ROphex Old Q1000+

Question 1

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

Answer 1

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

Question 2

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

Answer 2

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.

Question 3

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

Answer 3

C. 2,500

%σ = σ/N x 100 = N^.5/N x 100 = 100/N^.5 = 2

50 = N^.5

N = 2500

Question 4

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

Answer 4

D. 890, 30

900/1 min - 100/10 min = 890/1 min

590^.5 = 30 = 1 standard deviation

Question 5

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.

Answer 5

C. Means there is a 99% chance that the claim is true.

Question 6

Q1018. Concerning the poisson distribution (answer A for true and B for false):

1. It is another name for the normal distribution
2. It is due to random variations
3. Photon distribution on an x-ray film is a poisson distribution
4. Radioactive decay as a function of time is a poisson distribution
5. The standard deviation increases as the number of measurements increases
6. The percent standard deciation increases as the number of measurements increases

Answer 6

1. It is another name for the normal distribution - B, FALSE; Poisson is binomal distribution whether it either occurs or not. Normal is much larger.
2. It is due to random variations - A, TRUE
3. Photon distribution on an x-ray film is a poisson distribution - A, TRUE
4. Radioactive decay as a function of time is a poisson distribution - A, TRUE
5. The standard deviation increases as the number of measurements increases - A, TRUE; σ=(N)^1/2, so yes
6. The percent standard deciation increases as the number of measurements increases - B, FALSE

Question 7

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

Answer 7

D. RAM - random access memory

ROM is read only memory

Question 8

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.

Answer 8

A. ROM stands for Random Order Memory.

• stands for read only memory

Question 9

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

Answer 9

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 2^k-bit

k-bit of 2 = 4 shades of grey

k-bit of 8 = 256 shades of grey

Question 10

Q1026. The computational speed of a computer is measured in units of

A. MB

B. MIPS

C. RVU

D. BAUD

E. BPI

Answer 10

B. MIPS

–MIPS is Millions of Instructions Per Second.

Question 11

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

Answer 11

B. Freely read, but not written to

Question 12

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.

Answer 12

B. Using multiple processors to increase speed

Question 13

Q1032. There are approximately __ bits in a megabyte.

A. 1024

B. 2048

C. 8000

D. 2,000,000

E. 8,400,000

Answer 13

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

Question 14

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

Answer 14

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

Question 15

Q1036. The mainframe of a digital computer contains:

1. core memory
2. central processing unit (CPU)
3. only random access memory
4. software
5. connectors for peripheral devices

A. 1, 2, 5

B. 1, 2

C. 2, 4

D. 1, 4

E. 1, 2, 3, 4, 5

Answer 15

A. 1, 2, 5

Question 16

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

Answer 16

D. 65536

2^16

Question 17

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

Answer 17

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.

Question 18

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.

Answer 18

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)

Question 19

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

Answer 19

E. internal radiation

–NATURAL IS KEY, as other products are artificial.

–About 40 mrem/yr is contributed by radionuclides within the body, most 40K.

Question 20

1053. 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

Answer 20

D. diagnostic radiological examination

Nuclear med is 2nd after x-ray

Question 21

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

Answer 21

C. 100

–Radon adds another 230 mrem/yr. Man-made radiation, mostly diagnostic x-rays, is about 54 mrem/yr.

Question 22

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.

Answer 22

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)

Question 23

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.

Answer 23

C. Lung dose from alpha emission.

Question 24

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

Answer 24

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

Question 25

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 1. Average dose to a member of the population from radon 2. Maximum recommended dose to a radiation worker 3. Average dose to a member of the population from medical uses of radiation 4. Maximum recommended dose to a member of the public (infrequent exposure)

Answer 25

1. Average dose to a member of the population from radon (C. 2 mSv) 2. Maximum recommended dose to a radiation worker (A. 50 mSv) 3. Average dose to a member of the population from medical uses of radiation (E. 0.5 mSv) 4. Maximum recommended dose to a member of the public (infrequent exposure) (D. 1 mSv)

Question 26

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

Answer 26

B. 3.5

Question 27

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

Answer 27

E. 20

Question 28

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

Answer 28

C. 1 in 10,000

Question 29

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

Answer 29

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

Question 30

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

Answer 30

D. 20-30

Question 31

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.

Answer 31

E. Linear, no threshold.

Question 32

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

Answer 32

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%

Question 33

Q1086. The latent period for radiation-induced solid tumors is about ___ years. A. 1 B. 5 C. 10 D. 20 E. 50

Answer 33

D. 20 –According to BEIR V - Biological Effects of Ionizing Radiation

Question 34

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

Answer 34

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

Question 35

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)

Answer 35

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

Question 36

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

Answer 36

D. Leukemia

Question 37

Q1094. The following effects are: A. Stochastic B. Deterministic C. Both D. Neither 1. Induction of cancer from exposure to radiation 2. Skin burns from prolonged fluoroscopic exams

Answer 37

1. Induction of cancer from exposure to radiation - (A. Stochastic) 2. 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

Question 38

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

Answer 38

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%

Question 39

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).

Answer 39

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

Question 40

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).

Answer 40

D. The higher oxygen tension of the embryo/fetus results in a higher oxygen enhancement ratio (OER). –Just wrong.

Question 41

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.

Answer 41

D. Most congenital abnormalities occur during the period of organogenesis.

Question 42

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

Answer 42

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.

Question 43

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

Answer 43

C. 100 –Or 10 cGy.

Question 44

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

Answer 44

D. 20 –100 mrem = 1 mSv. –Dose equivalent (H) = Dose (D) x Quality Factor (Q) –Q for neutrons is 20.

Question 45

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.

Answer 45

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.

Question 46

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

Answer 46

E. 1.0 mSv –50 mSv/yr divided by 52 wks is 1.0 mSv/week

Question 47

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

Answer 47

D. 0.5

Question 48

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

Answer 48

E. 50 500

Question 49

Q1128. Filters are used in film badges to: (answer A for true and B for false) 1. Convert x-ray or y-ray energy to visible light to expose the film 2. Shield a portion of the film for a base density reading 3. Discriminate between different types of radiation 4. Discriminate between different radiation energies 5. Reduce the exposure to the user

Answer 49

1. Convert x-ray or y-ray energy to visible light to expose the film - (B, FALSE) 2. Shield a portion of the film for a base density reading- (B, FALSE) 3. Discriminate between different types of radiation - (A, TRUE) 4. Discriminate between different radiation energies - (A, TRUE) 5. Reduce the exposure to the user - (B, FALSE)

Question 50

Q1134. The NRC requires personnel to wear a radiation monitor if they are likely to receive \_\_\_% of the annual dose limit. A. 90 B. 50 C. 25 D. 10 E. 1

Answer 50

D. 10

Question 51

Q1136. Regulations in the U.S. limit the total dose equivalent to the fetus of a declared pregnant radiation worker to\_\_\_mSv. A. 50 B. 25 C. 5 D. 2.5 E. 0.5

Answer 51

C. 5

Question 52

Q1138. The recommended annual effective dose limits for members of the public (NCRP 116) are ___ mSv for infrequent exposure, and\_\_\_mSv for continuous exposure. A. 5, 1 B. 1, 5 C. 10, 5 D. 5, 5 E. 10, 20

Answer 52

A. 5, 1

Question 53

•If the workload of a diagnostic x-ray room is increased by a factor of 4, (all other factors, e.g., energy, use factor, etc., remaining unchanged), the shielding at the console should be increased by: A. 1 HVL B. 2 HVLs C. 4 HVLs D. 2 TVLs E. 4 TVLs

Answer 53

B. 2 HVLs –2 HVLs will reduce the dose by a factor of 2^2 = 4.

Question 54

Q1144. When calculating radiation barrier thickness requirements, the use factor "U" refers to: A. The weekly dose delivered at 1 m from the radiation source. B. The fraction of operating time during which the area on the other side of the barrier is occupied. C. The fraction of operating time during which the radiation is directed towards the barrier. D. The fraction of the work week during which a particular individual is in the area of interest.

Answer 54

C. The fraction of operating time during which the radiation is directed towards the barrier. –The use factor can be difficult to calculate, so standard fractionations can be used for the walls and floors.

Question 55

Q1147. A "controlled area" is defined as: A. Any area around a radiation facility where the exposure rate is above background. B. An area that one cannot enter unless one is wearing a film badge. C. An area where workers will not receive more than 5 mrem/week. D. An area where the exposure of workers is under the supervision of the Radiation Safety Officer (RSO).

Answer 55

D. An area where the exposure of workers is under the supervision of the Radiation Safety Officer (RSO). –Workers can receive up to 100 mrem/wk.

Question 56

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

Answer 56

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.

Question 57

If one lead apron attenuates 95% of an x-ray beam, two aprons will transmit approximately \_%. A. 10 B. 5 C. 2.5 D. 1.25 E. 0.25

Answer 57

E. 0.25 –One lead apron transmits 5%. –Two will transmit 0.052=0.0025 or 0.25%

Question 58

Q1156. When designing shielding for an x-ray machine, regulations require that the barrier for a fully-occupied non-controlled area must be about ___ HVLs thicker than that for a controlled area. A. 50 B. 10 C. 6 D. 2 E. 1

Answer 58

C. 6 –The current annual MPDs are 100 mrem for non-controlled areas and 5 rem (5000 mrem) for controlled areas. This factor of 50 requires about 6 HVLs more shielding (26=64).

Question 59

Q1160. The exposure rate at 1 m from a radioactive source is 100 mR/hr. A 0.06 mm lead shield is placed around the source. The exposure rate is now ___ mR/hr. (HVL in lead is 0.02 mm.) A. 75 B. 50 C. 25 D. 12.5 E. 5

Answer 59

D. 12.5 0.06 mm/ 0.02 mm = 3 HVLs So 100 x (1/2)^3 = 100/8 = 12.5

Question 60

Q1162. The exposure rate constant of a radionuclide is 12.9 R-cm^2/ mCi-hr. How many HVLs are required to reduce the exposure rate at 1 meter from a 10 mCi soure to 2 mR/ hr? A. 1 B. 2 C. 3 D. 6

Answer 60

C. 3 12.9 R-cm^2/ mCi-hr x 10 mCi x (1 m/ 100 cm)^2 = 12.9 R x 10 x m^2 / (10^4) hr = 129 x 10^-4 Rm^2/hr = 12.9 mR x m^2/ hr To get this down to 2 mR/hr would be (1/2)^3 or 3 HVLS

Question 61

Q1164. In the event of a 137Cs "dirty bomb" explosion, regarding the use of potassium iodide pills, which of the following is true? A. Kl should be taken as soon as possible. B. Kl should be taken if the thyroid has the potential of receiving a dose greater than 15 rem. C. A dose of 130 mg per day is suggested for adults. D. KI will offer no protection.

Answer 61

D. KI will offer no protection. –KI only good for nuclear bomb, reactor explosion. –Only good for Iodine. –I-131 has short half-life, not good for dirty bomb.

Question 62

Q1166. All of the following are used in x-ray machine room shielding calculations in the United States except: A. Occupancy factor and use factor B. Weekly dose limit to workers or public C. Workload D. Beam energy E. Instantaneous dose rate

Answer 62

E. Instantaneous dose rate

Question 63

Q1168. A shielding design for a diagnostic or therapy installation, when there is no restriction on the beam direction, must (answer A for true and B for false): 1. Consider all wals as primary barriers 2, Assign all walls a use factor (U) of 1 3. Assign any area which people may frequent an occupancy factor (T) of 1 4. Shield all areas to a radiation level of 0.1 rem per week 5. Shield such that unrestricted environments will not receive greater than 2 mR in any one hour

Answer 63

1. Consider all wals as primary barriers - TRUE 2. Assign all walls a use factor (U) of 1 - FALSE - In general walls are assigned use factors (U) of ¼ or 1/16. The walls could have all have a U of 1 at the same time. 3. Assign any area which people may frequent an occupancy factor (T) of 1 - FALSE - For areas such as corridors or toilets recommended occupancy factors (T) are ¼ or 1/16. Occupancy factors are based on the time that an individual is exposed to radiation in the area. A busy, one person office could be assigned a T of 1. 4. Shield all areas to a radiation level of 0.1 rem per week - FALSE - Controlled areas are allowed levels of 0.1 rem per week. Non-controlled areas are required to be shielded to levels of 0.002 rem per week. In practice most areas are shielded well below the maximum permissible level. 5. Shield such that unrestricted environments will not receive greater than 2 mR in any one hour - TRUE - This is an additional restriction on non-controlled areas in order to insure that any combination of use and occupancy cannot raise the total dose to greater than the MPD.

Question 64

Q1170. The occupancy factor (T) is changed from 1/16 to 1/2 and the activity (A) is doubled for a radiation source whose HVL is 0.3 mm Pb. In order to maintain the same level of protection ___ mmPb must be added. A. 0.3 B. 0.6 C. 0.9 D. 1.2 E. 1.5

Answer 64

D. 1.2 1/2^4 = 1/16 so you need 3 further HVLs due to changing occupancy factor In addition, activity is doubled so you need another HVL to counteract that 4 HVLs = 4 x 0.3 mm Pb = 1.2 mm Pb

Question 65

Q1176. A 0.5 mm lead equivalent protective apron is an effective protection device when working with (answer A for true and B for false): 1. A high energy beta source, such as P-32 2. Diagnostic x-rays 3. A Cs-137 implanted patient 4. Milking a Mo-99 - Tc-99m generator 5. High levels of I-125 6. High levels of positron emitters

Answer 65

1. A high energy beta source, such as P-32 (B, FALSE; 1.7 MeV) 2. Diagnostic x-rays (A, TRUE) 3. A Cs-137 implanted patient (B, FALSE; 662 keV requires 6.5 mm Pb) 4. Milking a Mo-99 - Tc-99m generator (B, FALSE; would get the 140 kev of 99mTc, but not the 740 kev of 99Mo) 5. High levels of I-125 (A, TRUE; 30 keV) 6. High levels of positron emitters (B, FALSE; 511 keV) --- Remember, Pb attenuates at 2 MeV/ mm

Question 66

Q1179. An ionization chamber is calibrated at 22°C and 760 mm Hg. If a measurement is made at 18°C and 750 mm Hg, what correction factor must be applied to the chamber reading? A. 0.81 B. 0.98 C. 1.00 D. 1.02 E. 1.19

Answer 66

C. 1.00 –The Temp-Pressure-Factor Correction factor is –[(273 + T)/295] x [760/P] --- So measurement is colder and less pressure compared to standard. Cold temperature artificially increases your readings by concentrating your particles, less pressure artifically decreases your readings due to decreasing concentration of particles for interaction. To correct for this: (760/750) x (18+273)/(22+273)

Question 67

Q1183. Geiger counters: A. Operate on the principle of scintillation. B. Are less sensitive than ionization chambers. C. Can be used in high-intensity radiation fields. D. Can detect individual photons or particles.

Answer 67

D. Can detect individual photons or particles. –The advantage of a geiger counter is its sensitivity. The ionizination of a single event is magnified due to gas multiplication, into a measurable signal. –Its disadvantage is that it can read zero in a high-intensity field.

Question 68

Q1185. In general, which of the following detectors has the greatest energy dependence for x- and gamma rays? A. Thin window geiger tube. B. Air equivalent wall ionization chamber. C. LiF thermoluminescent dosimeter. D. NaI scintillation detector.

Answer 68

D. NaI scintillation detector. –High Z of iodine makes it highly sensitive to low-energy radiation. This is because of the Z dependence of photoelectric interactions.

Question 69

Q1187. Which of the following detectors would be the best one to locate a dropped iodine-125 seed? A. Large volume ionization type survey meter. B. Thermoluminescent dosimeter. C. Air-equivalent wall "thimble' ionization chamber. D. Film badge. E. NaI scintillation probe.

Answer 69

E. NaI scintillation probe. –Need a portable detector with an instantaneous readout. –The scintillation probe will give a much higher reading (as would a geiger counter) than an ionization chamber, due to its greater sensitivity.

Question 70

Q1189. Match the most appropriate instrument to the procedure in each question. A. Liquid scintillation counter B. NaI well counter C. Geiger-Mueller (GM) counter D. Thermoluminescent dosimeter (TLD) E. Ionization chamber survey meter 1. Gamma ray sealed source wipe test 2. Contamination survey for 99mTc. 3. Radiation survey of a diagnostic x-ray installation 4. Personnel monitoring

Answer 70

1. Gamma ray sealed source wipe test - B. NaI well counter - NaI detects the low level gamma rays 2. Contamination survey for 99mTc - C. Geiger-Mueller (GM) counter - GM has a fast response, and get the low level gamma 3. Radiation survey of a diagnostic x-ray installation - E. Ionization chamber survey meter - minimal energy dependence 4. Personnel monitoring - D. Thermoluminescent dosimeter (TLD) 5. Contamination survey for tritium H-3 - A. Liquid scintillation counter - gets the low level beta rays form tritium.

Question 71

Q1195. Personnel monitors must have all of the following features except: A. No energy dependence over the range of radiation encountered B.. Neglible loss of stored signal over 1 month of wear and time taken to process detector C. Must be unaffected by normal fluctuations in the environment (temperature, pressure, humidity) D. Ability to differentiate between different energies and types of radiation

Answer 71

A. No energy dependence over the range of radiation encountered

Question 72

Q1197. Quenching gasses are usually used in: A. ionization chambers B. propotional counters C. Geiger Mueller counters D. Scintillation detectors E. A and C

Answer 72

C. Geiger Mueller counters Quench the avalanche of charge released by the passage of the initial charge particles. This decreases dead time and prevents the total paralyzation of the counter.

Question 73

Q1199. The attached graph shows the response curve for a gas filled detector. Given the appropriate region for each of the following descriptions: 1. The region where current is passed in the absence of radiation 2. The pulse heights are proportional to the incident particle energy 3. An ionization chamber is operated in this region 4. A Geiger survey meter is operated in this region 5. The saturation region where all charges are collected without multiplication

Answer 73

1. The region where current is passed in the absence of radiation - E; too high voltage will cause conduction of gases 2. The pulse heights are proportional to the incident particle energy - C; proportional counters operate in this region 3. An ionization chamber is operated in this region - B 4. A Geiger survey meter is operated in this region - D; event will trigger the avalanche 5. The saturation region where all charges are collected without multiplication - B

Question 74

Q1201. The quanitity that an ionization chamber actually measures is: A. Roentgen B. Gray C. Charge D. Kerma E. Voltage

Answer 74

C. Charge Radiation produces ionization in the chamber gas which is measured as an electric current.

Question 75

Q1203. Electron equilibrium is said to exist in a small mass when: A. the number of x-rays entering the mass is equal to the number of x-rays leaving it B. the build up of dose begins to slow down C. absorption of the radiation becomes exponential D. the number of electrons entering the mass is equal to the number of electrons leaving it

Answer 75

D. the number of electrons entering the mass is equal to the number of electrons leaving it

Question 76

Q1204. An instrument used for absolute measurements of exposure is: A. Baldwin-Farmer ionization chamber B. Thermoluminescent dosimeter C. Free-air chamber D. Survey meter

Answer 76

C. Free-air chamber - the standard

Question 77

Q1205. Calorimetry is a technique used for: A. routine exposure measurements B. routine dose measurements C. absolute exposure measurements D. absolute dose measurements

Answer 77

D. absolute dose measurements

Question 78

Q1206. ion-recombination can be a problem when using a: A, calorimeter B. Geiger counter C. ionization chamber D. TLD

Answer 78

C. ionization chamber

Question 79

Q1215. A patient who had an iodine 125 seed prostate implant 3 years ago is admitted for a transurethral resection of the prostate (TURP). The activity implanted was 25 mCi. The half life of iodine-125 is 60 days. Which of the following is true? A. radiographs should not be attempted as radiation from the seeds would fog the film B. the physician performing the TURP should wear a Pb apron and gloves C. the TURP should not be attempted because of the radiation hazard to OR staff

Answer 79

D. none of the above are true Remember that: 1.44 x 1/2 life = average life Total dose for LDR = average life x initial activity

Question 80

Q1219. After installation of a chest x-ray unit, which agency regulates its operation? A. NRC B. OSHA C. HICFA D. State

Answer 80

D. State –Once installed, x-ray units are regulated by the state. –Mammography units are also regulated by the FDA, under MGQSA standards. –The FDA regulates the manufactrue and installation of x-ray devices. –The NRC regulates the use of “man-made” radioactive materials such as Co-60 units and brachytherapy sources.

Question 81

Q1225. A "controlled area" is defined as: A. any area around a radiation facility where the exposure rate is above background B. an area which one cannot enter unless wearing a film badge C. an area where the workers will not recieve more than 10 mrem/ week D. an area where the exposure of workers is under the supervision of a radiation safety officer

Answer 81

D. an area where the exposure of workers is under the supervision of a radiation safety officer

Question 82

Q1226. A "High Radiation Area" sign must be posted at the entrance of all areas where the exposure rate can be: A. \>100 mR/hr B. \>100 mR/hr but \<1000 mR/hr C. \>10 mR/hr D. \>100 mR/week E. None of the above

Answer 82

A. \>100 mR/hr

Question 83

Q1227. All of the following are true regarding Percentage Depth Dose (PDD) except: A. Increases with increasing energy. B. Depends on field size. C. Is the dose at depth expressed as a % of the dose at dmax. D. Decreases with increasing SSD. E. Decreases as depth increases.

Answer 83

D. Decreases with increasing SSD. –PDD increases with increasing SSD because it has two components, attenuation and inverse square. - The inverse square component decreases as distance increases. * PDD= 100 (Dose at depth along central axis) dose at depth of maximum dose along central axis * Dependent upon: –Beam Quality •As beam quality (energy) increases, so does PDD –Depth in Material •At depth increases, the PDD decreases (except if in buildup region) –Size of Treatment Field •As field size increases, PDD increases (more scatter) –SSD •As SSD increases, PDD increases secondary to inverse square.

Question 84

Q1230. Match the formula with the quantity: A. (dose rate at dmax)/ (dose rate at depth) at SSD B. (dose rate at depth)/ (dose rate at dmax) at SSD C. (dose rate at dmax)/ (dose rate at depth) at SAD D. (dose rate at depth)/ (dose rate at dmax) at SAD E. (dose rate at dmax)/ (dose rate in air) at SAD 1. TMR 2. BSF 3. PDD/100

Answer 84

1. TMR - (D; (dose rate at depth)/ (dose rate at dmax) at SAD) 2. BSF - (E; (dose rate at dmax)/ (dose rate in air) at SAD) 3. PDD/100 - (B; (dose rate at depth)/ (dose rate at dmax) at SSD)

Question 85

Q1232. A single posterior spine field is treated at 130 cm SSD. Compared with treatment at 80 cm SSD the exit dose will be: A. greater B. smaller C. the same

Answer 85

A. greater As SSD increases PDD, and thus exit dose, increase (pet Mayneord’s f factor).

Question 86

Q1234. Choose the statement that is true: A. TAR increases as SSD increases B. BSF increases as beam energy increases above 1MV C. PDD increases with increasing SSD D. TMR cannot be measured for cobalt-60

Answer 86

C. PDD increases with increasing SSD –TAR is independent of SSD or SAD –BSF increases with increasing energy up to about 1.0 mm Cu HVL, then decreases as energy increases. –Although historically TARs were measured for cobalt-60 and TMR were introduced for higher energy beams, TMRs can be measured and used for calculating timer settings at any megavoltage energy.

Question 87

Q1236. For a 10 MV photon beam at 10 cm depth, the TMR will be ___ the PDD at 100 cm for the same collimator setting. A. Greater than B. Less than C. Approx. equal to

Answer 87

A. Greater than –TMRs are a measure of attenuation only, whereas PDDs comprise attenuation and inverse square components.

Question 88

Q1238. Tissue-Maximum Ratio (TMR) depends on: A. Energy, SAD, depth, and field size. B. Energy, SAD, and field size. C. SAD, depth, and field size. D. Energy, depth, and field size. E. SSD only.

Answer 88

D. Energy, depth, and field size. –TMR is independent of SAD; it simply only needs to be a SAD setup –It is the radio of the dose rate at depth (d) to dose rate at depth dmax, both measured at the same distance.

Question 89

Q1242. Back scatter factor: A. is the TAR at dmax B. increases as energy increases over 1 MV C. is the PDD at dmax D. is the ratio of dose in air to dose in tissue E. all of the above

Answer 89

A. is the TAR at dmax TAR is the dose rate at depth divided by the dose rate in air AKA Tissue Air Ratio The TAR at dmax is called the back scatter factor (BSF). BSF _decreases_ as the energy increases above 1 MV.

Question 90

Q1245. TMR: A. Stands for tumor maximum ratio B. Is the ratio of dose at dmax to dose at depth C. increases as SSD increases D. Cannot be measured on a cobalt-60 unit E. None of the above

Answer 90

E. None of the above TMR = tissue maximum ratio. It is the ratio of two dose rates measured in phantom, at the same distance from the source; one with a chosen thickness of overlaying phantom, the other with only the thickness required to attain dmax. It is independent of SSD and can be measured on any megavoltage x-ray or gamma ray unit.

Question 91

Q1246. Which of the following statements is false about TMR? A. It is equal to TAR/BSF B. It is approximately related to the percent depth dose by an inverse square factor C. It is the ratio of the dose at depth divided by the dose at dmax, both measured at the isocenter D. It is dependent on SSD E, It increases with increasing field size

Answer 91

D. It is dependent on SSD TMR is independent of SSD since the dose at depth and the dose at dmax are measured at the same distance from the source.

Question 92

Q1249. TAR is: A. equal to the backscatter factor (BSF) at dmax B. independent of SAD C. used in calculations of timer settings for rotational therapy D. all of the above E. none of the above

Answer 92

D. all of the above The timer or MU setting for rotation uses the average TAR, averaged over all depths for the area of rotation.

Question 93

Q1253. All of the following are independent of SSD except: A. TMR B. TAR C. PDD D. BSF

Answer 93

C. PDD PDD increases with increasing SSD, since it contains an inverse square component as well as attenuation. TMR, TAR, and BSF (TAR at dmax) measure attenuation only and are independent of SSD.

Question 94

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 1. BSF 2. PDD/100 3. TAR 4. TMR

Answer 94

1. BSF - (E. D1/D4) 2. PDD/100 - (D. D3/D2) 3. TAR - (A. D5/D4) 4. TMR - (B. D5/D1)

Question 95

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

Answer 95

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.

Question 96

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 1. Superficial, 2.5 mm Al HVL 2. Co-60 3. 10 MV x-rays

Answer 96

1. Superficial, 2.5 mm Al HVL - (E. 1.26) 2. Co-60 - (C. 1.035) 3. 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.

Question 97

Q1263. For megavoltage photons, TMR has replaced TAR because: A. TMR is independent of field size B. TAR is difficult to measure at high energies C. TMR is preferable to TAR for rotation calculations D. TMR is independent of depth of maximum dose

Answer 97

B. TAR is difficult to measure at high energies As the beam energy increases, the depth of dmax increases and the size of the chamber build-up cap for the in-air measurement will also increase. A large size build-up cap will start to act as a “mini” phantom.

Question 98

Q1265. Given a square and rectangle of the same area, which would you expect to have the greater percent depth dose for Co-60? A. the square B. the rectangle C. they have the same depth dose

Answer 98

A. the square Scatter from the corners of the rectangle has farther to travel than from the periphery of the square and will contribute less to the depth dose.

Question 99

Q1267. The TAR for a 10x10 cm SAD for 4 MV photons is 0.8 at a depth of 7 cm. What change would you expect in the TAR by extending the SSD from 93 cm to 193 cm (200 cm from the source), for the same field size at SAD? A. 5% incease in TAR B. 10% increase C. 15% increase D. no change in TAR

Answer 99

D. no change in TAR TAR is independent of SSD

Question 100

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

Answer 100

A. TAR = TMR x BSF

Question 101

Q1275. Percentage depth dose in photon beams: 1. Increases with increasing SSD 2. Increases with increasing field size 3. Increases with increasing beam energy 4. Decreases exponentially (not including inverse square effect and scattering beyond dmax A. 1 only B. 1, 2, 3 C. 2, 4 D. 4 only E. all are correct

Answer 101

E. all are correct

Question 102

Q1277. Match the following quantities with the units in which they are measured. A. Sievert B. Coulombs/kg C. Gray D. Bequerel E. cGy/hr 1. Activity 2. Absorbed dose 3. Exposure 4. Dose equivalent

Answer 102

1. Activity - (D. Bequerel) 2. Absorbed dose - (C. Gray) 3. Exposure - (B. Coulombs/kg) 4. Dose equivalent - (A. Sievert)

Question 103

Q1280. The depth of maximum dose for a photon beam is approximately equal to: A. The depth at which dose and kerma are equal. B. The maximum range of the secondary electrons. C. The depth at which electronic equilibrium occurs. D. All of the above. E. None of the above.

Answer 103

D. All of the above.

Question 104

Q1281. The TMR for a 10 x 10 cm² 4 MV photon field at 100 cm SAD is 0.73 at 10 cm depth. If the SSD were changed from 90 cm to 95 cm, the expected change in the TMR would be: A. Increase of 5%. B. Increase of 7.5%. C. Decrease of 5%. D. Decrease of 7.5%. E. Zero.

Answer 104

E. Zero. –TMR is independent of SSD, since it represents attenuation of a given thickness of tissue. –PDD has an inverse component and does depend on SSD.

Question 105

Q1283. Sievert (Sv) is A. 100 ergs per gram B. The SI unit of dose equivalent C. Absorbed dose/RBE D. The SI unit of exposure

Answer 105

B. The SI unit of dose equivalent –Formerly rem –Takes into account RBE of different types of radiation. i.e. the neutron effect

Question 106

Q1285. All of the following are true regarding Percent Depth Dose (PDD) except: A. The value can be adjusted for extended SSD using Mayneord's F factor. B. It comprises an attenuation component and an inverse square component. C. At a given depth it increases as field size increases. D. For a given field size and depth it increases as energy increases. E. It is defined as: (dose rate at dmax/dose rate at depth) x 100%.

Answer 106

E. It is defined as: (dose rate at dmax/dose rate at depth) x 100%. –Equation is inverted

Question 107

Q1288. Collecting all the negative ions produced by a beam of photons in a small volume of air, under conditions of electronic equilibrium, is a direct measure of: A. Dose equivalent B. LET C. Absorbed dose D. Exposure E. Specific ionization

Answer 107

D. Exposure

Question 108

Q1289. The greatest backscatter factor in soft tissue is associated with: A. 30 kVp x-rays B. 4 MV linear accelerator x-rays C. cobalt-60 teletherapy y-rays D. 2 mm Al HVL x-rays E. 1 mm Cu HVL x-rays

Answer 108

E. 1mm Cu HVL x-rays –Scatter is associated with compton interaction. At 2 mm Al HVL there is a considerable amount of photoelectric interactions. Above about 1 mm Cu HVl, photoelectic interaction are neglible and compton predominates. –As the photon energy increases, more energy is transferred to the electron, and less to the scattered photon. –The maximum backscatter is at about 0.7 mm Cu HVL.

Question 109

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

Answer 109

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

Question 110

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

Answer 110

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.

Question 111

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

Answer 111

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

Question 112

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

Answer 112

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

Question 113

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

Answer 113

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

Question 114

Q1306. A patient is to be irradiated to a depth of 7 cm using a field size of 10 x 10 cm at 80 cm SSD. The PDD at this point is 65% and the dose rate at dmax is 100 cGy/min. The time to deliver 150 cGy is ___ min. A. 0.65 B. 1.25 C 1.5 D 2.3 E. 2.7

Answer 114

D 2.3 MU rate = Dose at depth/(Output at dmax x PDD at depth) MU rate = 150 cGy / (100 cGy/min x 0.65) MU rate = 2.31 min

Question 115

Q1307. A patient is simulated with a 10x10 cm field at 80 cm SAD to deliver 200 cGy at 8 cm depth on a Co-60 unit. If technique is now changed to 80 cm SSD, what values are required to calculate the timer setting? 1. PDD for a 10x10 cm field, 8 cm depth, 80 cm SSD 2. Dose rate in air at 80 cm for a 10x10 cm field 3. Dose rate in air at 80.5 cm for a 10x10 cm field 4. Dose rate at 80.5 cm in air for a 9x9 cm field 5. TAR for a 10x10 cm field at 8 cm depth 6. Backscatter factor for a 10x10 cm field 7. Backscatter factor for a 9x9 cm field 8. PDD for a 9x9 cm field, 8 cm depth, 80 cm SSD A. 4,7,8 B. 3, 6, 1 C. 2, 6, 5 D. 4, 5 E. 2, 8

Answer 115

D. 4, 5 A field size of 10 cm at the tumor (8 cm depth) requires a field size of 10 x (80)/(88) = 9 cm on the surface at 80 cm SSD. At treatment SSD, Dose Rate = DR air at 80.5 cm x SBF for field size at the skin. PDD =PDD for field size at the surface and depth to prescription point. Time = D/(DR x PDD)

Question 116

Q1310. A single direct field is set up at 80 cm SSD. The prescribed dose is 3000 cGy in 10 fractions at 5 cm depth. The dose rate at dmax is 100 cGy/min. The PDD at D5cm = 78%. The timer setting and dose at dmax are: A. 3.00 min and 385 cGy B. 3.00 min and 300 cGy C. 3.85 min and 300 cGy D. 3.85 min and 385 cGy

Answer 116

D. 3.85 min and 385 cGy Timer setting = dose at depth/(dose rate at Dmax x PDD) =300/(100 x 0.78) = 3.85 min Given dose at 5 cm depth = 78% of dose at dmax Dose at depth/PDD = 300/0.78 = 385 cGy

Question 117

Q1311. Which of the following would be a correct formula to calculate the timer setting for the following? A single posterior spine field is treated on 80 SSD on a Co-60 unit, dose per fraction is 300 cGy at 3 cm dpeth; field size on skin 6x12 cm (assume equivalent square = 8 cm). A. t=30/((Dose rate in air at 30 cm) x TAR (8x8, d3cm)) B. t=300/(Dose rate at dmax, 80 cm SSD) C. t=300/((Dose rate in air at 80 cm) x PDD (8x8, d3cm)) D. t=300/((Dose rate at dmax at 80 cm SSD) x PDD (8x8, d3cm)) E. t=300/((Dose rate at dmax at 80 cm SAD) x BSF (8x8))

Answer 117

D. t=300/((Dose rate at dmax at 80 cm SSD) x PDD (8x8, d3cm))

Question 118

Q1316. The dose rate in air for a 10x10 cm Co-60 field is 100 cGy/min at 80 SAD. The BSF = 1.035. The dose rate in tissue at depth=dmax, 80 cm SSD is: A. 103.5 cGy/min B. 102.2 cGy/min C. 100 cGy/min D. 115 cGy/min E. 125 cGy/min

Answer 118

B. 102.2 cGy/min 100cGy/min x 1.035 x (80/80.5)² = 102.2 cGy/min

Question 119

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

Answer 119

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

Question 120

Q1326. If 4000 cGy is delivered to a patient's mediastinum via AP/PA fields, the lowest total dose at dmax is obtained using: A. Co-60, 80 cm SSD. B. 6 MV photons, isocentric fields. C. 6 MV photons, 100 cm SSD. D. 18 MV photons, 100 cm SSD. E. 18 MV photons, isocentric fields.

Answer 120

D. 18 MV photons, 100 cm SSD. –Dose at dmax, expressed as a percent of dose at midplane for parallel opposed fields _decreases_ with _increasing energy_ and _increasing SSD_.

Question 121

Q1328. The equivalent square of a 6x30 cm photon beam is a square CxC cm, where C is: A. Closer to 30 than 6. B. Equal to the square root of (6x30). C. The side of the square field which has the same PDD as the rectangular field. D. Is approximately equal to the (area / perimeter) for the rectangle.

Answer 121

C. The side of the square field which has the same PDD as the rectangular field.

Question 122

Q1330.