RT IRRSP STATE CARD 289 CARDS HAROLD BENNETT Flashcards
Industrial Radiography Radiation Safety Personnel (IRRSP)
Radiation Absorbed Dose (RAD) is measured in:
Tissue
The acronym RAD Means:
Radiation Absorbed Dose
The symbol R means:
Roentgen
The symbol mR means:
milliRoentgen (lower case m stands for milli or thousandth)
One Roentgen or 1R is equal to:
1000 milliroentgens
The term REM stands for?
Roentgen Equivalent Man
What two terms have the same relationship:
Quality Factor (QF) & Relative Biological Effectiveness (RBE)
The Quality Factor for x-rays and gamma rays is essentially:
1
The Quality Factor for Alpha Particles is essentially:
20
Specific activity of Radioactive Material is measured in:
Curies per Gram
Decay of a radioactive material is influenced primarily by its:
Half Life (the longer the half-life the longer it will take to decay to one half)
Becquerels and Curies are units of measurement of:
Activity and Decay Rate
Becquerel and Curies are units of measurement of:
Activity and Decay Rate
A state in which atoms have excess energy and are unstable is known as:
Radioactive
An elementary particle with a unit negative electrical charge and a mass approximately equal to 1/1840th that of a proton:
Electron
A positively charged elementary particle with a mass approximately equal to 1840 times that of an electron or 1 A.M.U.is known as:
Proton
Uncharged elementary particles with a mass nearly equal to that of a proton are known as:
Neutron
The area known as the center of an atom is called the:
Nucleus
Any source that disrupts the electrical balance of an atom and results in the production of ions is known as:
Ionization
Any byproduct material that is encased in a capsule designed to prevent leakage or escape of the byproduct material is known as a:
Sealed Source
A tube through which the radioactive source travels when inside a radiographic exposure device is called:
S-Tube
Survey meters must be calibrated at intervals not to exceed:
6 months and / or after instrument servicing except for battery changes
A survey meter must be capable of measuring a range of:
2 mR to 1000 mR/Hr
An analog survey meter must be calibrated on each scale at two points approximately:
At 33% and 66% of the scales potential
The process that results in the removal of orbital electrons from atoms resulting in the formation of ion pairs is called:
Ionization
The fact that gasses bombarded by ionizing radiation become conductors of electrical current make them useful in:
DETECTION EQUIPMENT
Radiation detection instruments measure exposure to radiation based on the principal that:
Ionizing gasses conduct electrical current in proportion to the amount of radiation
Radiation detection instruments measure exposure to radiation energies based on the principal called:
Ionization
A dosimeter must be capable of reading a range of:
0-200 mR
Pocket Ion chambers must at a minimum be capable of reading:
200 mR
An exposure rate is measured by a:
Survey meter
A survey meter reads radiation levels as?
An exposure rate
What dosimeters are least affected by moisture ?
TLD
Full deflection of a survey meter while on battery check means :
The Meter may or may not be working properly
Dosimeters must be calibrated at intervals not to exceed:
Annually
Rate alarms must be calibrated at intervals not to exceed:
Yearly
A TLD measures what?
Dose
A Film badge / TLD records what ?
Total dose history for the Month
When performing radiography in a permanent radiographic installation the technician is not required to:
Wear a rate alarm
A direct reading ionization chamber (i.e. dosimeter) has the advantage of providing:
An immediate dose history since last charged
A direct reading dosimeter (i.e. a pocket ionization chamber) has a disadvantage of:
Cannot provide total dose history that a Film Badge can
A radiographer reads 7.5 on the 10x scale of his survey meter what is the current dose rate at his position?
75 mR/Hr
A radiographer’s survey meter is showing 0 near an exposed source he can safely assume:
The meter is an in-operable state the radiographer should retreat from the source until and operational meter has replaced the suspect meter.
The half-life of an Ir 192 source is:
74.3 days
The half-life of Co 60 is:
5.3 years
The half-life of Selenium 75 is:
120 days
After 6 half-lives what is the percentage of a sources original value?
1.5625% (1/64) of the original
ON A BASIC CALCULATOR:
1. Enter the original activity value.
2. Press the division key
3. Enter “2” and press the equals key
4. Repeat steps 2 and 3 a total of 6 times.
The final result will be approximately 1.5625% of the original activity.
After 6 half value layers what percentage of radiation are you receiving :
1.60%
What is the percentage of radioactive material remaining after 6 half-lives?
1.60%
The use of 4 half value layers will reduce the exposure by a factor of :
16 times
Here’s a step-by-step explanation using a basic calculator:
1. Start with the number 1 (this represents the full exposure).
Divide by 2 for each HVL.
2. Using a Basic Calculator:
Step 1: Enter 1 (full exposure).
Step 2: Press ÷ (divide).
Step 3: Enter 2 (for the first HVL).
Step 4: Press = (equals). You should see 0.5 (half the exposure).
Step 5: Press ÷ again.
Step 6: Enter 2 (for the second HVL).
Step 7: Press =. You should see 0.25 (a quarter of the exposure).
Step 8: Press ÷ again.
Step 9: Enter 2 (for the third HVL).
Step 10: Press =. You should see 0.125 (an eighth of the exposure).
Step 11: Press ÷ again.
Step 12: Enter 2 (for the fourth HVL).
Step 13: Press =. You should see 0.0625 (one-sixteenth of the exposure).
3. Interpret the Result:
The final result, 0.0625, represents the fraction of the original exposure remaining after 4 HVLs.
TO FIND THE REDUCTION FACTOR, TAKE THE RECIPROCAL OF 0.0625.
PRESS 1 ÷ 0.0625 = (YOU SHOULD SEE 16)
Assuming a source has a half-life of 20 years how old would the source be in 4 half-lives?
80 years
If you have 80 mR/Hr at the surface of the exposure device what would the reading be after 2 half-lives :
20 mR
1. Half-Life =(80 Ci’s) ÷ (2 ) = 40
2. Half-Lives= (40 Ci’s) ÷ (2 ) = 20
How many curies of Ir 192 would you have after 148 days:
25%
You have 98 curies of Ir 192 after 148 days how many curies would you have ?
24.5 curies
Iridium192 has a half-life of about 74 days
148 days have passed
Divide 148 by 74 to find out how many half-lives have passed.
148 Divided by 74 = 2 So, 2 half-lives have passed.
Started with 98 curies Divided by 2 half Lives = 49 curies.
After the second half life (another 74 days), the amount is halved again:
49 Divided by 2 = 24.5 curies.
So, after 148 days, you would have 24.5 curies of Ir192 left.
A source of Ir192 has undergone 3 half-lives by what factor has that source been reduced ?
8
To calculate the reduction factor after 3 half-lives using a basic calculator, you can follow these steps:
1. Enter the initial quantity.
2. Divide by 2 (this represents one half-life).
3. Press the equals (=) button to get the result.
4. Repeat steps 2 and 3 two more times (for a total of three divisions by 2).
Here’s a step-by-step example:
1. Enter the #
2. Divide by 2 (# ÷ 2 = 0.5).
3. Divide by 2 again (0.5 ÷ 2 = 0.25).
4. Divide by 2 one more time (0.25 ÷ 2 = 0.125).
So, after 3 half-lives, the remaining quantity is 0.125, which means the source has been reduced by a factor of 8
After 2 half-lives which of the following characteristics remains unchanged :
Source size
A Sealed source emits what ?
Gamma Rays
A Cobalt source has decayed from its original activity 3 half-lives originally it was 88 curies it’s current activity is:
11 curies
1. Half-Life =(88Ci’s ) ÷ (2 ) = 44Ci’s
2. Half-Lives= (44Ci’s ÷ (2 ) = 22Ci’s
3. Half-Lives= (22Ci’s ) ÷ (2 ) = 11Ci’s
Your Cobalt source of 73 curies is exposed for a full 60 minutes. Assume a 14.0 R per curie factor shooting through a 3-half value collimator where would your restricted area be on the cold side of the collimator ?
253 feet
STEP BY STEP WITH A BASIC CALCULATOR:
1st. Calculate the initial dose rate at one foot:
a. Source strength: 73 curies
b. Dose rate factor: 14.0 R per curie at one foot
c. Initial dose rate = 73 Ci X 14.0 R = 1022 R/hr
2nd. ON YOUR CALCULATOR:
a. Enter 73
b. Multiply by 14
c. Result: 1022
3rd. Account for the collimator attenuation:
a. Collimator attenuation: 3 half-value layers (HVL)
b. Each HVL reduces the dose rate by half.
c. Dose rate after collimation = 1022 R/hr 2^3 = 1022 R/hr 8 = 127.75 R/hr
4th. ON YOUR CALCULATOR:
a. Enter 1022
b. Divide by 8(since 2 to the Power of 3 )(2^3 = 8)or (2 x 2 x 2 = 8)
c. Result: 127.75
5th. Calculate the dose received in 60 minutes:
a. Exposure time: 60 minutes (1 hour)
b. Dose received = 127.75 R/hr TIMES 1 hr = 127.75 R
6th. ON YOUR CALCULATOR:
a. Enter 127.75
b. Multiply by 1 (since it’s 1 hour)
c. Result: 127.75
7th. Determine the restricted area using the inverse square law:
a. Let’s assume the safe dose rate is 2 mR/hr (0.002 R/hr).
b. Using the inverse square law: ( I_1 times d_1^2 = I_2 times d_2^2 )
c. ( 127.75 times 1^2 = 0.002 times d_2^2 )
d. ( d_2 = sqrt {127.75}{0.002}
e. ( d_2 = sqrt {63875}
f. ( d_2 approx . 252.74) feet
8th. ON YOUR CALCULATOR:
a. Enter 127.75
b. Divide by 0.002
c. Result: 63875
d. Find the square root of 63875 (use the square root function, often labeled as √)
e. Result: 252.74
So, the restricted area on the cold side of the collimator would be approximately 252.74 feet from the source.
The ball on a source assembly that prevents the source from exiting through the back of the camera is called:
Stop ball
Emission of gamma rays, alpha rays and beta rays is considered what :
Characteristics of source decay
To penetrate a thicker or more dense material you would need:
A source with more energy (shorter wavelength)
Will a shorter or longer wavelength penetrate thicker or denser material?
A Shorter
An Isotope may be:
1. A Stable atom
2. An Unstable atom
3. And Radioactive
An elements weight is the:
Atomic Number
An element is identified by the number of _________in its nucleus
Protons
The first indication of a localized personal radiation over exposure is?
Reddening of the skin
Gamma radiation has a shorter wavelength than visible light therefore making it?
More penetrating
The restricted area is located at?
2 mR/HR
The radiation area is located at?
5 mR/HR
A radiation area is defined as an area accessible to individuals in which they could receive?
5 mR per hour
A high radiation area is defined as an area accessible to individuals in which they could receive?
100 mR per hour
A very high radiation area is defined as an area accessible to individuals in which they could receive?
500 R per hour
A member of the public or un-monitored individual is allowed to receive up to but not to exceed?
2 mR in any one hour or 100 mR/ Year
When performing radiography in the field regulations require the radiographer to post
1. THE RADIATION AREA and
2. THE HIGH RADIATION AREA
When performing radiography 10 CFR part 20 requires that?
1. A BOUNDARY BE POSTED FOR THE RADIATION AREA and
2. A BOUNDARY BE POSED FOR THE HIGH RADIATION AREA
A restricted area is defined as?
An area which access is restricted for controlling radiation exposure.
While performing radiographic operations the radiographer is required to post boundaries to prevent individuals from receiving an exposure which could lead them to receive a dose of more than 2 mR in any one hour or 100 mR in one year. The boundary would be located at?
2 mR/hour
A high radiation area must be posted at?
100 mR/Hr
A very high radiation area is?
500 R/Hr
A restricted area is defined as?
An area to which is limited by the licensee for protecting individuals against undue risks from exposure to radiation.
An un-restricted area is defined as?
An area access to which is neither limited nor controlled by the licensee
A Radiographer and assistant are standing in a 2 mR/Hr field what would their total dose be after 4 hours (Remember Dose is in Rems (Body)and Dose Rate is in Roentgens (measured in air) :
8.0 mRem
2mR an hour times 4 hours is 8 mR
You have 24 exposures to make your shot time is 5 minutes per exposure and youre showing 30 mR/ Hr what will be your total dose at the end of the shift?
60 mRem
A Radiographer is receiving 100 mR/Hr at the crank assembly; the crank assembly is 25’ in length. What distance would the Radiation Area be Designated?
111.8 feet
STEP-BY-STEP ON A BASIC CALCULATOR:
1. Square the initial distance (25 feet):
Enter 25 Press the multiplication key (×)
Enter 25 again Press the equals key (=) to get 625
2. Multiply the initial intensity (100 mR/hr) by the squared distance:
Enter 100 Press the multiplication key (×)
Enter 625 (the result from step 1) Press the equals key (=) to get 62500
3. Divide the result by the new intensity (5 mR/hr):
Enter 62500 Press the division key (÷)
Enter 5 Press the equals key (=) to get 12500
4. Take the square root of the result:
Enter 12500 Press the square root key (√) to get approximately 111.8
So, the Radiation Area should be designated at approximately 111.8 feet from the crank assembly
The radiographer is cranking and picking up 100 mR/Hr using 25 ft cranks .How long will it take for them to pick up a total dose of 50mR ?
30 minutes
USING A BASIC CALCULATOR:
1. Enter the total dose you want to calculate, which is 50.
2. Press the division (÷) button.
3. Enter the dose rate, which is 100.
4. Press the equals (=) button.
The display should show 0.5, which means it will take 0.5 hours (or 30 minutes) to pick up a total dose of 50 mR.
An assistant can receive up to a maximum of?
5 REM per year
A Radiographer or assistant my receive no more than?
5 R per year
A declared pregnant Radiographer or assistant may receive no more than_ during their entire pregnancy?
500 mRem
The un-born fetus may receive no more than during their entire pregnancy. :
500 mRem
You have Cobalt with 61 curies exposed for one 30-minute exposure. Using a 14-R per curie factor shooting through a 3-half value collimator standing at 100 feet what will your total exposure be ?
5.3 mRem
Here’s how you can calculate the total exposure using a basic calculator:
1st. Calculate the initial exposure rate:
a. Multiply the number of curies by the R per curie factor.
b. Enter 61 (curies) × 14 (R/curie) = 854 (R/hour).
2nd. Adjust for the collimator:
a. Divide the result by 8 (since a 3-half value collimator reduces the exposure by a factor of 8).
b. Enter 854 ÷ 8 = 106.75 (R/hour).
3rd. Adjust for distance:
a. Divide the result by 10,000 (the square of the distance, and the distance is 100 feet).
b. Enter 106.75 ÷ 10,000 = 0.010675 (R/hour).
4th. Calculate the total exposure for 30 minutes:
a. Total exposure = Exposure rate at 100 feet × Time
b. Multiply the result by 0.5 (since 30 minutes is half an hour).
c. Enter 0.010675 × 0.5 = 0.0053375
So, your total exposure would be approximately 0.0053 Roentgen [R] = 5.3 Milliroentgen [mR]
Remember 1 Roentgen = 1,000 Milliroentgen so just multiply 0.0053 by 1,000 = 5.3
You have 56 curies of Cobalt shooting for 60 minutes. Using a 14.0 R per curie factor shooting through a 3-half value collimator what would your dose rate be at 225 feet?
2 mR/Hr
1ST CALCULATE THE INITIAL DOSE RATE :
Dose rate (R/hr) = Activity (curies) × Dose rate factor (R/hr per curie)
Enter 56 (curies) on your calculator.
Multiply by 14 (R/hr per curie).
The result is 784 R/hr.
2ND ACCOUNT FOR THE COLLIMATOR ATTENUATION:
A 3-half value collimator reduces the dose rate by a factor of (2^3) (2 to the power of 3) or (2x2x2=8) (since each half-value layer reduces the dose rate by half).
Enter 784 (initial dose rate) on your calculator.
Divide by 8 (attenuation factor, which is (2^3).
The result is 98 R/hr.
3RD CALCULATE THE DOSE RATE AT 225 FEET: (Inverse Square Law)
Enter 98 (dose rate after attenuation) on your calculator.
Divide by 225 (distance in feet).
Press = button
Divide the result by 225 again.
The result is approximately 0.00194 R/hr or 1.94 mR/hr.
So, the dose rate at 225 feet would be approximately 1.94 mR/hr
The definition of ALARA is:
As low as reasonably achievable
You have IR-192 with 88 curies using an R factor of 5.2 R where would you place your HIGH Radiation boundary?
68 feet
1st. Convert the Desired Dose Rate: Convert 100 mR/h to R/h: 100 mR/h = 0.1 R/h
Use the Inverse Square Law: The formula to calculate the distance is:
D = A times R Sqrt {D} WHERE:
(A) is the SOURCE ACTIVITY : 88 (CI)
(R) is the R FACTOR: 5.2 R
(D) is the DESIRED BOUNDARY DOSE RATE : 100 (MR/H)
2nd. Calculate the Distance: numerator: 88 times 5.2 = 457.6
THEN, DIVIDE BY THE DESIRED DOSE RATE:
3rd. 457.6 divided by 0.1 = 4576
FINALLY
4th. D = the square root of {4576} = 67.64 feet
You have IR-192 with 88 curies your meter shows 76 mR/Hr at 100 feet away what would the exposure rate be standing 200 feet away?
19 mR/h
USING A BASIC CALCULATOR
1. SQUARE THE INITIAL DISTANCE (100 FEET)
Enter 100 ( × TIMES) 100 = You should get 10,000
2. SQUARE THE NEW DISTANCE (200 FEET)
Enter 200 ( × TIMES) 200 = You should get 40,000
3. MULTIPLY THE INITIAL INTENSITY (76 MR/HR) BY THE SQUARED INITIAL DISTANCE (10,000)
Enter 76 ( × TIMES) 10,000 = You should get 760,000
4. DIVIDE THE RESULT BY THE SQUARED NEW DISTANCE (40,000)
Enter 760,000 (÷ DIVIDED BY ) 40,000 = You should get 19
so, the exposure rate at 200 feet away would be 19 mr/hr
You have Cobalt with 54 curies your meter reads 140 mR/Hr at 80 feet away what would your exposure rate be at 40 feet?
560 mR/h
USING A BASIC CALCULATOR
1. Square the initial distance (80 feet)
Enter 80 × 80 = You should get 6400
2. Square the new distance (40 feet)
Enter 40 × 40 = You should get 1600
3. Multiply the initial intensity (140 mR/hr) by the squared initial distance (6400)
Enter 140 × 6400 = You should get 896,000
4. Divide the result by the squared new distance (1600)
Enter 896,000 ÷ 1,600 = You should get 560
So, the exposure rate at 40 feet is 560 mR/hr
A monitored person may receive up to 5000 mR per year. What would be considered an excessive amount of radiation exposure to that individual :
Any unnecessary exposure to radiation is excessive
What material would be considered the best shielding :
Lead
What material would be considered the best shielding :
Material with a high atomic density
Which is the best protection against radiation over exposure :
1. Time
2. Distance
3. Shielding
The primary form of shielding provided by modern exposure devices is:
Depleted uranium (DU)
The Pig is composed of:
Depleted Uranium
Pb is the chemical symbol for Lead 0.19 of lead will reduce the exposure of Ir 192 to one half its original intensity this is known as the:
HALF VALUE LAYER of lead for the energies associated with Ir 192
Assuming that 0.19 of lead is one half value layer how many half value layers would you have with a sheet of lead .57 in thickness :
3 HVL
steps on a basic calculator:
1. Enter the thickness of the lead sheet: 0.57 ÷
2. Enter the thickness of one HVL: Press 0.19 =
The display should show 3, which means there are 3 half-value layers in the 0.57-inch-thick lead sheet.
0.57 ÷ 0.19 = 3
Assuming 0.19 of lead is one half value layer and you have a total of 3 HVLs of lead between you and 100 mR/h what would your exposure rate be :
12.5 mR per hour
USING A BASIC CALCULATOR:
1. Start with the initial exposure rate 100 mR/h)
2. Divide by 2 for each HVL:
100 Divided by 2 = 50 Divided by 2 again = 25 Divided by 2 again = 12.5.
So, after passing through 3 HVLs of lead, the exposure rate is 12.5 mR/h.
Assuming 0.61” of steel equals one half value layer for Ir 192. How many HVLs would you have with 2” of steel :
3.28 HVLs
USING A BASIC CALCULATOR:
Enter the thickness of the material: (2) divided by the thickness of one HVL (0.61)=
Your calculator should display approximately 3.278, which is the number of HVLs in 2 inches of steel for Ir-192.
(2 ÷ 0.61= 3.278)
Assuming you have 5 HVL’s of steel you are using 100 curies of Ir 192 where would you place your radiation area boundary Assume a R factor of 5.2R :
90.13 feet
Using a Basic Calculator
1. CALCULATE THE EXPOSURE RATE AFTER 5 HVLS:
Enter 520 (initial exposure rate) Divide by 2 =
Repeat the division by 2 four more times to get 16.25
1st.HVL (520 ÷2)=260
2nd.HVL (260 ÷2)=130
3rd.HVL (130÷2)=65
4th.HVL (65÷2)-=32
5th.HVL (32.5)÷2=16.25
2. CALCULATE THE DISTANCE:
Enter 16.25 (exposure rate after 5 HVLs)
Divide by 0.002 (desired exposure rate)
Press = to get 8125
Press the square root button (√) to get approximately 90.13 feet
16.25 ÷ 0.002 = 8,125 √ = 90.13 feet**
Assuming 0.19 of lead is one half value layer a piece of lead 0.38 thick would reduce the exposure rate by:
75%
Using a Basic Calculator:
1. Enter the initial exposure rate (let’s assume it’s 100 for simplicity).
2. Divide by 2 (for the first HVL): 100 ÷ 2 = 50.
3. Divide by 2 again (for the second HVL): 50 ÷ 2 = 25.
Therefore, 25% minus 100% is 75% of the original rate.
.38 ÷.19= 2 HVL’s
(So ÷100% by 2 HVL’s)
100% ÷2÷2=25% it was reduced by 75%
100%-25%=75%**
Assuming 0.19”of lead is one half value layer how much lead would you need to reduce an exposure rate of 100 mR per hour to 25 mR per hour :
0.38”
1. Determine the number of HVLs needed:
You want to reduce the exposure rate from 100 mR/hr to 25 mR/hr.
Each HVL reduces the exposure rate by half.
So, the first HVL reduces 100 mR/hr to 50 mR/hr.
The second HVL reduces 50 mR/hr to 25 mR/hr.
Therefore, you need 2 HVLs to reduce the exposure rate to 25 mR/hr.
2. Calculate the total thickness of lead required:
Given that 1 HVL is 0.19 inches of lead.
Multiply the number of HVLs by the thickness of one HVL.
HERE’S HOW YOU CAN DO IT ON A BASIC CALCULATOR:
2 × 0.19” = 0.38”
1. Enter the number of HVLs:
Press 2.
2. Multiply by the thickness of one HVL:
Press the × button. Enter 0.19.
3. Get the result: Press the = button.
The display should show 0.38, which means you need 0.38 inches of lead to reduce the exposure rate from 100 mR/hr to 25 mR/hr.
What is the maximum curie strength of a cobalt 60 source in a type A exposure device
10.8 Cl :
What is the maximum curie strength of an Iridium 192 source in a type A exposure device :
27.0 Cl
What regulations cover transportation of a source :
(DOT) Department of Transportation
When transporting a source it must:
1. Be secured in the vehicle
2. Not carried in the drivers compartment
3. Be less than 2 mR/h at the exterior of the vehicle
A source is being transported with a surface reading of 25 mR/h and 3 mR/h at 1 meter which transport label must be used :
Yellow III
A shipping container is surveyed and found to be 3 mR/h at 1 meter what label would be required :
Yellow III
this was a mistake
so i just filled in with this
A control panel in an x-ray system must have clearly visible the following statement:
Caution X-rays produced when energized
X-ray systems are required to have two indicators indicating when x-rays are produced one of which may be the milliampere meter and the other a red light this light must be labeled:
X-RAY ON
At the boundary of an area that exceeds 100 mR/HR (1 MSV/HR) posting with signs is necessary. Which of the following must be used?
Caution High Radiation Area
