Physics Facts

Question 1

Force =

Answer 1

M * A (F = force; M = mass; A=acceleration)

Question 2

1 Joule =

Answer 2

1 kg m²/s² = 1 Newton-meter

Question 3

Kinetic Energy

Answer 3

(m * v²)/2 (m = mass in Kg; v = velocity (m/s))

Question 4

Potential Energy

Answer 4

m * g * h (m = mass in kg; g = gravitational acceleration; h = height above floor)

Question 5

Atomic Number (Z)

Answer 5

of protons in nucleus of atom

Question 6

Atomic mass number (A)

Answer 6

of protons + neutrons in nucleus of atom

Question 7

Compare the following particles in terms of Mass, Rest Energy (MeV), and Charge

Answer 7

Particle Mass (kg) Rest Energy (MeV) Charge (e) *
Electron 9.11 x 10^-31 0.511 -1
Positron 9.11 x 10^-31 0.511 +1
Proton 1.67 x 10^-27 938.2 +1
Neutron 1.68 x 10^-26 939.5 0
Alpha 6.65 x 10^-27 3729 +2
* unit of charge is 1.6 x 10^-19 coulomb

Question 8

Electron volt (eV):

Answer 8

kinectic energy gained by electron after being accelerated through a potential difference of one volt

1eV = 1.6021 * 10^-12 ergs = 1.6021 * 10^-19 joules = 3.83 * 10^-20 cal

(remember that joules = (Kg * m²)/ s²))

Question 9

Alpha Particle:

Answer 9

helium nuclei (2 protons + 2 neutrons)

Question 10

Beta minus particle

Answer 10

electron emitted by nucleus

Question 11

Beta plus particle

Answer 11

positron emitted by nucleus (same mass as electron but with a +charge)

Question 12

Neutron

Answer 12

part of nucleus of atom (has no charge)

Question 13

Gamma ray

Answer 13

X-ray emitted from nucleus

Question 14

Speed of light (in vacuum):

Answer 14

3 x 10⁸ m/s

Question 15

What is the relationship between wavelength, frequency (v) in Hz, speed of light (c):

Answer 15

Speed of light (c) = wavelength * frequency (v)

or Wavelength = speed of light/frequency

  or     Frequency (v) = speed of light (c)/ wavelength

Example: What is the frequency of a photon with a wavelength 15 * 10^-11m
Frequency (v) = (3 * 10⁸ m/s) / (15 x 10^-11m) = 2 * 10¹⁸ Hz

Example: What is the wavelength of the typical EM waves used to accelerate electrons in a linear accelerator (typically frequency of 3000 MHz)?

Wavelength = (3 * 10⁸ m/s)/(3 * 10⁹ Hz) = 0.1m or 10cm

Hz is cycles per second.

Question 16

Energy of Photon:

Answer 16

E= hv (E = energy of photon; v= frequency of photon, h = Planck’s constant = 6.626 x 10^-34 joules-second)

or **v = E/h **

or E= hc/wavelength (since Frequency (v) = speed of light (c)/ wavelength)

Example: What is the energy of a photon with wavelength 15 * 10^-11m?
E = (1.24 * 10^-12 MeV*m) / (15 x 10^-11 m) = 0.0083 MeV

(remember that hc = 1.24 * 10^-12 MeV*m because h = 6.626 *10^-34 joules * second or 4.136 * 10^-15 eV * second and c = 3 * 10⁸ m/s; remember to multiply eV * 10^-6 to get to MeV!!!)

Question 17

Wavelength formula

Answer 17

Wavelength = speed of light (c)/frequency (v)
or Wavelength = speed of light (c)/(E/h)
so Wavelength = hc/E (hc = 1.24 * 10^-12 MeV*m)

                                                        Example: What is wavelength of 100 keV photon?
                                                         Wavelength = (1.24 \* 10<sup>-12 </sup>MeVm)/0.1 MeV) = **1.24 \* 10<sup>-11</sup>**

Question 18

Characteristic x-ray

Answer 18

photon given off when outer orbit electron moves to fill a inner electron shell vacancy caused by an incoming electron causes ionization of an atom

Question 19

Bremsstrahlung (braking radiation):

Answer 19

produced when incoming electron interacts with positive charge of the atomic nucleus

source of most x-ray production

energy lost by incoming electron is given off as a photon

Bremsstrahlung production efficiency = (electron energy in eV) * (atomic # of target) * 10^-9

At diagnostic energies (<120keV) bremsstrahlung efficiency is ~1%, At therapeutic MV energies it is 15-50%!

Question 20

Thoraeus filter

Answer 20

Tin (closest to x-ray tube) →copper → aluminum

used to harden orthovoltage beams (characteristic x-ray of aluminum is only 1.5keV and is absorbed by air before hitting patient)

Question 21

Calculate electron energy conversion to photon energy

Answer 21

F = 3.5 * 10^-4 ZE (Z = atomic #; E = maximum energy of electrons in MeV)

Question 22

Calculate Beam Intensity Transmission

Answer 22

T = I/I₀

Question 23

Beam Attenuation (single energy source)

Answer 23

I = I_oe^-µx (x = thickness of filter; µ = linear attenuation coefficient (a function of filter material and energy of photon beam))

Example: Initial intensity = 2000 photons, µ= 0.2mm^-1, thickness = 3mm
I = 1098 photons

Question 24

Linear attenuation coefficient

Answer 24

= µ

• probability that a given photon will be attenuated in a unit thickness of a particular attenuator
• varies with photon energy and filter composition (atomic number, Z)

-a larger number (closer to 1.0) indicates a higher probability of photon attenuation

The CT number (in Hounsfield units) = 1000 x [(u_material - u_water)/u_water where u = the linear attenuation coefficient

Question 25

Describe beam hardening

Answer 25

only occurs with a polyenergetic or heterogeneous beam (mono-energetic beams cannot be “hardened”!!)

Question 26

Homogeneity coefficient (HC):

Answer 26

-ratio of first HVL to second HVL (a larger value (closer to 1.0) means the beam is more uniform)

Question 27

Half value layer

Answer 27

HVL = 0.693/µ (ln 2 = 0.693; µ = linear attenuation coefficient)

• typically mm Al for superficial x-rays; mm Cu for orthovoltage; mm Pb for megavoltage
• beam shielding blocks are typically 5 HVL of shielding (<5% transmission)

As attenuation coefficient decreases, HVL increases. For photon energies from 100keV to 20 MeV, HVL initially increases (due to Compton interactions) and then decreases (as pair production becomes more prevalent at higher energies)

Question 28

Mass attenuation coefficient

Answer 28

(µ/ρ); linear attenuation coefficient divided by mass density of the material (measured in cm²/g)

Question 29

Coherent Scatter

Answer 29

incoming photon is completely absorbed by electrons of an atom which causes the electrons to vibrate and radiate a photon with identical energy in a different direction

-only occurs at very low photon energies

-can cause blurring of diagnostic X-ray images

-of no importance in high energy XRT

Question 30

Photoelectric Effect

Answer 30

incoming high-energy photon ejects an inner shell electron with subsequent filling of vacant electron shell causing emission of characteristic x-rays. Incoming photon is completely absorbed (it disappears!).

-the ejected electron has the kinetic energy of the incoming photon minus the binding energy of the electron
(Incoming photon - binding energy of electron = kinetic energy of ejected electron)

• characteristic x-ray energy equals the energy difference between electron shells (outer shell electron fills the empty spot in the inner shell)
• the probability of photoelectric interaction: Z³/E³ (Z=atomic #; E = energy of photon)
• more likely to happen in high Z material (like bone)
• less likely to happen with increasing photon energy

The photoelectric effect is predominant for photon energies in the range of 10 keV (For soft tissue, the PE and Compton effects are equal at 25 keV.For bone, the PE and Compon effects are equal at 40 keV)

In imaging (such as kV CB CT), High Z-material will cause alot of artifact due to dominance of the photoelectric effect at kV energies!!

Question 31

Compton Effect

Answer 31

high energy photon knocks a loosely held outer shell electron from an atom (leaving a positively charged ion), with the remaining energy forming a lower energy photon traveling in a different direction

• depends only on electron density (electrons per unit mass) and is independent of the atomic number of the absorbing material! (electron density is directly proportional to the physical density)
• Energy change is related to the angle of scatter (Θ); calculate using the change in the photon’s wavelength (λ); remember units of λ are in meters

The scattered photon is always less energetic than the incoming photon; the most energetic 90° photon is .511 MeV

The Compton photon can be scattered at any angle, but the Compton electron is emitted at an angle limited to 0-90° with respect to the direction of the incident photon.

Maximum energy is transferred to the electron when the photon is backscattered (so less energy is transferred to the scattered photon; 180° (from original direction) backscatter photon has the least energy, 255.5keV or less), the Compton electron then goes off in the direction exactly opposite to the backscattered photon (0° relative to the direction of the initial photon)

Example: Calculate Energy change of incoming 1 MeV photon that produces a scattered photon at 90° from incoming direction:
Remember: E = hc/λ so λ= hc/E (hc = 1.24 * 10^-12 MeVm)
∆λ = 2.4 * 10^-12 (1-cos Θ)m (Θ is angle between the paths of initial and scattered photon) → ∆λ = 2.4 * 10^-12 (1-cos 90°)m = 2.4 * 10^-12m
λf= λi + ∆λ
λi = hc/E = 1.24 * 10^-12 (m * MeV)/ 1 MeV = 1.24 * 10^-12 m
λf =λi + ∆λ = 1.24 * 10^-12m + 2.4 * 10^-12 m = 3.64 * 10^-12 m
So the Scattered photon energy (E) = hc/λ = 1.24 * 10^-12 MeVm)/3.64 * 10^-12 m = 0.3407 MeV (340.7 keV)

For imaging, such as MV CB CT, the artifact from high-Z materials will be relatively less than with a kV CB CT since most interactions with MV energies is Compton and is not affected by Z nearly as much as the photoelectric effect (such as in kV imaging)

Question 32

Pair Production

Answer 32

Incoming photon interacts with electric field of nucleus (electric field strength depends on Z), creating an electron and a positron (positive electron).

-Incoming photon is completely absorbed.

-most important above 10MeV

• requires at least 1.02 MeV incoming photon (since each particle in the pair has a rest mass of 0.511 MeV); excess energy above 1.02 MeV is divided as kinetic energy
• Likelihood of interaction is directly proportional to the log of Z
• positron then undergoes Pair Annihilation with a different free electron (producing two annihilation photons, each of 0.511 MeV.

Question 33

Pair Annihilation

Answer 33

Positron (created by pair production) interacts with a nearby (different) free electron, causing the annihilation of both and creating two photons with energy of 0.511MeV each, which travel in opposite directions

Question 34

Photonuclear Interaction (gamma-n interaction):

Answer 34

-only occurs with photon energy above 15MeV

-high energy photon is absorbed by nucleus → forms unstable nucleus → to regain stability, a neutron is emitted from the nucleus; neutron then interacts with another nucleus making it radioactive

-Can result in transiently (~10 min) radioactivity of parts of the linear accelerator (beam flattening filters, wedges; especially if made of copper)

Question 35

Mass energy absorption coefficient

Answer 35

**μ_en/ρ ** (μ_en= linear energy coefficient; ρ = density of material in the attenuator

Question 36

Temperature/Pressure Correction

Answer 36

C = (760/P) * (273.2 + t/295.2) where P and t are the given pressure and temperature

Higher temperature thins the air (rarefies it), and the TC offsets this

Higher Pressure compresses the air, and the PC offsets this

Question 37

Ion Chamber collection efficiency (f)

Answer 37

charge collected / charge liberated by initial ionization

Question 38

TLD dose to patient

Answer 38

calibration dose * (patient reading/calibration reading)

Question 39

Calculating a Half Life or Remaining Activity

Answer 39

**A(T) = A₀ (1/2)^T ** (when time is measured in half lives)
Example: T_1/2 = 3 hours, T = 10 hours, A₀ = 100Bq → A(T) = 100 (0.5)^3.333 = 100 (0.099) = 9.9 Bq

or

A(t) = A₀e^-λt (when time is measured in a unit of time, e = 2.71828 and λ=0.6931/T_1/2 and t= time (days/hours etc in same unit as T_1/2)

Example: A(t) = 100e^-2.31=100 (0.099) = 9.9Bq when T_1/2 = 3 hours, t = 10 hours)

Question 40

Gas Ionization Detectors

Answer 40

Geiger-Mueller (G-M) counters (can detect very amounts of radiation such as from a lost I¹²⁵ seed), proportional counters (can discriminate type of radiation and measure exposure rates; example is a portable ionization chamber called a cutie pie), ionization chambers (used to calibrate linacs and measure treatment beam characteristics)

In general, ionization chambers are filled with a gas and have two electrodes; ionizing radiation causes ionized ion pairs to travel to the positive and negative electrodes)

Electrodes must be polarized, which requires voltage (supplied by battery or other voltage source)

Question 41

Geiger-Mueller counters

Answer 41

Used to detect low intensity radiation

Usually measures exposure in counts per minute (cpm) or exposure per hour (mR/hr or R/hr); 5mR/hr is about 250cpm

G-M counters are very sensitive since one ionization event creates a pulse of current in which all of the chamber volume ionizes at once (the chamber must then reset before it can register another ionization event); thus they have a dead time; so in a high-intensity radiation field the G-M counter can read zero (above 4R/hr)

G-M counters can under-respond at energies <30keV because of beam attenution in the walls (high Z material, usually metal), and over-respond at moderate energies (30-100keV) because of the photoelectric effect of high Z material walls

Question 42

Neutron Dosimeters

Answer 42

Linacs with energy >10MV produce neutrons; low Z moderating detectors are required to detect (hydrogen or boron detectors)

Question 43

Scintillation Detectors

Answer 43

scintillation crystal (sodium iodide, calcium floride, bismuth germinate) absorb a photon which produces ionization and in turn produces light (amount of light is proportional to energy of the absorbed photon)

More sensitive than G-M counters (because they have a photomultiplier tube)

Can discriminate one isotope from another by evaluating differences in pulse strength (energy)

Can be used to measure surface contamination or brachytherapy source leakage (via wipe tests) or bioassay thyroids after exposure to I¹³¹ dosing

Question 44

Linear Energy Transfer (LET)

Answer 44

LET: energy deposited per unit path length (keV/µm). Linear energy transfer (LET) is the average amount of energy a particular radiation imparts to the local medium per unit length; ie: Energy per Length. For radiotherapy, we are normally concerned about small amounts of energy over small distances, so the units we use are keV/μm.

Examples:

Orthovoltage Photons (250keV): 2.0 keV/μm
60Co Photons (1.17 - 1.33 MeV) 0.3 keV/μm
Linac based Photons (3 MeV): 0.3 keV/μm
Neutrons (14 MeV) Track Average: 12 keV/μm
Neutrons (14 MeV) Energy Average: 100 keV/μm
Protons (10 MeV) Average: 4.7 keV/μm
Protons (10 MeV) On Entering Phantom: 0.5 keV/μm
Protons (10 MeV) At Bragg Peak: 100 keV/μm

Key Facts about LET:

1. LET is proportional to the charge density of a medium
2. LET is proportional to the charge (squared) on the particle moving through a medium
3. LET is inversely proportional to the speed (squared) of the particle
4. LET is related to teh density of ionization along the particle’s track
5. Maximum cell killing occurs at an LET of approximately 100 keV/μm
6. RBE decreases with increasing LET above about 100 keV/μm (“overkill effect”)
7. RBE shows the greatest changes for LET values between roughly 20 and 100 keV/μm
8. OER decreases slowly with increasing LET for low LET values, but falls rapidly after LET exceeds about 60 keV/μm

Question 45

Thermoluminescent Dosimeters

Answer 45

Measure skin dose and radiation dose abutments; also used in ring badges

Lithium Flouride (LiF) is material most commonly used since it has an effective Z similar to tissue and air

CaF₂:Mn is also used for radiation badges

TLD’s are heated to release trapped energy in the form of light (thermoluminescence); this is done in a TLD reader which heats the TLD to 300C; light is measured by the photomultiplier tube in the TLD reader

Crystals are “zeroed” in an annealing oven for 1 hour @ 400C and 2 hours @ 100C

Before reading the TLD dose, a pre-anneal step of heating @ 100C for 10 minutes is used to dump low temperature traps

Dose is calculated by multiplying by a calibration factor known to the TLD

Question 46

Diode Detectors

Answer 46

Solid state devices that measure dose and/or dose rate

Capable of measuring dose at a surface (like a TLD)

Capable of reading/displaying dose immediately (unlike a TLD there is no extra steps necessary)

Diodes can be used for photon or electrons, but sperate detectors are designed for each

Question 47

Roentgen

Answer 47

1 Roentgen = 1R = 2.58 * 10^-4 coulomb/kg (of dry air)

Roentgen is a unit of exposure (ionizations)

1 Roentgen (exposure dose) = ~0.873cGy (absorbed dose)

Highest energy that can be measured is 3MeV

Is only defined for X-ray and gamma-ray radiation (not for particles like electrons, neutrons, protons)

R/min = (Exposure reading) * (Temp in Kº) * (760mmHg) * (Calibration Factor)/(Exposure time) * (295ºK) * (Pressure in mmHg)

Question 48

Beam Output

Answer 48

Output decreases with filtration (exponentially with monoenergetic beams, complex with polyenergetic beams)

Output increases linearly with increased tube current (mA); so to compare output with an initial mA and a different mA multiply the R/min * (mA_new/mA_initial) –> make sure the answer makes sense (larger if more mA!!)

Output increases with increased kVp (at 100 kVp, output increases with the square of the kVp so to compare a given kVp to a different kVp multiply the R/min * (kVp_new/kVp_initial)² –> make sure the answer makes sense (larger if more kVp!!)

If the distance from the source changes, a closer distance will have more output (calculate by multiplying R/min * (D_initial/D_new)²

Question 49

KERMA

Answer 49

• Kinetic Energy Released in Matter in Gy (not the energy Absorbed (dose; also in Gy) in matter)
• KERMA is measured in joules/kg (Gy)
• In the buildup region (e.g. 1cm depth for a 15MV photon beam), the KERMA is greater than the absorbed dose!
• Air KERMA is the kerma released in air

Question 50

rad

Answer 50

rad = radiation absorbed dose

rad = 100 ergs of energy absorbed per gram of matter

1 rad = 1 cGy (SI unit equivalent)

rad (and cGy) can be used for photon, charged particles, and neutron dosimetry

Question 51

Gy

Answer 51

1Gy = 100 cGy = 100 rad

Gy is a unit of absorbed dose

1Gy = 1 joule of absorbed energy/1 kilogram of material = 1J/kg

Question 52

Sievert and rem

Answer 52

1 Sv = 100 rem

Sv is an SI unit, rem is not

Question 53

f factor (also known as the f_medium factor)

Answer 53

relates dose in air to dose in tissue (exposure dose in air x f factor = absorbed dose)

also referred to as the roentgen to cGy (or rad) conversion factor (be careful that they give you the correct units as they may give you mGy or something else and you will get tricked!!!)

f_med = ((μ_en/ρ)medium / (μ_en/ρ)air) * 0.873 cGy/R

D_med = X * f_med (X is the Roentgen, D_med is the cGy or rad)

For high-energy beams the chamber correction factor is required (A or A_eq) which for ⁶⁰Co is 0.985; thus the equation changes to D_med = A_eq * f_med * X = D_med = C_{<strong>λ</strong>} *** X ** (A_eq * f_med is sometimes called C_λ)

At lower photon energies (~100keV) the higher the Z (e.g. bone vs. water) the higher the f_med (because of substantial photoelectric absorption that is proportional to Z^₃)

A similar process is used for electrons using the equation: D_med = C_e * X cGy

From >100keV to 2 MeV the f_med of higher Z materials falls due to pair production

At high energies (>2MeV) the f_med of higher Z materials once again starts to rise (but is lower than water until ~10MeV)

Soft tissue f_med is fairly constant

Question 54

Bragg-Gray cavity theory

Answer 54

Used to determine absorbed dose for x-rays above 3 MeV (since roentgen can’t be used for these higher energies)

Assumes that the ionization chamber acts ike a tiny cavity (hole) inside a uniform phantom

Question 55

Quality Factor

Answer 55

Q; compares the biological effectiveness of a particulate radiation to a statndard X-ray radiation (usually orthovoltage or ⁶⁰Co)

Q is basically RBE

H = dose equivalent

H(Sv) = D(Gy) x Q

or H(rem) = D(rad) x Q

remember that 1Sv = 100 rem

Question 56

Quality Factor (Q) vs. Beam Energy

Answer 56

Radiation Source Q

X-ray, γ-ray 1

Electron 1

Proton 2

Neutron 10

Alpha Particle 20

Heavy recoil nuclei 20

(Neutrons have a higher Quality Factor (Q) than electrons becuase they transfer energy to protons, which have a large mass and are densely ionizing at the end of their tracks (Bragg Peak)

Question 57

NCRP Radiation Exposure Limits (and key ICRP differences)

Answer 57

Effective Dose: Sum of Absorbed dose * W_R * W_T

(W_R = radiation weighting factor; W_T = tissue weighting factor)

Deterministic Effects:

1. 150 mSv/year for the lens of the eye
2. 500 mSv/year for localized areas of the skin, the hands, the feet

Stochastic Effects:

1. No occupational exposure until 18 years of age
2. Effective dose in any year should not exceed 50mSv (5 rem)/year
3. Individual worker’s lifetime effective dose should not exceed age in years x 10mSv (ICRP is 20 mSv/y averaged over 5 years)

Embryo/Fetus Exposure (once pregnancy declared): 0.5 mSv/month (ICRP is a total of 1mSv to abdomen surface)

Public Exposure (annual): Does not include medical imaging dose as this is assumed to confer a “benefit” to the person

• Effective dose limit: continuous or frequent exposure: 1 mSv/year
• Effective dose limit (infrequent exposure): 5 mSv/year (ICRP is still 1mSv/year…no infrequent vs. frequent designation!!!)
• Dose equivalent limits - Lens of eye: 15mSv/year
• Skin/extremities: 50 mSv/year

Education/Training Exposure (annual): (ICRP has no statement on these dose constraints!!)

• Effective dose limit: 1 mSv/year
• Dose equivalent limits - Lens of eye: 15mSv/year
• Skin/extremities: 50 mSv/year

Negligible Individual Dose (annual): the dose below which further expenditure to improve radiation protection is unwarranted: 0.01 mSv/year

Lifesaving Events: exposure up to 0.5 Sv (choose older workers if possible. Exposure over 0.5 counsel regarding short/long term consequences

Average Annual equivalent dose to radiation workers: 2 mSv (total detriment of 1/10000; this is equal to the risk of a fatal accident in a safe industry)

Question 58

Relative Biological Equivalent = RBE

Answer 58

• Relative Biological Equivalent = RBE
• RBE = D_X/D_R (where DX is a reference absorbed dose of radiation of a standard type X (generally an X-ray beam with 250 KeV photons), and DR is the absorbed dose of radiation of type R that causes the same amount of biological damage (e.g. 50% cell kill). Both doses are quantified by the amount of energy absorbed in the cells)
• RBE is essentially Q (quality factor)
  
  • RBE of 250 keV X-ray = 1
  • RBE of 1-10 MV X-ray = 0.8-0.85

Question 59

Radioactive Decay equation

Answer 59

A = A₀e^-λt

λ (lambda) = probability per unit time of decay (sometimes called the disintegration constant)

t = time from zero

A = the radioactivity in units of disintegrations per unit time

If you do not have actual activity numbers, you can use:

A = A₀ x 0.5^{(days elapsed/half life of source)}

Question 60

Half Life (T_1/2)

Answer 60

A = A₀e^-λt —> A/A₀ = 1/2 = e^-λT_1/2

so taking the log of both sides —> ln 1/2 = -λ * T_1/2

and –> T_1/2 = ln 1/2 / -λ = -0.693/-λ = 0.693/λ

thus —> T_1/2 = 0.693/λ

(remember that λ is the disintegration constant)

Question 61

ß- decay

Answer 61

Nucleus with too many neutrons converts one of its neutrons to a proton /ß- and expels a ß- particle; leaving the atom more stable

⁶⁰Co decays by beta decay to the stable isotope ⁶⁰Ni The activated nickel nucleus emits two gamma rays with energies of 1.17 and 1.33 MeV, hence the overall nuclear equation of the reaction is:

⁶⁰Co –> ⁶⁰Ni + ß- + anti-neutrino (v) + gamma rays

(gamma rays are from the ⁶⁰Ni nucleaus with energies 1.17 MeV and 1.33 MeV for average 1.25MeV)

(the anti-neutrino is almost massless; energy released is in form of kinetic energy sharted by the three resulting particles –> so not all ß- have the same energy!!)

Question 62

ß+ decay

Answer 62

Nucleus with too many protons and too few neutrons converts one of its protons to a neutron and expells a ß+ particle, leaving the atom more stable

¹²N —> ¹²C + ß+ + neutrino + energy

ß+ particle interacts with matter like an electron, but once it slows down it combines with a nearby electon in mutual annihilation producing 2 photons of 0.511 MeV each!!!

For B+ decay, it is a NEUTRINO!!!

Question 63

Electron capture

Answer 63

An orbiting electron is captured by the nucleus and a proton is converted to a neutron

125I + e^- –> 125Te + v_e

(v_e is an electron neutrino)

Question 64

Isomeric transition (IT)

Answer 64

Repositioning of nuclear particles within nuclear shells to obtain a more stable configuration with no particulate emission.

Excess energy is released in the form of a gamma ray.

Example is Technetium-99 used in nuclear medicine imaging

Technetium-99m (^99mTc) –> ⁹⁹Tc + 144keV

Question 65

Percent Depth Dose (PDD)

Answer 65

PDD = 100 * (Dose at depth (d) / dose at depth (dmax))

• Fraction of the maximum dose remaining at a particular depth in a phantom or patient (e.g. PDD at 10cm)
• PDD is a measure of beam penetration
• Two components: patient attenuation and inverse square dose fall-off

PDD:

• Field size: increases with increasing field size due to increased dose due to scatter
• Beam Quality: Increases with increasing beam quality due to the increased penetration of the primary (unscattered) photons
• Depth: decreases as depth increases; also realize that PDD increases between the surface to d-max, and then decreases exponentially
• Elongation of field: for a given field area, an elongated field has a lower PDD than a square field because the number of scattered photons is less; the PDD for a rectangular field is the same as that for its equivalent square)
• SSD: Increases… because by ratio, the dose at dmax (aka output or dose/MU) decreases faster than dose at depth (refer to Mayneords f-factor)

D2 = D1 * (PDD2 / PDD1)

By energy at 100 cm SSD, 10x10 field, and depth of 10cm

• Co-60 56%
• 4MV 61%
• 6MV 67%
• 10MV 73%
• 20MV 80%
• 25MV 83%

To calculate PDD from electron percent depth ionization, multiply by (L/ρ) (water to air)

Question 66

Geometric Penumbra

Answer 66

P = S * (SSD - SDD)/SDD

Penumbra is measures in cm

S = source diameter or focal spot size (in cm)

SSD = source skin distance (in cm)

SDD = source diaphragm distance (or source collimator distance (in cm)

A larger source size will have a larger penumbra

The smaller the penumbra, the more the radiation is confined to just the tumor

A smaller source size will have a less penumbra

Decreasing the SSD will decrease the penumbra!

Increasing the SDD will decrease the penumbra!

Question 67

Physical Penumbra

Answer 67

Physical Penumbra = Geometric Penumbra + side scatter photons/electrons

Often noted as the distance between the 80% and 20% beam intensities on the beam profile (or 90% and 20% as per Khan)

Question 68

Recordable Event

Answer 68

Define

Question 69

Medical Event/Misadministration

Answer 69

(a) A licensee shall report any event, except for an event that results from patient intervention, in which the administration of byproduct material or radiation from byproduct material results in—

(1) A dose that differs from the prescribed dose or dose that would have resulted from the prescribed dosage by more than 0.05 Sv (5 rem) effective dose equivalent, 0.5 Sv (50 rem) to an organ or tissue, or 0.5 Sv (50 rem) shallow dose equivalent to the skin; and
(i) The total dose delivered differs from the prescribed dose by 20 percent or more;
(ii) The total dosage delivered differs from the prescribed dosage by 20 percent or more or falls outside the prescribed dosage range; or
(iii) The fractionated dose delivered differs from the prescribed dose, for a single fraction, by 50 percent or more.
(2) A dose that exceeds 0.05 Sv (5 rem) effective dose equivalent, 0.5 Sv (50 rem) to an organ or tissue, or 0.5 Sv (50 rem) shallow dose equivalent to the skin from any of the following—
(i) An administration of a wrong radioactive drug containing byproduct material;
(ii) An administration of a radioactive drug containing byproduct material by the wrong route of administration;
(iii) An administration of a dose or dosage to the wrong individual or human research subject;
(iv) An administration of a dose or dosage delivered by the wrong mode of treatment; or
(v) A leaking sealed source.
(3) A dose to the skin or an organ or tissue other than the treatment site that exceeds by 0.5 Sv (50 rem) to an organ or tissue and 50 percent or more of the dose expected from the administration defined in the written directive (excluding, for permanent implants, seeds that were implanted in the correct site but migrated outside the treatment site).
(b) A licensee shall report any event resulting from intervention of a patient or human research subject in which the administration of byproduct material or radiation from byproduct material results or will result in unintended permanent functional damage to an organ or a physiological system, as determined by a physician.

Question 70

Skin Gap

Answer 70

Skin gap = (L1 / 2)*(d/SSD1) + (L2 / 2)*(d/SSD2)

or

d/2 (L1/SAD + L2/SAD)

L = length of the field, d = depth of match, SSD = source to
surface distance; for isocentric setups substitute SAD for SSD

Question 71

Diagnostic tube output calculation

Answer 71

Output = kVp² x mA

(dose rate increases proportionally with an increase in mA)

Question 72

Fractional Depth Dose (FDD)

Answer 72

Fractional Depth Dose (FDD) = (Dose rate at depth) / (dose rate at d_max)

FDD represents attenuation and inverse square falloff between d_max and depth d

Question 73

Mayneord F-Factor

Answer 73

** Mayneord F-factor = PDD₂/PDD₁**

Mayneord F-factor = [(SSD₁ + d)/(SSD₁ + d_max)]² * [(SSD₂ + d_max)/(SSD₂ + d)]²

F > 1 if SSD₂ > SSD₁ (i.e. PDD₂ > PDD₁)

F = 1 if SSD’s are equal

F < 1 if SSD₂ < SSD₁ (i.e. PDD₂ < PDD₁ )

Question 74

Photon D_max values

Answer 74

• Co-60 0.5 cm
• 4MV 1.2 cm
• 6MV 1.5 cm
• 10MV 2.5 cm
• 15MV 3.0 cm
• 18MV 3.3 cm
• 20MV 3.5 cm
• 25MV 4.0 cm

D_max is = the depth at which dose and kerma are equal, the maximum range of secondary electrons, the depth at which electronic equilibrium occurs

Question 75

Photon Attenuation

Answer 75

• Co-60 ~4.0% per 1 cm depth
• 6MV ~3.5% per 1 cm depth
• 20MV ~2.0% per 1 cm depth

Question 76

Probability of Interaction

Answer 76

Coherent scattering ≈ Z
Photoelectric absorption ≈ Z³/E³
Compton scattering ≈ independent of Z, ≈ 1/E, ≈ electrons/gram
Pair production ≈ Z² (directly proportional to the log of Z)

Question 77

Heterogeneity Correction Factors

Answer 77

Air: 10 cm of air ≈ 3 cm of tissue (air attenuates ~1/3 of tissue, so if no correction there is a higher dose at the prescription point)

Bone: 10 cm of bone ≈ 16 cm of tissue (bone attenuates ~1.6 times normal tissue, **so if no correction there is a lower dose at the prescription point)**

With higher energy, less correction necessary (since Compton effect is 1/E)

With higher energy, slower build-up at lung/tumor interface, and thus possibly underdosing

Question 78

Skin Dose

Answer 78

Varies with energy (decreases as photon energy increases)

SSD (decreases as SSD increases)

field size (increases as field size increases)

bolus (increases with increased bolus)

oblique incidence (increases with increasing oblique incidence)

Question 79

Tissue air ratio (SAD setup)

Answer 79

TAR = Dose at depth / Dose in air

Question 80

Tissue phantom ratio (SAD setup):

Answer 80

TPR = Dose at depth / Dose at reference depth

Question 81

Tissue maximum ratio (SAD setup):

Answer 81

TMR = Dose at depth / Dose at dmax

Question 82

Total dose calculation of a seed implant

Answer 82

Total dose = initial dose rate * 1.44 * half-life

(1.44 * half-life is called the mean-life)

Example: ¹²⁵I with initial dose rate of 0.1Gy/hr

Total dose = 0.1Gy/hr * 1.44 * 60 days * 24hr/day = 207 Gy

Question 83

Inverse Square Law

Answer 83

Inverse Square Law:

I₁/I₂ = D₂² /D₁²

Thus… I₂ = I₁ * D₁² / D₂²

I₁ = Intensity @ Distance 1

I₂ = Intensity @ Distance 2

D₁ = Distance 1

D₂ = Distance 2

Question 84

Thermal Enhancement Ratio (TER)

Answer 84

RT dose without heat/ RT dose for equivalent effect with heat

Question 85

TG-51

Answer 85

• Applies for photon and electron beams with energies between Co-60 and 50 MeV
• Provides a way to measure absorbed dose in water, in Gy, at the point of measurement of the ion chamber, when it is absent

Ionization chamber must have a calibration factor (traceable to the NIST standard) obtained in water, at an Accredited Dosimetry Calibration Laboratory (ADCL)

Reference dosimetry must be performed in a phantom with dimensions of at least 30x30x30cm

Requires yearly full calibration in a water tank for all photon and electron beams in clinical use

Allows the use of plane-parallel chambers in a high-energy electron beam, calibrated against a cylindrical chamber having an ADCL calibration (so you do NOT need to have a chamber having an electron calibration factor provied by an ADCL).

Standard temperature and pressure is usually 22°C (273+22 = 295K) and 760 mmHg (or 101.33kPa)

Beam quality for MeV electron beams is based on the beam’s R₅₀ (depth in water at which the absorbed dose is 50% of the maximum dose for the beam.

Question 86

Homogeneity index

Answer 86

Degree of dose uniformity in the target volume

Question 87

Conformity Index

Answer 87

Tissue volume receiving the given dose / target volume encompassed by the same dose

CI = TV / PTV

Question 88

Wedges and Wedge Angle

Answer 88

The angle through which the isodose at 10cm depth is rotated from its position in the open beam (don’t be fooled into thinking its the angle of the physical wedge!!!!)

Wedge Angle = 90° - (0.5 * hinge angle)

• Example: hinge angle = 120*° ; wedge angle = 30°
• or*
• Hinge Angle = 180° - (2 * wedge angle)*

Thick end of wedge = heel; thin end of wedge = toe

Effective wedge angle (aka “universal wedge”): use of a single wedge, with varying proportion of the mu setting that the wedge is in the beam (dynamic wedge or virtual wedge)

Question 89

Brachytherapy Seed options and energies

Answer 89

Radionuclide Average Photon Energy (keV) Range Half-Life HVL (mm Lead) Type
Cesium-137 662 - 30y 5.5 γ-ray
Iridium-192 380 136-1060 74.2d 2.5 photon
Iodine-125 28 3-35 60.2d 0.025 photon
Gold-198 412 - 2.7d 2.5 γ-ray
Americium-241 60 - 432y 0.125
Palladium-103 21 20-23 17d 0.008 photon
Samarium-145 41 38-61 340d 0.06
Ytterbium-169 100 10-308 32d 0.1

*Some planning systems account for anisotropy (for low energy sources like I-125 and Pd-103); others treat the source as an isotropic point source but add a dose reduction factor to account for anisotropy

Unsealed sources:

I-131: is NOT used for brachytherapy, it is an unsealed source administered systemically for diagnostic or therapeutic purposes. It has a half-life of 8 days. It is a medium-energy, mixed-spectrum β- and γ-emitter with a γ emission at 364 keV that can be detected using a gamma camera. Part of Bexxar (used for refractory NHL)

⁹⁰Y: is a pure β-particle emitter

• Decay product of Strontium-90 (also a pure B- emitter) which has a T_1/2 = 28.6 years
  
  • used as an applicator for treatment of pterygium of the eye (surface dose rate of 48cGy/second)
  • Do not confuse with Strontium-89 which is also a pure B- emitter with a T_1/2 = 52 days and a predilication to bone (used for bone mets!)
• relatively high energy (0.9 MeV average, 2.27 MeV maximum; average penetration of 2.5 mm.
• Part of Zevalin (used for refractory NHL) and also hepatic microsphere therapy (microspheres made of glass (Thera-Spheres; tx of HCC) and microspheres made of resin (SIR-Spheres; treatment of metastatic CRC))

Air Kerma Strength: air kerma rate (μGy/h) at a specified distance (usually 1 m) from the source center along the perpendicular bisector = μGym² h^-1

Activity = λ * Number of atoms present (remember that λ = 0.693/T_1/2 and that λ is the disintegration constant)

A = A₀e^-λt

1 Curie = 1 gram of radium = 3.7 * 10¹⁰ disintegration/second = 3.7 * 10¹⁰ Bq

Question 90

Hounsefield Units (HU) for typical materials

Answer 90

The CT number (in Hounsfield units) = 1000 * [(u_material - u_water)/u_water

(where u = the linear attenuation coefficient)

• Lung: -700 to -900
• Fat: -50 to -100
• Muscle: +40
• Soft Tissue: +100
• Bone: +700 to +1000

Black is the most negative value (smallest) and white is the most positive number (largest)

Question 91

TG-142 (Quality Assurance)

Answer 91

Daily Linac QA:

Output for each photon energy

Accuracy of the distance indicator (ODI)

Laser alignment

Operation of the patient viewing system

Yearly Linac QA:

Output vs. Field Size

SBRT:

Mechanical tolerance of laser localization for SRS/SBRT: 1.0 mm

Imaging and treatment beam coincidence tolerance for SRS/SBRT: ≤1.0 mm

See TG-142 Table III for contents of requirements.

Question 92

Optical Density

Answer 92

Optical Density (OD) = Log(I₀/I_t) where I₀ is the incident intensity and I_t is the transmitted intensity

Example:

OD = 2 = Log (100/1) where I₀ = 100% and I_t = 1%

Question 93

V₉₅ vs. D₉₅

Answer 93

V₉₅: percent of the organ or OAR volume covered by 95% of the prescription dose (a VOLUME getting a certain dose (in this case 95% of the prescription dose)

• Example: V_20Gy specifies the volume of the structure receiving at least 20Gy (the bin at 20Gy contains the volumes of all of the area to the right on the Differential DVH
• the unknown or quantity to define is a Volume!!!

D₉₅: % of the prescription dose that covers 95% of the target volume (a DOSE covering a certain volume (in this case, the dose covering 95% of the target volume)

• Example: the D₉₅ specifies the minimum dose that covers 95% of the structure volume
• The unknown or quantity to define is a DOSE!!!

DVH Facts:

• a Differential DVH (bins)
• A Normalized (typical view) of the DVH allows us to look at different organs with widely differing volumes on a single chart
  
  • to “normalize” the data, the volume at each dose is divided by the total volume of the object of interest
    
    • so you see the percent of the object of interest vs. the actual volume of the object of interest (because different object can have vastly different total volumes!!!)

Question 94

Imaging Storage Calculation

Answer 94

GB of Storage = (bits/pixel) * (1 byte/8 bits) * (1GB/ 1.024*10⁹ bytes) * (# * # pixels/slice) * (cm of scan/cm of scan thickness) * # of phases scanned

• Example:16 bit gray scale, 512x512 pixels, 25cm long, 5mm slice thickness, 10 phases = ? GB
  
  • GB = (16bits/pixel) * (1byte/8bits) * (1GB/1.24*10⁹ bytes) * (512 * 512 pixels/slice) * (25cm/0.5cm per slice) * (10 phases) = 0.256GB

Voxel = pixel area * slice thickness = 3 D volume element

Question 95

Transport Index (TI)

Answer 95

Transport Index (TI) = highest radiation level (in mR/hr) at 1 meter from the surface of a package

Example: Exposure at 1 m from the surface of a package is measured to be 0.003 R/hr; the TI is 3 (0.003R/hr = 3mR/hr)

Question 96

TG-43 (Brachytherapy)

Answer 96

• Dose rate constant for the source, Λ, defined as the dose rate per unit air kerma strength at 1 cm along the transverse axis of the source.

Question 97

Equivalent Square

Answer 97

Equivalent Square = 2xy/x+y

Equivalent Square of a Circle: for a square with side a will be equivalent to a circle with radius r when they have the same area, a^2 = πr^2, so a = r √π, or a = 0.89 * D (D=diameter)

Question 98

Electron Dosimetry

Answer 98

• Probability of bremsstrahlung interaction: Z²
• Lead block thickness to attenuate 95%: t_Pb (mm) = Electron energy / 2
  
  • Cerrobend block thickness t_Cerr = 1.2 * t_Pb
• Range
  
  • Practical range in water: Rp (cm) = Electron energy / 2
  • R50: depth at which dose is 50% of maximum
    
    • R50 = Electron Energy at surface/2.33
• X-ray emission spectrum proportionate to kVp2 * mAs / d2, also depends on amount of filtration
• Depth of calibration
  
  • I50: Find depth of 50% ionization in water
  • R50: Calculate R50 = 1.029 * I50 - 0.06 if <10 cm depth, R50=1.059 * I50 - 0.37 if >10 cm depth
  • d_ref = 0.6 * R50 - 0.1
  • Energy is specified by the R50 parameter
• Typically treated as SSD setup
  
  • No physical source in accelerator head; clinical beams appears to emerge from a “virtual source”. Can be found by backprojecting beam profiles at different depths
  • Virtual SSD shorter than actual (photon) SSD
  • Inverse square corrections can be done on virtual SSD for large fields; for small fields effective SSD should be determined
  • Output Dose rate = Applicator Dose rate * Back scatter factor(cutout)/Back scatter factor(Applicator)/ (SSD/SSD+SO)^2 (SSD= Source to surface distance & SO= Stand Off

Question 99

isotopes, isotones, isobars, isomers

Answer 99

Isotope: nuclides with the same Z but different A

Isotone: same number of neutrons (A - Z) but different Z

Isobar: nuclides with the same A but different Z

Isomer: same A, same Z, but different internal nuclear energy, different configuration of the neutrons and the protons

Question 100

Radioactive equilibrium

Answer 100

• Secular equilibrium: occurs when T_1/2 parent >>> T_1/2 daughter
  
  • Equilibrium is reached in ~5-8 T1/2 of daughter (~1 month for Ra capsule).
  • Radium needles: When equilibrium is reached the activity of radon equals the activity of radium (i.e., the number of radon atoms decaying equals the number of radon atoms produced).
  • When equilibrium is reached the activity of radon equals the activity of radium (i.e., the number of radon atoms decaying equals the number of radon atoms produced).
  • 90 Sr applicators
• Transient equilibrium : occurs when T_1/2 of the parent is greater, but not much greater, than the T_1/2 of the daughter.
  
  • In transient equilibrium there is an initial buildup of the daughter nuclide on a time scale dependent on the daughter’s half life. After transient equilibrium is established the activities of both the parent and daughter decay with the half life of the parent.
  • Activity(daughter) = Activity(parent) × [T_1/2 parent⁄(T_1/2 parent −T_1/2 daughter)]
  • Nuclear medicine generators

Activity = λ × Number of atoms present

λ = 0.6931⁄T_1/2

Question 101

Standing wave guides vs. Traveling wave guides

Answer 101

Traveling wave guide:

• Microwaves are injected into one end of the accelerator wave guide. At this end electrons are also injected.
• Those electrons which encounter the accelerating portion of the electric field will travel down the tube in synchrony with the field.
• Microwaves are dumped at the end of the waveguide into a sink where they are absorbed and their energy is converted into heat.
• the length of the tube makes a vertically mounted accelerator impractical; therefore the accelerator is mounted horizontally in the gantry; the electron beam is then bent 90° or more often 270° after leaving the accelerator wave guide

Standing Wave guide:

• In a standing wave accelerator the microwaves are not dumped at the end but instead are
  reflected back.
• The standing wave pattern alternates in time between accelerating and decelerating phases and
  also alternates in space down the wave guide. Separating the accelerating/decelerating cavities
  are null cavities which do not provide acceleration at any time
• Since the null cavities (or coupling cavities) do not affect the electron beam a better design has them moved off to the side of the structure. This side coupled standing wave accelerator, is a highly efficient accelerator, capable of producing a 6 MeV electron beam in a length of 30 cm
• A 6 MeV accelerator of the standing wave design can be mounted vertically on a gantry with the X-rays emerging from a target at the end of the accelerator (CyberKnife is of this design)

Question 102

Dose Reporting (ICRU - International Commision on Radiation Units)

Answer 102

ICRU Dose Reporting Guidelines:

• ICRU recommends that the reported dose be the dose at a point, the ICRU reference point.
  
  • The dose at the point is clinically relevant and representative of dose in PTV
  • The point is chosen to satisfy the following:
    
    • point can be clearly and unambiguously defined
    • The dose can be accurately determined (physically)
    • There is no steep dose gradient at the point.
  • To satisfy these requirements the ICRU recommends the reference point be:
    
    • at the center of the PTV and where the tumor cell density is greatest
    • on or near the beam central axis (or axes for multiple beams)
    • ***If both conditions cannot be simultaneously satisfied, the first takes precedence.
• In addition to the dose to the ICRU reference point, the report should include the maximum and minimum dose in the PTV

Three levels of ICRU dose reporting:

• Level 1: Basic; reference point dose and estimates of maximum and minimum doses in the PTV using central axis dose tables
• Level 2: Advanced; Level 1 + dose distribution (isodose curves) in a plane or planes
• Level 3: 3D; Level 1 + volumetric dose distributions, non-coplanar beams, dose volume histograms (DVH’s)

Question 103

MRI

Answer 103

Larmor frequency (ω) = γ * B₀

• γ = gyromagnetic ratio (each nuclear spin has a different value for γ and therefore a different Larmor frequency)
• B₀ = Tesla
• For protons, the Larmor frequency (in Mhz) = ω = 42.58 * B₀

T1 vs. T2

• T1 is the longitudinal relaxation time; spin-lattice relaxation time
• T2 = transverse relaxation time; spin-spin relaxation time

Question 104

Ultrasound

Answer 104

**t (time) = 2d/v ** (d = depth of the object below the surface; v = velocity of sound in the tissue (assumed to be 1540 meters/second))

**d = vt/2 ** (to find depth of an object below the patient surface)

• Example: 65 micro-seconds after initial US pulse the transducer picks up the reflected sound wave, what is the depth of the object?
• d = 1540m/s * (65*10^-6) / 2 = 0.05 meters = 5 cm

Attenuation:

• α (dB/cm) = 0.5 * frequency (Mhz)
  
  • α = attenuation coefficient
  • in tissue, attenuation increases linearly with the ultrasound frequency
  • lower frequency results in greater penetration, higher frequency produces an image with better spatial resolution

Question 105

Brachytherapy numbers to know

Answer 105

• 1 Curie = 1 gram of radium = 3.7 * 10¹⁰ disintegration/second = 3.7 * 10¹⁰ Bq
  
  • Curie is a specification of activity (# of decays per second, NOT a measure of a mass of material)
• radium equivalent mass = milligram radium equivalent (mg-Ra eq)
  
  • 0.825 mR/hour exposure @ 1 meter in air; 8.25 R/h @ 1cm
  • NOT a measure of mass
  • NOT a measure of activity –> It is a measure of exposure rate
  • 1 mCi of a radionuclide X DOES NOT have the same exposure rate as 1 mCi of Ra
• Exposure rate (R/h) = Γ * A/r²
  
  • Γ = R-cm² / mg-h
    
    • Γ assumes encapsulation of 0.5mm platinum (exposure rate is reduced by 2% for every additional 0.1mm Pt
    • Γ for commonly used sources
      
      • ²²⁶Ra = 8.25
      • ¹³⁷Cs = 3.26
      • ¹⁹²Ir = 4.69
• Air Kerma Strength (cGy) = Exposure (R) * W/e
  
  • W/e = 0.876cGy/R
  • S_K= Γ * A * W/e

Question 106

Approaches to Brachytherapy implants

Answer 106

1. Manchester (Paterson-Parker)
   
   1. non-uniform distribution of Radium in order to produce a uniform distribution of dose (+/- 10%)
   2. More of the Radium is in the periphery of the implant than in the center
2. Memorial (Quimby) System
   
   1. uniform distribution of sources, spaced 1 cm apart; leads to a non-uniform dose distribution with the maximum dose at the center of the implant and the minimum at the periphery
   2. If the dose is specified at the maximum, then fewer milligrams of Radium are used than in the P-P system; if the dose is specified at the minimum, then more Radium is used than in the PP system!
3. Paris System
   
   1. activity per cm is uniform for all sources
   2. spacing between sources is constant but not necessarily 1cm
   3. treated length = 0.7 * active length
   4. presciption isodose (reference isodose) is a percentage of the basal dose rate
      
      1. BD = BD₁ + BD₂ + …/ # of BD’s

Implant Verification

• Magnification Factor = (source to film distance) / (source to implant distance)

Isodose Distribution issues

• Some planning systems account for anisotropy (for low energy sources like I-125 and Pd-103); others treat the source as an isotropic point source but add a dose reduction factor to account for anisotropy
  
  • Anisotropy factors
    
    • ¹²⁵I = 0.95
    • ¹⁰³Pd = 0.90

Question 107

Radiation Protection

Answer 107

• Nuclear Reactors and reactor product radionuclides are regulated by the NRC (Nuclear Regulatory Commission)
• Naturally occuring radionuclides (Radium) and X-ray units and accelerators are regulated by state radiation protection offices
  
  • these state radiation protection offices generally follow NCRP (National Council on Radiation Protection and Measurements) recommendations
• Dose Equivalent:
  
  • Dose Equivalent (Sievert or rems) = Q * Dose (Gray or cGy)
    
    • Q = quality factor (like W_R)
      
      • Photon: 1
      • Electron:
      • Proton: 2
      • Charged Pion: 2
      • Alpha Particle: 20
      • Fision fragments/heavy ions: 20
      • Neutrons (10 keV): 5
      • Neutrons (10 keV - 20 MeV): 10 to 20
      • Neutrons (>20 MeV): 5
• Radiation Protection: 2 D’s + 1 B
  
  • Distance
  • Duration
  • Barrier
• Exposure (Roentgens) = (Γ * A * t * B)/ d^₂
  
  • Γ = exposure rate constant in R-cm²/mCi-hr
  • A = activity in mCi
  • t = time in hours at the location
  • d = distance (in cm) from the source to the individual
  • B = barrier transmission factor (the fraction of the incident radiation that gets transmitted)

Primary Protective Barrier: (shielding for direct beam)

• DE = (W * U * T * B)/ d²
  
  • DE = the weekly maximum permissible dose equivalent (in rems)
  • W = weekly workload in cGy per week @ 1 meter from the source
  • U = use factor, the fraction of time the beam is directed to the location in question
    
    • U = 1 for the floor
    • U = 1/4 for the walls
    • U = 1/4 to 1/2 for the ceiling
  • T = occupancy factor, the fraction of time an individual spends at the location in question
    
    • for occupational exposure occupancy factor is always 1 (offices, work areas, etc)
    • for public exposure the value depends on the type of room adjacent to the treatment room
      
      • partial occupancy = 1/4 = for corridors, rest rooms, etc
      • occasional occupancy = 1/16 = for waiting rooms, toilets, stairways, etc
  • d = distance (in meters) from the source to the point
  • B = barrier factor
    
    • for high energy photons (Compton is dominant) –> barriers are lead, steel, or concrete (most often used for cost considerations)

Secondary Protective Barrier: (shielding from scatter radiation and head leakage)

• Scattered Radiation
  
  • DE = (W * T * B * α) / (d₁ * d₂)²
    
    • d₁ = distance from the X-ray source to the patient
    • d₂ = distance from the patient to the individual
    • α = fraction of the incident radiation that is scattered at the appropriate angle
    • for 6 MV X-rays, α = 6 * 10^-4
• Head Leakage
  
  • DE = (0.001W * T * B)/d²
  • For leakage, the use factor (U) is = 1
  • Cobalt machines have constant leakage (must be no more than 2 mR/hr @ 1 meter from the source)

Frequent public exposure limit is 1 mSv/year = 0.1 rem/year = 0.002rem/week

Radiation Level Verification:

• Geiger-Mueller counters are more sensitive (a single particle traversing the chamber causes all the gas in the chamber to ionize, resulting in a large signal)
  
  • good for quickly finidng a radiation hot spot
• “Cutie Pie” counters are large volume portable ionization chamber used to more to accurately measure exposure level
• Neutron Survey Meters are necessary when using energies above 10MV
  
  • filled with BF₃ gas enriched with ^₁₀B (has a high probability of absorbing a thermal neutron and emitting an alpha particle)
  • to detect fast neutrons, the tube is centered in a polyethylene sphere (slows fast neutrons via collisions with the hydrogen in the sphere)