PHYS5404 Radiation Physics and Dosimetry RADIATION ASSIGNMENT 3 STUDENT NO

PHYS5404 Radiation Physics and Dosimetry
RADIATION ASSIGNMENT 3
STUDENT NO: 22364461
QUESTION 1:
(a). What is the primary standard generally used for photon beams below 300kv?
Answer: The primary standard for 10-300 kv x-rays is the FREE ION CHAMBER which is used world widely as primary standard for calibration of x-ray exposure. Standard free air ion chambers measure air kerma in air according to its definition by collecting ions produces by radiation beams that result from the direct transfer of energy from photons to primary electrons in defined volume of air. It measures all ionizations produced by collision interaction sin air resulting from x-ray interactions in known mass of air related to exposure. different designs are used in standardized laboratories like cylindrical and some of plane parallel geometry.

some alternate designs are also used like attix designs which have some properties like no voltage divider and no guarding electrode systems and errors generated from field non uniformity are eliminated and other one is PTB designed s which is coaxial cylindrical design and is smaller for same photon energy than standard design and potentially transportable.

REFRENCES:
https://www.aapm.org/meetings/09SS/documents/15McEwen-PrimaryStandardsfinalforVL.pdf
module 16 and 17 class notes and lecture slides available.

(b). Can this primary standard be used for MV beams, if not why and which device(s) are generally used?
Answer: No,this primary standard can not be used for MV beams because of range of charge carriers in MV in air would be large enough to handle. the multiplication from this range of charged would be large so output signal that is to be detected at signal should too large to be get collected.Generally the GRAPHITE CALORIMETERS are used as primary standard for MV beams.

REFRENCES:
https://www.arpansa.gov.au/our-services/testing-and-calibration/calibration/radiotherapy- calibration/australian-primary-standards/graphite-calorimeter
Module 16 and 17 lecture notes and lecture slides available.

(c). This apparatus is generally constructed of graphite but we generally specify dose to water? Why is graphite used and not water?
ANSWER: Graphite is generally used instead of water due to its specific heat capacity which is one sixth of that of water and therefore graphite’s temperature rise is six times more than that equivalent in water and measurements are more accurate .Graphite has low atomic no. and all absorbed energy can be reappeared as heat without any loss of hear in other mechanisms like heat defects .In graphite calorimeter the average temperature is measured in a body that is generally thermally insulated from surrounding bodies by evacuated vacuum pumps and dose transfer procedures are done in conjunction with graphite calorimeters to allow for transfer of absorbed dose from graphite to water.

REFRENCES:
https://en.wikipedia.org/wiki/DosimetryQUESTION 2: Both the proportional counter and Geiger counter are based on internal gas multiplication. Discuss the use of quenching gas in proportional counters versus Geiger Muller tubes and its impact on dead time?
ANSWER: In proportional counters gas filled in chamber is an inert gas which is ionized by an incident radiation and a quench gas to ensure that each pulse discharge terminates. Quench gas is a mixture of a small amount of gas molecules and molecules having vibrational mode to absorb energy and not radiate UV photons. They will tend to collect the charge exchange reactions and are less likely to emit electrons from cathode on neutralization .However these molecules tend to get damaged flow gas. A common mixture of gases used is 90% argon and 10% methane known as P-10.———-1
Quenching gas is needed for as the number of atoms in excites state will grow proportional to the multiplication . the excited states tend to decay by photon emission rather than by ionization between neutral atoms and molecules. Photons often strike with chamber walls i.e cathode .Cathode we chosen will be a metal so it will also tend to emit photoelectron so the collision of an atomic ion with cathode release an electron during process of neutralization so successive formation of pulses takes place in a long time and hence have a large dead times. So quenching gas will absorbs the UV photons which is added to primary photomultiplier tube and hence will decrease the dead time.———–2
In GM counters the multiplication and a continuous output of multiple pulses is much severe than that of proportional counters for this reason quenching is done in GM counters to prevent the possibility of excessive multiple pulsing as the output. The GM should give single pulse for every single ionizing event due to radiation so that it should come easily in passive state and should be ready for next event .As the gas filled in GM tube is a mixture of inert gases such as helium ,argon and neon which is ionized by ionizing radiation and a quench gas of 5% of a halogen gas or an organic vapor to prevent multiple pulsing by quenching electron avalanches when positive ion tend to reach cathode and gain electrons behave as neutral atom and can be excited to higher energy levels and tend to come back to ground state by emitting photons ad produce further ionization and therefore secondary discharges takes place. these prolonged avalanches will increase dead time so no new events can be detected and it should become continuous and can damage the tube so quenching has to be done to reduce the dead time and prevent the damage of GM tube. Quenching can be done in two ways one by external and other by internal quenching in external quenching some electronics circuitry is used like use of high resistor value for constant charge collection in circuit is of millisecond and it produce very small pulsing rates. but common method that is used is internal quenching to prevent multiple pulsing which is attained by adding quench gas to primary gas and there is mostly an addition of polyatomic gas or vapor originally such as butane or ethanol. So this type of quenching in GM counters will prevent to produce the spurious pulses and hence have a short dead time.———–3
REFRENCES:
https://www2.chemistry.msu.edu/courses/CEM988Nuclear/lectures/Chem988_S09-Ch6.pdfhttp://pulsar.sternwarte.uni-erlangen.de/wilms/teach/astrospace/spacechap3.pdfhttps://en.wikipedia.org/wiki/Geiger%E2%80%93M%C3%BCller_tube218 page no. of RADIATION DETECTION AND MEASURMENT BY Knoll
QUESTION 3. A source of 116mIn (half-life =54.3min) is counted using a G-M tube. Successive 60 seconds readings gave 131,340 counts at 12:00 noon and 93,384 counts are 12:40. Neglecting background, calculate the true interaction rate in the G-M tube at 12:00 noon.
ANSWER: Let n be true interaction rate, m be recorded count rate , ? be dead time of system.

So for measuring dead time losses two types of models are considered one is paralyzable and other is non paralyzableRate at which true events lost is given by =mn?——-{1}
Rate of losses=true interaction rate – recorded count rate —–{2}
From {1} and {2},we get;
n-m=nm?n=m1-m?——–{3}
the rate at which such events occur by true rate n
m=ne-n? ——{4} for paralyzable model —————–1
by taking log on both sides of above equation we get :
logm0=logn0-n?———{5}
in case of zero or negligible background
n=n0e-?t—————-2
using this value in equation {3}:
n0e-?t=m1-m?On solving this by multiplying with factor e?t,we get
me?t=n0-n0m?———{6}
let m0 be measured rate at time ‘0’ and m1 be measured rate at time ‘1’
taking log on both sides of above{6} equation and after rearranging terms we get :
n0?=e?t(n0ln-m1ln-?t)
using this value in equation {5},we get;
ln(m0)=ln(n0)- e?t(n0ln-m1ln-?t)
by rearranging all terms while solving above equation we get;
ln(n0)= e?tlnm1+?t-ln?(m0)e?t-1——–{7}
from given statement In question :
m0=131340 at 12:00
m1=93384 at 12:40
half life period=54.3
?=0.69354.3=0.0127
time t=12:40-12:00=40 minutes
using all above given values in equation {7}
ln(n0)=e0.51049 ln93384+0.51049-ln?(131340)-1+e0.51049 minute-1
ln(n0)=8.131340.6666minute-1
n0=e12.19182=197169.68 minute-1
for paralyzable model:
by using equation;n-m1=nm1?
using equation n=n0e-?t in above equation :
n0=m1(1-m1?)e-?twe get n0?=n0e-?t -m1e-?t m1using this value of n0? in equation {5} and after rearranging terms and solving it we get:
n0=(e?t-1)m0m1m0-m1put given values of all quantities as stated above we get:
n0=e0.51049 -1*1.2265*1010131340-93384n0=-1+1.66107*1.2265*101037956n0=215244.514minute-1
REFRENCES:
1.equation (4.27) at page no. 137 in book of RADIATION DETECTION AND MEASURMENT BY Knoll
2.equation (4.34) at page no. 137 in book of RADIATION DETECTION AND MEASURMENT BY Knoll
QUESTION 4.On the 1st of April 2017, a worker spent 5.5 hours in a room with a radio isotope which has a reference activity of 50mCi Curie on the 10th of July 2014. The isotope has a half-life of 4 years. The worker has a mass of 80kg and a surface area of 1.8m2 and generally works 3m away from the radioactive source. The isotope decays and produces gamma-rays that carry 2.6MeV of energy. Only 40% of the x-rays are absorbed by the workers body.

( a) What is the effective dose the worker experiences each day?
ANSWER: intial activity of person on the 1oth july,2014=50mCi
Half life period of given isotope=4years
?=0.6934years=0.17325 year-1
As we know :A=A0e-?tA=50*e-0.17325 year-1*996 daysA=50*e-0.17325*1*996days365daysA=50*0.62328mCi
=31.16mCi
=115292*104 disintergrations per second
d=3m
Decrease in radiation per unit area=1/4?d2=8.83*10-3metre-2
Energy absorbed by worker =1.8*A*energy*time of exposue4?d2Energy of gamma rays =2.6Mev
Time of exposure for worker=5.5 hours
So
E=1.8*2.6*5.5*60*60*7*115292*1044*22*9E=7.478*1014792Mev
E=9.4419*1011Mev
E=9.4419*1011 *1.602*10-13 joules
E=0.15125 joules
Mass of worker=80kg
40% of absorbed dose=0.4*Em=0.4*0.15125 J80kg =7.562*10-4 J/kg
Effective dose =Absorbed dose *weighing factor for gamma rays
=7.562*10-4J/kg*1
=7.562*10-4*103msV
=0.7562msV
( b). How long before the worker reaches their annual limit in WA?
ANSWER: current annual limit of radiation exposure for workers =50mSv ——–1
Radiation exposed by worker in one day is = 1/0.75 mSv per day
Time before which the worker reaches the annual limit=50/0.75=66.1 days
REFRENCES:
https://www.arpansa.gov.au/our-services/monitoring/australian-national-radiation-dose-register/information-for-workers
QUESTION 5: A totally depleted silicon detector with 0.1 mm thickness is operated with large overbias so as to saturate the carrier electrons everywhere within the wafer. Estimate the maximum electron and hole collection times.
ANSWER:
Thickness d=0.1mm=0.01cm
As we know that ;For large over bias voltage; tc,np=dµn,pE1——–1
And vd=µE———1(a)
So ; tc,np=dvd———–1(b)
Where at high fileds the constant drift velocity is given by =107cm/s for E;105V/cm——1(c )
tc,np=0.01cm107cm/s=1*109 per second
by obtaining value of drift velocity for holes and electrons from graphs :

drift velocity for holes at 300k is given by 1.4*106cm/s
charge collection time for holes=0.01cm1.4*106=7.142*10-9per second
drift velocity for electrons =1.9*106cm/s
charge collection time for electrons=0.01cm1.9*106cm/s =5.2*10-9 per second
REFRENCES:
1.(a),(b),(c) http://www.desy.de/~niebuhr/Vorlesung/Detektor/Vorlesung_5.pdfGraphs for drift velocity from book of book of RADIATION DETECTION AND MEASURMENT BY Knoll at page 372
QUESTION 6:
A scintillator absorbs an incident flux of 5 MeV alpha particles that totals 106 particles/s. The scintillation efficiency for these particles is 5%, and the average wavelength of the emitted light is 420 nm. If the scintillator is coupled to a photodiode with an average quantum efficiency of 70% for the scintillation light and the light collection efficiency is 80%, calculate the expected signal from the photodiode when operated in current mode. Can this level of current be measured?
ANSWER: given that:
Energy of alpha particles =5Mev
Rate=106 particles/second
Average wavelength =420nm
Scntillaion efficiency(S)=5%,quantum efficiency (QE)=70%,light collection efficiency=80%
The output signal in form of current signal is given by:
Energy absorbed per second by scintillator=5%*5*106*106eV
=0.05*5*1012eV=4.005*10-8Joules
Energy absorbed by charged particle per second =4.005*10-8Joules*1.6*10-19coloumb
=6.408*10-27joule coloumbEnergy of photons =hc?=6.623*10-34*3*108JOULE METRE PER SECOND 420*10-9 meter=4.7*10-19J second
Current signal=Energy absorbed by charged particle per second*QE*light collection efficiencyEnergy of photons =6.408*10-27joule coloumb4.7*10-19Jsecond *0.7*0.8
=7.635*10-9coloumb/second=7.635*10-9Amperes=7.653nA
This level of current is small but detectable.

QUESTION 7.Explain why the existence of electron trapping sites in a crystalline material is desirable if the material is used as a thermoluminescent dosimeter, but undesirable if the material is used as a conventional scintillator.

ANSWER:A TLD is used often used instead of film badges. These dosimeters can measure low dose rate the only limitation for using TLD’S is a time required for readout and store the information. In TLD’s there is a imperfect lattice structure is used which can easily absorb and store ionization radiation and when heated up and reemission is done inform of light and that reemitted light is detected and and output form of light is correlated to absorbed dose that is received earlier. when a T LD material is exposed to ionizing radiation at some certain value of temperatures then radiation interacts with crystal and deposit all incident energy in material. some of atoms get ionized by energy that get absorbed and free electrons will produce and holes will also be created in TLD material.so now electrons will move free in conduction band and may be get trapped at sites that are some imperfections in crystal lattice and release trapped electrons which fall down to ground state and will release energy in form of light .Released light is counted by PM tubes and counted number of photons are proportional to quantity of ionizing radiation .The number of resulting events are measured by repeated process again and again so total number of events in total time often repeating can be get integrated.——–1(a),1(b)
The scintillator materials posses the property of scintillation when excited by ionizing radiation .when an ionizing radiation struck the luminescent materials absorb energy and absorbed energy get reemitted in form of light .In in organic scintillators the scintillation processes are possible to electron band structure of crystals .so an incoming particle excite an electron from valence band to conduction band and excitation band that is located between the conduction band and valence band and holes are formed in valence band. Impurities in crystal lattice structure causes electronic levels in forbidden gap. These excitations are loosely band pairs of electron holes pairs and moved in crystal lattice until they are captured in impurity center of crystal lattice structure, rapidly de excited in form of an emission of light scintillating .The activator impurities are used for reemission to be done in visible light and PMT tubes response is effective. The holes that are associated with conduction band electrons can move freely and these holes and electrons are captured by impurity centers under some metastable states .The delayed de excitation of theses impure metastable states results in slow scintillation light emission mechanisms.——2(a)
The scintillators are used because of the properties like high density and low cost fast operation speeds and durability parameters .High operating speeds will posses the good resolution of spectra. Particles deposit their energy that is directly proportional to response to scintillators and short decay times are important for measurement of time intervals and for output results that coincide or occurring at same time interavls.—–2b
REFRENCES:
1.(a) https://www.nde-ed.org/EducationResources/CommunityCollege/RadiationSafety/radiation_safety_equipment/thermoluminescent.htm(b) 19.2 article from class module notes
(c )page no. 746,747 from book of of RADIATION DETECTION AND MEASURMENT BY Knoll
2a https://en.wikipedia.org/wiki/Scintillator2bpage no. 246 from book of RADIATION DETECTION AND MEASURMENT BY Knoll
QUESTION 8:
Discuss the feasibility of operating a BF3 or 3He tube in the ionization or Geiger regions rather than as a proportional tube.

ANSWER:
With a very small amount of voltage or no threshold voltage most of the ions then tend to recombine and no electrical signal output signal is produced .when we apply positive voltage across anode then electrons will move towards it and positively charged ions move towards anode .The output signal magnitude generated depends upon applied voltage ,geometry of chamber and filled gas all these parameters help to determine whether in which region the detector would work in ionization ,proportional or Geiger Muller region .In ionization region we apply voltage enough so as to collect all electrons before they get recombined and plateau is formed at this point and when we increase the voltage further then collected charge is proportional to energy deposited initially in detector systems the generated pulses at output in this region is small.——1(a),(b)

In next proportional region electric field strength is very large enough for ionization of gas and produce secondary ionization and further increased voltage will ionize the molecules of gas by secondary produced electrons.it will lead to large multiplication of formed ion pairs primarily so in proportional region charge collected is linearly proportional to energy get deposited in gas .the output pulses formed will have initial fast time rise because of electrons motion an slow rise time due to positive ions motion which are formed near anode so pulse amplitude is large due to drifting of positive ions. the pulse have full amplitude when fully positive ions get collected.——-1©
Further increase in applied voltage will be responsible for loss of proportionality between primary charge deposited and output signals. This loss is because f saturation regions near anode wire the avalanches process begins and grow upto maximum value as secondary electrons create an additional avalanche and multiplication stops and prevent secondary electrons acceleration so now in this region detector operates in G.M region.——-1(d)The spectrum formed by neutrons is independent of neutrons energy but simply depends upon construction of detector. As gamma rays produce very small pulses from neutrons and can be easier to distinguish pulses by increasing the chamber size The operating voltage to work in GM region is quite high the electronic noise is generated and pulses generated due to background gamma rays exceed and result into the spurious counts.so neutron detectors operated in either ionization or proportional mode can give us average output current which depends upon electronics that used.BF3 gas filled detectors basically used in proportional regions because in this gas multiplication is linear and proportional to charge collected initially .——1(e )
REFRENCES:
1(a) http://www.conceptualphysicstoday.com/2012/09/boron-tri-fluoride-10bf3-gas-filled.html(b),(c),(d) http://www.lanl.gov/orgs/n/n1/panda/00326408.pdf(e) https://www.orau.org/PTP/collection/proportional%20counters/bf3info.htmQUESTION9:
A survey meter, whose time constant is 4 seconds reads 10 mR (100 ?Sv) per hour while measuring the radiation from a dental X-ray exposure of 0.08 second.
What was the actual exposure rate?
ANSWER: Actual exposure rate=dose rate1-e-tRCGiven :Dose rate =10mR per hour
t=0.08second
RC=4 seconds
Using given values in above equation:
=10mR per hour1-e-0.02=10mR per hour1-0.98019=504.795mR/hour
1mR=0.001R
So actual exposure rate=504.795*0.001=0.50479 R/hour
What would have been the dose to the dental hygienist if she had been at the point of measurement?
ANSWER:Exposure dose =dose rate *time ——-1
=0.50479R/hour*0.08second
1R/hour=2.777*10-6Gy/s——–2
=0.00000140219Gy/s*0.08 second
=1.12*10-7Gy
REFRENCES:
1 https://www.nde-ed.org/EducationResources/CommunityCollege/RadiationSafety/safe_use/time.htm
2 https://www.translatorscafe.com/unit-converter/en/radiation/26-1/QUESTION 10.What is the basic radiation physics principle for bone densitometry?
ANSWER:
Bone densitometry is done to find the amount of bone mineral in bone tissue.so as per name indicates the bone densitometry is the amount of mass of mineral per volume of bone. But clinically is it dine by measuring optical density per square centimeter of bone surface upon which imaging is done .this treatment is done by small radiation exposure in clinics for risk of osteropsis (a disease in which bone weakness increase risk of bone broken and fracture risk)and results are generally given with gram per square centimeter and interpretation of results is done with T-score and Z-score .different types of bone densitometry tests are done like DEXA(DUAL ENERGY X-RAY ABSORPTIOMETRY),QUS(QUALITATIVE ULTRA SOUND),SPA(SINGLE PHOTON APSORPTIOMETRY),SEXA(SINGLE ENERY X-RAY ABSORPTIOMETRY) and more treatments techniques are used.——–1
DEXA is most commonly used but QUS Is cost effective . In DEXA two gamma ray energies were used to measure bone mineral density. this procedure is also called dual photon absorptiometry and 153 Gd isotope is used to emit gamma rays up to 100 kev range. now a days gamma ray exposure from radioisotope has been replaced by the X –ray tube .two different techniques are used to generate two x-ray beams energies in first approach KV and filtrations are rapidly switched during imaging in other approach a single x-ray energy with two different filters are used to generate and harden beams .In DEXA treatment scintillation counter and x-ray tube are mounted on c-arm that allows exposure of pencil x-ray beam in rectilinear way used for scanning .scintillators used are usually consist of CdWO4 and NaI(Tl) in PMT’s are coupled .pencil beams will reduce the scattering of radiation exposure .The c-arm is moved in rotating manner for CT imaging and outputs can be recorded by computer for data processing for dual energies .——-2
In densitometry the way that how photons are absorbed like a mass attenuation coefficient which decreases with photon energy and rises with atomic number and mass attenuation coefficient related to dual energies will help us to distinguish between tissue and bone because photon absorption will be more in bone than fats and equivalent in tissues and muscles and water.——3
REFRENCES:
1 https://en.wikipedia.org/wiki/Bone_density2https://en.wikibooks.org/wiki/Basic_Physics_of_Nuclear_Medicine/Dual-Energy_Absorptiometry3 http://hoiloangxuonghcm.vn/AbstractSBA2013/Worksop-SBA2013/Radiation%20Physics%20Principles%20of%20DXA%20Basic%20Statistics%20for%20DXA%20Essential%20Anatomy%20for%20DXA.pdf