Rin)un - R(out)un + (Rin)ch - R(out)ch
● For charged particles energy loss is directly ……..
● Energy transfer is the energy imparted to a …….. ……….
● Energy absorbed is the ………. loss experienced by a charged particle, described by the …….. stopping power.
○ Defined as the energy ……….. per unit …….
● Energy deposited is the energy …….. or …….. in a single …….
● Energy imparted is the sum of … …………. … …….. minus … …………. … ……..
● For charged particles energy loss is directly absorbed
● Energy transfer is the energy imparted to a secondary electron by a photon.
● Energy absorbed is the collisional loss experienced by a charged particle, described by the collisional stopping power.
○ Defined as the energy absorbed per unit mass.
● Energy deposited is the energy transferred or absorbed in a single interaction
● Energy imparted is the sum of all energies (charged and uncharged particles) entering the volume of interest minus the energies leaving the volume.
What is the restricted collisional stopping power?
What fundamental metric does it contribute to?
The Restricted Linear Collision Stopping Power is the energy lost by a charged particle through hard and soft collisions over a unit path length, minus the total kinetic energy of delta rays with Ek over a set limit.
This is the amount of energy absorbed by a material.
If scaled by the Energy fluence will give DOSE - I.e the energy ABSORBED
Define KERMA
How does it relate to fluence?
Kinetic Energy Released per unit mass
K=(d._Etr)/d.mass
It is the energy transferred to a mass.
For mono-energetic beams:
K = ENERGY fluence(Utr/density) i.e.
Fluence x mass linear energy Tf coefficient,
Where the LET coeff Utr = Etr/hv (Etr=Eabs + E rad)
Units = J/Kg (Gy)
Difference between Gy and sievert:
1 Gy is the deposit of a joule of radiation energy per kg of matter or tissue. 1 Sv = 1 joule/kilogram – a biological effect.
Define collisional KERMA:
Like restricted LET, Kcol ignores average energy lost through radiative processes from excited electrons (Bremsstrahlung or annihilation).
Kcol = KERMA(1-_g)
where _g is the Average fraction of energy lost through radiative process
Units J/Kg
Another way to think about it:
Ktot = (Rin)uncharged - (Rout)uncharged
Kcal = (Rin)uncharged - (Rout)uncharged - Rradcharged
Rradcharged = Radiative losses from charged particles released in the volume (V). THIS is Regardless of whether radiative loss happens outside of V, so long as the secondary e came from V.
When is Kcol approximately = KERMA?
When radiative losses from charged particles is negligible.
Ktot = (Rin)uncharged - (Rout)uncharged
Kcal = (Rin)uncharged - (Rout)uncharged - Rradcharged
Rradcharged = Radiative losses from charged particles released in the volume. (V). THIS is Regardless of whether radiative loss happens outside of V, so long as the secondary e came from V.
In Low Z most loss is collisional
For polyenergetic photons:
Collisional kerma is related to fluence by
KERMAcol is related to energy fluence by mass energy Tf coeff (Uen, which = Utr(1-_g)), such that:
KERMAcol = Uen*ENERGY FLUENCE
Similarly for polyenergetic photons,
but replace Uen with _Uen,
_Uen = the mass energy absorption coefficient averaged over the energy fluence spectrum
Why is collisional KERMA usually more applicable than Dose?
Since most bremsstrahlung photons escape from the volume of interest, collision kerma is more applicable
Compare dose to collisional KERMA
Kcol: indirectly ionising (uncharged particle) Energy Tf to electrons (charged) in a volume.
Dose: uncharged and charged (direct and indirect ionisation) Energy Tf to a volume.
Kcol = (Rin)uncharged - (Rout)uncharged - Rrad (Rrad = Bremsstrahlung released by e liberated within V)
D = ((Rin)un - (Rout)un) + ((Rin)ch - (Rout)ch)
I.e D considers charged particles leaving V.. ((Rout)ch)
When does D = Kcol (what is this called)?
Charged particle Equilibrium (CPE):
If Kcol considers uncharged E entering = (Rin)un, and uncharged E leaving = (Rout)un:
Kcol = (Rin)un - (Rout)un - Rrad
And D considers both charged and uncharged particles:
D = ((Rin)un - (Rout)un) + ((Rin)ch - (Rout)ch)
Then D= Kcol when energy in of charged particles ((Rin)ch) = (Rout)ch. Therefore, (Rin)ch - (Rout)ch = 0:
D = ((Rin)un - (Rout)un) + 0 = Kcol
Besides (Rin)ch) = (Rout)ch, for true CPE to exist KERMA is constant (in real life KERMA attenuates, but transient CPE does)
Why is it important to be able to relate Kcol to D
Using CPE, a quantity can be calculated (Kcol) from a quality that can be measured (Dose).
Under CPE:
D = Kcol
D = Energy Fluence*(Uen/density)
How would you graph the relationship between Kcol and Dose
X-axis - Depth in medium (arb units), show Dmax
Y-axis - Relative Energy per unit Mass!
Show:
D build up, start a bit above y=0 (i.e electrons from air), end build up at Zmax the graph D as linear decline.
Kerma linear decrease with depth intersects D at Dmax.
Mark this point as “CPE”, then cont Kcol just below D
Beta = D/Kcol, mark beta<1, beta=1
What is Equivalent Dose:
Units?
● Radiation-weighted dose quantity, taking into account the type of radiation that produces the dose. It is tissue independent.
● To account for different biologic effect of different radiation qualities.
● absorbed dose multiplied by a radiation weighting factor
● Unit is J/Kg or Sievert
What is Effective Dose:
Units?
Accounts for tissue(s) irradiated
● the summation of all tissue equivalent dose, each multiplied by a tissue weighting factor.
● Accounts for the relative difference in sensitivities of different tissues.
● Represents stochastic effects (eg. cancer and hereditary)
● Eg. difficult to produce hereditary effects from irradiation of hands and feet.
● Unit is Sievert
Outline how to get from dose (e.g to 2 tissues) to Effective dose
1) convert to equivalent dose by multiplication with corresponding radiation weighting factor. (this is based on whole body radiation)
2) Convert this product(s) (i.e dosexWr) to Effective dose by multiplication with relevant tissue weighting factor Wt
3) If multiple areas irradiated sum up the above products to arrive at Effective dose.
Define Exposure:
Radiation exposure is a measure of the ionization of air due to ionizing radiation from photons. Units C/Kg (or R)
dQ/dm
Where dQ is the sum of ion charges| produced by electrons or positrons created or liberated within a mass of air (dm) by photons, divided by that mass.
Exposure (X) can be related to Kcol and K (air) how?
If we know the amount of coulombs of charge created per unit of energy deposited (electron charge/Wair)
Then this number x the collisional Energy deposited (Kcol) will give exposure X = kcol*(e/Wair)
e=1.602 x 10^-19 C
Wair = 33.97 × 1.602 × 10^-19 J/ion pair:
e/Wair = 1/33.97 J/C
X=Kcol1/33.97
Ktot = “Air KERMA” = (X/(1-_g))(Wair/e)
= 33.97(Exposure/(1-g))
Superficial and orthovoltage refer to what range of energies
Photon kV
Superficial: 50-150 kV
Ortho: 150-500 kV
Superficial XR therapy units typically deliver 90% of surface dose to ….. mm deep.
Superficial XR therapy units usually delivers 90% of surface dose to 5mm deep
Orthovoltage XR therapy units typically deliver 90% of surface dose to ….. mm deep.
Orthovoltage XR therapy unit usually deliver 90% of surface dose to 2cm deep
For superficial and ortho voltage machines beam hardening is expressed as:
Define this.
Degree of hardening is expressed as the Half Value Layer
The thickness (in cm) of a specified material that when introduced into the path of the beam reduces the EXPOSURE rate by half.
Essential components of a kilovoltage treatment machine:
1) kV generator:
○ supplies low voltage to the filament, enabling thermionic emission
○ Also supplies voltage across the anode and cathode to accelerate electrons.
2) X-ray tube
3) Beam collimation (primary and secondary/interchangeable) and applicators
4) Support stand
5) Safety features: eg. interlocks, emergency stops.
Clinical requirements of a linear accelerator (actual dumbass exam question):
Clinical requirements of a linear accelerator: ○ Compact ○ Reliable ○ Safe ○ High accuracy and precision ○ Well defined and uniform beam ○ Stable
Outline the essential components of a kilovoltage treatment machine XR tube
Kv Tx machine X-ray tube:
○ cathode and anode separated by space within a vacuum.
○ electrons are accelerated from cathode (filament) to anode (target) producing x-rays
○ electron energy >200 kV can eject secondary electrons from the anode, and hooding prevents buildup of electrons in the wall, forming electrostatic charge.
The basic components of beam collimation and application in kV treatment machines:
Consideration of source to surface is needed why?
Beam collimation and applicators:
○ Primary collimator is a conical hole in a block of lead.
○ Interchangeable applicators are used for secondary collimation- different sizes and shapes are available
○ As SSD decreases, dose rate increases.
Essential components of a linear accelerator:
1) Power supply and pulse modulator: flat topped pulses of ~50 kV delivered to simultaneously to magnetron/klystron and electron gun.
2) Electron gun
3) Microwave power source (Magnetron or Klystron)
4) Accelerator Waveguide (Travelling or standing)
5) Bending magnet - typically 270deg
6) Machine Head:
■ Target
■ Primary collimator
■ Flattening filter
■ Scattering foil to scatter the pencil beam
■ Monitor chamber
■ Field defining light and range finder
■ Secondary collimators and MLCs
What does the bending magnet in a linac do?
1) Beam bent through 270 (or some magnets 120) deg
2) Focuses beam
3) Removes chromatic aberrations (hence achromatic magnet) - filter low and high energy electrons
Compare a magnetron to a klystron
1) Magnetron generates microwaves (high power oscillator), Klystron amplifies them (has lower power microwave input)
2) Both exploit electron deceleratio to produce radiofrequency photons.
3) Magnetrons produce lower Energy Rf waves
4) Magnetrons are smaller
5) Magnetrons are cheaper, but live half as long
What does the wave guide do?
What are the 2 types? Which is smaller?
1) Accelerates electrons using interaction with microwaves
2) Focusing magnets center the electrons
Either Travelling (larger) or standing waveguide (smaller and able to fit in treatment head)
1st 3 parts of linac machine head (assume photon output desired) and their role:
1) Target
2) Primary collimator
● Defines maximum field size
● Usually conical
● Tungsten
3) Flattening filter (as opposed to scatter foil)
● Usually steel
● Flattens pencil beam
● Attenuates and hardens beam.
1st 3 parts of linac machine head (assume electron output desired) and their role:
1) Primary collimator
● Defines maximum field size
● Usually conical
● Tungsten
2) Scattering foil to scatter the pencil beam (not too thick or end up with bremsstahlung photons
3) Monitor chamber
● monitor photon and electron beam output
● monitors transverse beam flatness
● Backscatter plate reduces backscattered radiation into the monitor chambers
Parts of linac machine head (both for photon and electron output):
1) Target
2) Primary collimator
3) Flattening filter
4) Scattering foil to scatter the pencil beam
5) Monitor chamber
* ** Fixed wedge may be placed after chamber
6) Field defining light and range finder
7) Secondary collimators and MLCs
Besides the primary linac collimator, what are the other collimators that may be employed?
1) Secondary Collimators (typically tungsten): upper and lower jaws, provide rectangular fields
2) MLCs
3) Cerrobend
How does the physical penumbra differ between cerobend and MLCs
The physical penumbra with MLC is larger than that of Cerrobend blocks.
In addition to functioning as a tertiary collimator. MLCs can also function as:
How do varian and Electra differ?
Can function as upper jaw, lower jaw
Electra replaces one set of secondary jaws with an MLC - Advantage is field is smaller closer to the source so more responsive IMRT. Also smaller treatment heads
Varian add the MLC as the third pair of Jaws. Advantage thinner MLCs + keep secondary jaws (allows leakage reduction between tertiary leave - tertiary leaves have les leakage)
Outline the design of MLCs
○ two facing banks of tungsten leaves (usually 40-60 each bank)
○ width at isocenter ~5-10mm.
○ individually motor controlled and independent
○ Collimates the beam to give an irregular outline.
\ie. can match field to tumour shape.
○ Tongue and groove design
- Intraleaf transmission <2%, interleaf transmission <3%
○ Leaf ends rounded (e.g. Varian) vs focused (Siemens). So that field edge penumbra constant at all field sizes.
For MLCs
Intraleaf transmission < ?
Interleaf transmission < ?
What is the tongue and groove effect?
Intraleaf transmission <2%,
interleaf transmission <3%
Tongue and groove effect: Under coding between leave during dynamic treatment.
2 Problems with the design of MLCs?
2 Issues:
○ Light field underestimates radiation field by up to 1.2 mm
○ Leaf end transmission ~40%
Focused collimators are:
This allows?
What do MLCs do?
Are moved along a circular arc to be orthogonal to the the beam. This means the beam is attenuated the same at all angles. Meaning a sharp field edge.
Unfocused collimators move horizontally, such that when the field is large, oblique beams at the field edge pass through less attenuating collimator material.
MLCs use rounded leaf so beams at the field edge path through the same amount of material regardless offend size.
Disadvantage of rounded MLC leaf ends
When closed completely less density where leafs meet implies increased transmission.
What are the 2 types of linac wedge?
Besides speed, what are some major benefits of the more modern approach?
Physical/variable and dynamic.
1) Less beam hardening with dynamic wedge
2) Scatter is minimised with dynamic wedges, as the beam is open for most of the treatment. If we were to use a universal wedge for a breast field the scatter to the contralateral breast would be about 1.5% higher than with the dynamic wedge relative to the central axis dose.
When might a wedge be used?
- Improve dose distribution where field edges overlap - preventing to prevent high dose areas
- Compensate for oblique body contour
- Sloping target volume
Watch is the Wedge Angle?
How would you draw it?
The Wedge Angle - is defined to be the angle through which an isodose curve is tilted at the central ray of a beam at the 10 cm depth.
x-axis: Distance to Central axis (cm).
Y axis: Depth (cm)
Draw: angled isodose curves at 5, 10, 15 cm. The one at 10cm draw a line parallel to x-axis to make a triangle between wedge line and drawn line. show the angle (e.g. 45deg)
Physical wedges come in 2-types?
What are physical wedges typically made out of
1) “Universal” - built into the machine head of Elekta machines
2) Individual - mounted in the machine head
Dense Z - steel or lead
What happens to the wedge isodose line with depth and why?
Angle of tilt decreases because of scatter
A competing source with linac for high-energy photons?
Why may this option be better for some health systems?
Where else do you see this system?
Cobalt-60 machine - less moving parts, can be cheaper to maintain, but typically less conformal (but consider gamma knife).
How is Cobalt-60 created?
Co-59 + n → Co-60
I.e C-59 is bombarded with neutrons
The cobalt-60 isotope has a half-life of:
What does it decay to?
The cobalt-60 isotope has a half-life of 5.3 years so the cobalt-60 needs to be replaced occasionally.
Stable isotope (ie. same number of nucleons) is Ni-60
Cobalt-60 emits?
Largely gamma ray emitter, emitting 1.17 and 1.33 MeV gamma rays with an activity of 44 TBq/g (about 1100 Ci/g).
But obviously some beta particles are also emitted (e.g will scatter and cause Bremmstahlung in the source housing).
Cobalt-60 Machines are comprised of:
1) Source (1-2cm) - The occupation of finite physical space implies a geometric penumbra (i.e. not a point source). Source sealed in stainless steel, then sealed in 2nd steel capsule.
2) Housing = Source head.
○ steel shell filled with lead
○ device for bringing the source in front of an opening Source is returned automatically to the ‘off’ position in the event of power failure.
3) Beam collimation and penumbra:
- Like secondary collimators in linac. Typically hinged/circular trajectory to reduce transmission through corner edges when field large (=bigger TRANSMISSION penumbra)
In gamma decay a radioactive nucleus first decays by the emission of:
The daughter nucleus that results is usually left in an excited state and it can:
In gamma decay a radioactive nucleus first decays by the emission of an α or β particle.
The daughter nucleus that results is usually left in an excited state and it can decay to a lower energy state by emitting a gamma ray photon.
Sources of contamination in a Co-60 machine?
1) Beta particles: emission of electrons, electron interactions w/housing producing Bremmstahlung.
2) Gamma ray interactions w/: source, capsule, housing, collimators causing scatter and beam heterogeneity.
Define penumbra:
Types?
The region at the edge of the beam where the dose rate changes rapidly as a function of distance from the beam axis.
3 Types:
1) Geometric
2) Transmission
3) Physical
Define Physical Penumbra:
Physical penumbra is the combination of:
geometric, transmission and electron scattering at the beam edges.
■ Defined as the lateral distance two specified isodose curves at a specified depth.
Define Geometric Penumbra:
What is it independent of
Geometric penumbra: due to finite source size.
■ Related to source size, source to skin distance, source to diaphragm distance.
■ not influenced by field size (SDD is constant with increase in field size). I.e TRANSMISSION penumbra catches this.
How would you draw the calculation of geometric penumbra at depth:
Source - with length S
Collimators - With Source to Diaphragm Distance (SDD)
Skin line - SSD interval drawn
Depth line - d interval from skin to depth
Beams contra and ipsilateral to define interval pd
How would you calculate geometric penumbra (pd)?
pd = s(SSD+d-SDD)/SDD
where,
SDD = Source to Diaphragm (end of collimator)
s= horizontal length of source
d = depth below SSD (SSD usually = skin)
gamma ray sources are usually isotropic and produce ? photon beams.
X ray targets are non-isotropic sources producing?
gamma ray sources are usually isotropic and produce monoenergetic photon beams, while X ray targets are non-isotropic sources producing heteroge- neous photon spectra.
Superficial/kV machines produce what type of beam?
Heterogeneous spectrum with superimposed characteristic energy peaks at discrete energies (55Kev for Tungsten)
Describe the key components (including filtration) of a kilovoltage XR machine photon spectrum:
Y axis - Relative intensity
X Axis - Photon Energy (kev) 0 to 200 in 50 Kev steps
- The unfiltered spectrum (inherent beam) is a straight line hitting the x axis at the voltage difference between cathode an anode (i.e peak electron kinetic energy).
- Average intensity is at 1/3 max energy
- Filter removes all hv<10Kev then x^2 rise to peak.
- Intensity of each curve (e.g 50 kev, 100Kev) increases with energy
With increasing energy direction of X-ray emission becomes more?
What happens below 100keV?
With increasing energy direction of X-ray emission becomes more forward
Below 100keV there is approx 50% backscatter.
Describe the key features the photon spectrum of linacs:
● Maximum photon energy is equal to the maximum incident electron energy.
● Intensity reaches a peak at ⅓ of the maximum, dropping off below peak energy due to attenuation of low energy photons by the target material
● X-rays emitted are forward directed, hence the use of transmission targets (where x-rays are obtained on the other side) in megavoltage linacs.
