ROphex Old Q1000+ Flashcards
Q1000. A radioactive sample is counted for a ten minute interval many times, yielding a mean count rate of 100 cpm. The most probable distribution is:
A. 68% of the measurements fall between 990 and 1010 cpm
B. 68% of the measurements fall between 936 and 1064 cpm
C. 68% of the measurements fall between 900 and 1100 cpm
D. 95% of the measurements fall between 936 and 1064 cpm
E. 95% of the measurements fall between 800 and 1064 cpm
A. 68% of the measurements fall between 990 and 1010 cpm
–If a large number of measurements(N) are made, approximately 67% will fall between +σ, and 96% between +2σ of the mean. The standard deviation σ=sqrt(N).
In this case, 1000 cpm x 10 min = N = 10000 counts. σ=sqrt(N) so it is 100. 100 averaged over 10 minutes is 10 cpm.
Therefore 1000+/-(10)= 990-1010
Q1002. A series of measurements has a mean of 100 counts. A range of +σ is ___ .
A. 95-105
B. 90-110
C. 68-137
D. 50-150
E. 33-167
B. 90-110
–The standard deviation s is the square root of the mean, in this case sqrt(100) = 10.
–68% fall within σ of the mean.
–95% fall within 2 σ of the mean.
Q1004. To achieve a standard deviation of 2%, ___ counts must be collected.
A. 400
B. 1,414
C. 2,500
D. 10,000
E. 40,000
C. 2,500
%σ = σ/N x 100 = N^.5/N x 100 = 100/N^.5 = 2
50 = N^.5
N = 2500
Q1006. A radioactive sample is counted for 1 minute and produces 900 counts. The background is counted for 10 minutes and produces 100 counts. The net count rate and net standard deviation are about ___,___ counts.
A. 800, 28
B. 800, 30
C. 890, 28
D. 890, 30
E. 899, 30
D. 890, 30
900/1 min - 100/10 min = 890/1 min
590^.5 = 30 = 1 standard deviation
Q1012. In a chi-square test, looking for a statistically significant difference between two experimental results, claims of such a difference with a p value of 0.01:
A. Means there is unquestionably a difference between the two results.
B. Allows the experimenter a wider latitude of error than would a p value of 0.05.
C. Means there is a 99% chance that the claim is true.
D. Means there is a 99% chance that the claim is incorrect.
C. Means there is a 99% chance that the claim is true.
Q1018. Concerning the poisson distribution (answer A for true and B for false):
- It is another name for the normal distribution
- It is due to random variations
- Photon distribution on an x-ray film is a poisson distribution
- Radioactive decay as a function of time is a poisson distribution
- The standard deviation increases as the number of measurements increases
- The percent standard deciation increases as the number of measurements increases
- It is another name for the normal distribution - B, FALSE; Poisson is binomal distribution whether it either occurs or not. Normal is much larger.
- It is due to random variations - A, TRUE
- Photon distribution on an x-ray film is a poisson distribution - A, TRUE
- Radioactive decay as a function of time is a poisson distribution - A, TRUE
- The standard deviation increases as the number of measurements increases - A, TRUE; σ=(N)^1/2, so yes
- The percent standard deciation increases as the number of measurements increases - B, FALSE
Q1020. Information will be destroyed in/on a ___ when the computer power is turned off.
A. Floppy disk
B. Hard disk
C. Magnetic tape
D. RAM
E. ROM
D. RAM - random access memory
ROM is read only memory
Q1022. Concerning digital computers, all of the following are true, except:
A. ROM stands for Random Order Memory.
B. A Word is a set of consecutive bits treated as an entity, and occupying one storage location in memory.
C. A byte contains 8 bits.
D. A modem is a device that converts a digital signal into a frequency-coded signal for transmission over a telephone line.
A. ROM stands for Random Order Memory.
- stands for read only memory
Q1024. The number of binary bits that are required to represent all CT numbers from -1024 to 3096 is bits.
A. 8
B. 9
C. 10
D. 11
E. 12
E. 12
3096+1024 = 4120
2^12 = 4096
k-bit is the number of bits per pixel, the grey scale of an image is equal to 2k-bit
k-bit of 2 = 4 shades of grey
k-bit of 8 = 256 shades of grey
Q1026. The computational speed of a computer is measured in units of
A. MB
B. MIPS
C. RVU
D. BAUD
E. BPI
B. MIPS
–MIPS is Millions of Instructions Per Second.
Q1028. ROM is memory that can be
A. Used and changed freely
B. Freely read, but not written to
C. Repeatedly used to store output from an input device
D. Randomly accessed
B. Freely read, but not written to
Q1030. Parallel processing refers to:
A. Running multiple tasks simultaneously.
B. Using multiple processors to increase speed.
C. Computer networking.
D. Sharing peripheral devices between computers.
B. Using multiple processors to increase speed
Q1032. There are approximately __ bits in a megabyte.
A. 1024
B. 2048
C. 8000
D. 2,000,000
E. 8,400,000
E. 8,400,000
1 byte = 8 bits
1 kilobyte = 1024 bytes (computers based on binary (base 2)
1 megabyte = 1024 x 1024 bytes = 1048576 bytes
so 1048576 bytes x 8 bits/ 1 byte = 8388608
Q1034. A CT image consists of 200 slices each 512 x 512 pixels each pixel having a 16 bit pixel depth. The size of the file is ____
A. 500 kB
B. 5 MB
C. 10 MB
D. 50 MB
E. 100 MB
E. 100 MB
200 slices x 512 x 512 pixels x 16 bits/ pixel x 1 byte/ 8 bits x 1 megabyte/ (1024 x 1024 bytes) = 100 MB
Q1036. The mainframe of a digital computer contains:
- core memory
- central processing unit (CPU)
- only random access memory
- software
- connectors for peripheral devices
A. 1, 2, 5
B. 1, 2
C. 2, 4
D. 1, 4
E. 1, 2, 3, 4, 5
A. 1, 2, 5
Q1043. A 16 bit word computer can directly address a maximum of how many different locations?
A. 16
B. 32581
C. 58325
D. 65536
E. 130036
D. 65536
2^16
Q1048. All of the following contribute about equally to the average annual dose equivalent received by a member of the US population, except:
A. Internal
B. Terrestrial, other than radon
C. Medical x-rays
D. Nuclear medicine
E. Cosmic
D. Nuclear medicine
–Out of a total of about 2.6 mSv, nuclear medicine contributs about 0.14 mSv, and the other all contribute 0.3 to 0.4 mSv each.
Q1050. The highest dose received by the population from natural background radiation is from:
A Cosmic radiation.
B. Radon.
C. Internal radiation.
D. Terrestrial, other than radon.
B. Radon
–Radon contributes 2 mSv to the average annual effective dose equivalent.
–Out of a total of about 2.6 mSv, nuclear medicine contributes about 0.14 mSv, and the other all contribute 0.3 to 0.4 mSv each (including medical x-rays, internal, cosmic, and terrestial radiation EXCLUDING radon)
Q1052. The average natural background is made up of cosmic radiation, terrestrial radiation and:
A. fallout
B. scattered medical radiation
C. nuclear plant releases
D. radioactive waste disposal contamination
E. internal radiation
E. internal radiation
–NATURAL IS KEY, as other products are artificial.
–About 40 mrem/yr is contributed by radionuclides within the body, most 40K.
- The most significant source of man made radiation dose to the population as a whole is from:
A. high altitude air travel
B. television recievers and other consumer products
C. fallout from nuclear weapons exploded in the atmosphere
D. diagnostic radiological examination
E. nuclear reactor effluents
D. diagnostic radiological examination
Nuclear med is 2nd after x-ray
Q1056. The annual average natural background radiation dose to members of the public in the United States, excluding radon, is approximately ___ mrem.
A. 10
B. 50
C. 100
D. 200
E. 400
C. 100
–Radon adds another 230 mrem/yr. Man-made radiation, mostly diagnostic x-rays, is about 54 mrem/yr.
Q1058. The largest contribution to the radiation exposure of the U.S. population as a whole is from:
A. Radon in the home.
B. Medical x-rays.
C. Nuclear medicine procedures.
D. The nuclear power industry.
A. Radon in the home.
–Radon, at 230 mrem/yr, is twice other natural background radiation, which in turn is about twice all man-made radiation put together.
Radon (~200 mrem) > other background natural (100 mrem) > man made AKA mostly x-rays (50 mrem)
Q1060. The principal hazard from indoor radon involves:
A. Whole body dose from gamma rays.
B. Skin dose from betas.
C. Lung dose from alpha emission.
D. Bone dose from deposited radionuclides.
C. Lung dose from alpha emission.
Q1065. About half the average effective dose equivalent received by the U.S. population is attributable to:
A. Radon
B. Medical procedures
C. Fallout
D. Cosmic radiation
E. Internal radionuclide
A. Radon
Radon (~200 mrem) > other background natural (100 mrem) > man made AKA mostly x-rays (50 mrem)
Remember radon is damaging via alpha to lung tissue
Q1066. Match the following annual radiation levels with the sources listed. Answers may be used more than once.
A. 50 mSv
B. 10 mSv
C. 2 mSv
D. 1 mSv
E. 0.5 mSv
- Average dose to a member of the population from radon
- Maximum recommended dose to a radiation worker
- Average dose to a member of the population from medical uses of radiation
- Maximum recommended dose to a member of the public (infrequent exposure)
- Average dose to a member of the population from radon (C. 2 mSv)
- Maximum recommended dose to a radiation worker (A. 50 mSv)
- Average dose to a member of the population from medical uses of radiation (E. 0.5 mSv)
- Maximum recommended dose to a member of the public (infrequent exposure) (D. 1 mSv)
Q1068. The average total annual dose to a member of the public from background and man-made radiation is ___ mSv.
A. 5.0
B. 3.5
C. 1.5
D. 0.4
E. 0.2
B. 3.5
Q1073. According to NCRP report #91, the quality factor for high LET radiations such as neutrons should generally be:
A. 1
B. 2
C. 5
D. 10
E. 20
E. 20
Q1076. Studies of the affects of the atomic bombs dropped on Japan during World War II indicate that the probability of inducing cancer in a large population that is irradiated with 1 rem is about ___ during that population’s lifetime.
A. 1 in 10
B. 1 in 1,000
C. 1 in 10,000
D. 1 in 100,000
E. 1 in 1,000,000
C. 1 in 10,000
Q1078. Current models of radiation dose v. effect assume that 100 mSv (10 rem) delivered over a year to 1 thousand people will cause ___ additional cancer deaths, compared to an unexposed group of the same size.
A 0.05
B. 0.5
C. 5
D. 50
E. 100
C. 5
–The current value is 5 x 10^-2 per Sv. (BEIR V)
1 rem = 1 in 10,000 additional cancer cases
10 rem = 1 in 1,000 additional cancer cases
Q1080. The latent period for radiation-induced carcinogenesis (solid tumors) is about ___ years.
A. 1
B. 5
C. 10
D. 20-30
E. 40-50
D. 20-30
Q1082. The currently accepted model of radiation dose versus effect used by regulatory agencies to determine dose standards is ___ .
A. Linear quadratic.
B. Exponential.
C. Cubic.
D. Linear, threshold.
E. Linear, no threshold.
E. Linear, no threshold.
Q1084. A whole body dose of 5 mSv/yr for 20 years would increase a radiation worker’s risk of cancer by approximately ___%.
A. 0.05
B. 0.5
C. 5
D. 10
B. 0.5
5 mSv/yr for 20 years = 100 mSv = 0.1 Sv
(5 x 10^-2)/ Sv is the increased risk; thus:
(5 x 10^-2)/ Sv x 0.1 Sv = 0.005 = 0.5%
Q1086. The latent period for radiation-induced solid tumors is about ___ years.
A. 1
B. 5
C. 10
D. 20
E. 50
D. 20
–According to BEIR V - Biological Effects of Ionizing Radiation
Q1088. For risk-benefit calculation purposes, NCRP Report 116 (1993) estimates a probability of developing fatal breast cancer from one 2-view mammographic exam to be about___%. (Assume a mean glandular dose of 4 mGy to each breast.)
A. 0.001
B. 0.1
C. 1
D. 10
A. 0.001
–0.05 per Sv x breast organ weighting factor (0.05) for a total of 0.0025 per Sv.
–For x-rays 1 Sv = 1 Gy
Q1090. The radiation protection quantity which has been used in attempts to estimate the cancer risk from x-ray irradiation of personnel is
A. Exposure (X)
B. Air Kerma (K)
C. Absorbed dose (D)
D. Dose equivalent (H)
E. Effective dose (E)
E. Effective dose (E)
–Attempts to weight the radiation dose to different organs by the relative cancer risk of each organ.
Equivalent dose = Radiation weighing factor
Effective dose = Tissue weighing factor
Q1092. Deterministic or non-stochastic effects of radiation include all of the following except:
A. Bone marrow damage
B. Skin damage
C. Cataract induction
D. Leukemia
E. Infertility due to gonadal irradiation
D. Leukemia
Q1094. The following effects are:
A. Stochastic
B. Deterministic
C. Both
D. Neither
- Induction of cancer from exposure to radiation
- Skin burns from prolonged fluoroscopic exams
- Induction of cancer from exposure to radiation - (A. Stochastic)
- Skin burns from prolonged fluoroscopic exams - (B. Deterministic)
–
Hint:
Deterministic - you can “determine” the extent of your consequences and change the severity of them occuring by actions
Stochastic - Be “stoic” and accept whatever fate throws at you, can can’t change the outcome that is up to chance
Q1096. According the BEIR V, the additional risk of cancer death from a 0.1 Sv (10 rem) exposure to a population of 100,000 people would be about ___%.
A. 0.01
B. 0.1
C. 1.0
D. 10.0
C. 1.0
BEIR V (1990) states that the excess cancer deaths in a population of 100,000 persons exposed to 0.1 Sv would be about 770 males or 810 females.
Remember that based off of the A-bomb studies from World War II, probability of inducing cancer in a large population that is irradiated with 1 rem is about 1 in 10,000 during that population’s lifetime.
1 rem = 0.01 Sv, so 0.1 Sv would lead to 1 in 1,000
1,000/ 100,000 = 1/100 = 1%
Q1098. Perinatal death (at or around the time of birth) is most likely to occur as a result of irradiation in humans which occurs in the gestational period of:
A Implantation of the embryo.
B. Major organogenesis (21-40 days).
C. Second trimester.
D. Just before birth (30-36 weeks).
B. Major organogenesis (21-40 days), AKA 3-8 weeks (deterministic effect, threshold is 0.1 Gy)
- Do NOT confuse this with PRE-natal death which is highest risk with radiation exposure in first 3 weeks of pregnancy; PERI-natal/NEO-natal only takes into effect at earlist after 3 weeks post implantation
–In early organogenesis, the organ buds consist of a few cells, and the lose of some of these can result in a major defect which may not be apparent during gestation, but after birth is too severe to permit independent life.
- In week 8-25, greatest risk is mental retardation
- In week 25+ it is risk of cancer
Q1101. In radiation protection the embryo/fetus is considered more vulnerable to radiation than an adult, for all of the following reasons except:
A. In a given volume, more embryonic cells are proliferating than adult cells.
B. In a given volume, more embryonic cells are differentiating than adult cells.
C. An embryo consists of fewer cells, making the loss of cells by radiation injury potentially more damaging.
D. The higher oxygen tension of the embryo/fetus results in a higher oxygen enhancement ratio (OER).
D. The higher oxygen tension of the embryo/fetus results in a higher oxygen enhancement ratio (OER).
–Just wrong.
Q1103. Regarding radiation effects in the embryo:
A. Prenatal death is equally likely at all stages of development.
B. Neonatal death is most likely for embryos irradiated in the last trimester.
C. Embryonic cells are resistant to radiation because of an efficient repair mechanism.
D. Most congenital abnormalities occur during the period of organogenesis.
D. Most congenital abnormalities occur during the period of organogenesis.
Q1105. The estimated increased risk of birth defect in a fetus receiving 1 rem (10 mSv) in the 12th week of gestation is about _%.
A 0.01
B. 0.1
C. 1.0
D. 10
C. 1.0
–The highest risk from 2 to 16 weeks post-conception is a maximum incidence of 1% per rem for small head size.
Q1107. According to NCRP there is a negligible increase in the risk of adverse effects to the fetus, compared with other risks of pregnancy, up to a total dose of___mGy.
A. 5
B. 20
C. 100
D. 500
E. 1000
C. 100
–Or 10 cGy.
Q1109. For radiation protection purposes, the dose equivalent of 100 mrem of neutrons is ___ mSv.
A 1
B. 2
C. 10
D. 20
E. 100
D. 20
–100 mrem = 1 mSv.
–Dose equivalent (H) = Dose (D) x Quality Factor (Q)
–Q for neutrons is 20.
Q1111. Which of the following is true about film badges?
A. Can measure total dose, but cannot distinguish between high- and low-energy x-rays.
B. Can measure exposures of 2 mR.
C. Are insensitive to heat.
D. Are used to determine exposure by measuring the optical density of the film.
E. None of the above is true.
D. Are used to determine exposure by measuring the optical density of the film.
Placing filters over parts of the film allows one to estimate the proportion of dose due to x-rays in different energy ranges
–Cannot measure below 20 mR.
–Are sensitive to heat and sunlight.
Q1118. The recommended weekly effective dose equivalent permitted for radiologists under current regulations is:
A. 10 μSv
B. 50 μSv
C. 100 μSv
D. 0.5 mSv
E. 1.0 mSv
E. 1.0 mSv
–50 mSv/yr divided by 52 wks is 1.0 mSv/week
Q1120. Regulations limit the dose equivalent to the embryo/fetus of a declared pregnant radiation worker to __ mSv/month.
A. 50
B. 10
C. 5
D. 0.5
E. 0.1
D. 0.5
Q1121. According to NCRP Report No. 116, the recommended maximum annual dose equivalent for radiation workers’ whole body is ___ mSv and for the hands is ___ mSv.
A 5 5
B. 5 50
C. 10 100
D. 50 50
E. 50 500
E. 50 500
A dose rate of 2 mrem/hr is measured in a non-radiation worker’s office, adjacent to a source storage room. How much shielding must be added to reduce this to an acceptable level?
A. No additional shielding is required.
B. 1 HVL
C. 2 HVLs
D. 1 TVL
E. 2 TVLs
E. 2 TVLs
–The acceptable level for continous or frequent exposure of members of the public is 100 mrem/yr = 2 mrem/wk = 0.05 mrem/hr, assuming a hr work week.
–To reduce 2 to 0.05, requires 2 TVL’s
–Tenth-Value-Layers (TVL)
TVL = Shield thickness needed to reduce exposure by 1/10th.
Q1199. The attached graph shows the response curve for a gas filled detector. Given the appropriate region for each of the following descriptions:
- The region where current is passed in the absence of radiation
- The pulse heights are proportional to the incident particle energy
- An ionization chamber is operated in this region
- A Geiger survey meter is operated in this region
- The saturation region where all charges are collected without multiplication
- The region where current is passed in the absence of radiation - E; too high voltage will cause conduction of gases
- The pulse heights are proportional to the incident particle energy - C; proportional counters operate in this region
- An ionization chamber is operated in this region - B
- A Geiger survey meter is operated in this region - D; event will trigger the avalanche
- The saturation region where all charges are collected without multiplication - B
Q1254. Match the parameter with its definition, according to the diagrams below:
A. D5/D4
B. D5/D1
C. D3/D4
D. D3/D2
E. D1/D4
- BSF
- PDD/100
- TAR
- TMR
- BSF - (E. D1/D4)
- PDD/100 - (D. D3/D2)
- TAR - (A. D5/D4)
- TMR - (B. D5/D1)
Q1256. A patient is treated with cobalt-60 radiation in the manner shown below. Point A is 1 cm from the exit surface. The dose at point A is calculated using PDD tables. Relative to the actual value, the calculated dose at point A is:
A. more than 15% high
B. slightly high
C. correct
D. slightly low
E. more than 15% low
B. slightly high
The dose to point A will be about 4% lower than that found from tables, due to the lack of total backscatter. The fact is generally ignored in treatment planning dose computer alogrithms.
Q1258. Match the backscatter factor for a 10x10 cm field with the energy of the radiation:
A. 1.00
B. 1.02
C. 1.035
D. 1.15
E. 1.26
- Superficial, 2.5 mm Al HVL
- Co-60
- 10 MV x-rays
- Superficial, 2.5 mm Al HVL - (E. 1.26)
- Co-60 - (C. 1.035)
- 10 MV x-rays - (B. 1.02)
The BSF is a function of beam quality and field size. It increase to a value of about 1.5 for large fields at about 1 mm Cu HVL, then falls to a negligible value above about 10 MV. It cannot have a value of 1.0 because this would imply that there is no difference between the dose rate in air that that in dmax in tissue.
Q1273. Which of the following is correct?
A. TAR = TMR x BSF
B. TAR = TMR/ BSF
C. TMR = BSF/ TAR
D. TMR = TAR x BSF
E. BSF = TAR x TMR
A. TAR = TMR x BSF
Q1291. Please see attached chart.
The MU setting to deliver 180 cGy at a depth of 5 cm, at 100 cm SSD, for a field of equivalent square 8x8 cm is ____ MU
A. 106
B. 153
C. 198
D. 204
E. 212
E. 212
SSD setup so need to use PDD; which accounts for both inverse square and depth
1 MU = 1 cGy at standardized conditions, so:
180 cGy / [(PDD 8x8 @5cm) x (Output 8x8 @ 100 cm SSD)]
= 180/ (0.862 x 0.985)
= 212
Q1295. Please see attached chart.
The MU setting to deliver 180 cGy at a depth of 5 cm, at 100 cm SSD, for a field of equivalent square 8x8 cm is 212 MU. The dose at 10 cm depth is ___ cGy.
A. 115
B. 139
C. 143
D. 170
E. 185
B. 139
Since you know the dose at d1, you can solve for dmax. The PDD @ 10 cm is given as 66.4%, so that’s 66.4% of dmax. (Dose at d1/ PDD1 = dmax)
Dose at d2 = (PDD2/PDD1) x Dose at d1
= (0.664/0.862) x 180
= 0.77 x 180
= 139
DO NOT do a simple inverse square (105/110)^2 ratio in this setting. PDD accounts for both inverse square AND attenuation so you would be artifically high using inverse square alone.
Q1298. Please see attached chart.
The MU setting per beam to deliver 250 cGy total to the isocenter via parallel opposed, 23x18 cm whole brain fields, for a patient separation of 15 cm is ___ MU.
A. 130
B. 156
C. 182
D. 208
E. 265
A. 130
First off calculate your equivalent square using Sterling’s formula, where the length of one side of your equivalent square = 4 x area/ perimeter
4 x (23x18)/ (2 x (23+18)) = 20; so you have a 20 x 20 equivalent square
Distance to deposit dose is in the middle of the patient, so depth is 15 cm/ 2 or 7.5 cm
Picking SAD: (250 cGy/2)/ (TMR 20x20 @ 7.5 cm) x (OF 20x20 @ 100 SAD)
= 125/(0.88*1.089) = 130
Q1300. Please refer to the attached chart
A 6 MV anterior R supraclavicular field is set up at 100 cm SSD to the center. The prescribed dose is 200 cGy to the supraclavicular point (102 cm SSD), at depth 3 cm. Ignoring off-axis factors the MU setting is ___:
A. 213
B. 209
C. 205
D. 201
E. 194
A. 213
SSD setup, so use PDD and take into account inverse square. Dmax is given to you at the PDD of 100 at 1.6 cm.
Account for the difference in the SSD setup originally for dmax (which is 101.6 cm) and the new distance to SSD on the actual treatment at 102 cm, which is 2 cm further than the original setup (so now your dmax is actually at 101.6 + 2 additional cm).
200/ [(PDD@3) x (inv sq) x (OF@ 100 cm SSD)]
= 200/ [0.956 x ((101.6)/(101.6+2))^2 x 1.019)
= 213
Q1303. Refer to attached chart. The Monitor Unit (MU) setting to deliver 180 cGy at 5 cm depth to a 6 x 28 cm2 field at 100 cm SSD is ___
A. 180
B. 191
C. 199
D. 207
E. 215
D. 207
SSD, so will use PDD chart
6 x 28 cm field equivalent square is 4x(6x28)/(2x(6+28)) = 2x6x28/(6+28) = 9.88
Use 10 x 10 cm field size at 5 cm depth; thus referring to the chart the PDD is 87.1
SSD output at depth 1.6, SSD 100 cm is 1.0 for 10x10
MU = 180/(0.871 x 1.0) = 207
Q1324. Use the attached chart. For parallel opposed 80 cm SAD fields, separation 18 cm, field size at midplane 12x12 cm, 200 cGy total dose per fraction, the timer setting per beam is:
A. 1.07 min
B. 1.11 min
C. 1.15 min
D. 1.39 min
E. 1.41 min
B. 1.11 min
Parallel opposed, 18 cm separation means 100 cGy per beam delivered to 9 cm depth each
SAD setup, use TAR for timer setting
100/(TAR@d9cm for 12x12)x(DR air @80 cm for 10x10)x(OF air 12x12))
= 100/(0.775 x 115 x 1.013)
= 1.11 min
Q1330.
