ASNT INDUSTRIAL RADIOGRAPHY
RADIATION SAFETY EXAM 2026/2027
WITH 220 QUESTIONS AND EXPERT-
VERIFIED CORRECT ANSWERS |
ALREADY GRADED A+ |
GUARANTEED PASS
Exposure devices using depleted uranium (DU) must be leak tested every: -
ANSWER-12 months
A leak test would be considered failed should the sample tested return a value
greater than: - ANSWER-0.005 μCi (185 Bq)
Should a source or exposure devise return a wipe resulting in more than 0.005 μCi
(185 Bq) the licensee must do what? - ANSWER-Immediately remove it from
service and have it decontaminated or disposed of
Gamma alarms (audible & visible), also known as area alarms, must be tested
every - ANSWER-day prior to use
,There are basically three methods of protection from ionizing radiation What are
they? - ANSWER-Dense atomic structures, time and distance
To determine the total dose received - ANSWER-Multiply the rate of dose by the
total time exposed
At a dose rate of 60 mR/h (600 11Sv/h), 1h of exposure would be? - ANSWER-60
mR (600 11Sv)
The specific gamma ray constant for Co-60 can be
expressed as: - ANSWER-14 R/h (140 mSv/h)
The specific gamma ray constant for Ir-192 can be
expressed as: - ANSWER-5.3 R/h/Ci at 1 ft (140 mSv/Ci at 1 ft)
A specific gamma ray constant can be stated as: - ANSWER-dose rate per curie at
a specific distance
The inverse square law is written as: - ANSWER-the intensity is indirectly
proportional to the distance squared
Mathematically, the inverse square law is written as: - ANSWER-I (sub 1) x I (sub
2) = D (sub 2) squared x D (sub 1) squared
Shielding can be expressed by the amount of shilding provided, so that the amount
of shielding that would reduce the original exposure rate to one-half would be
stated as: - ANSWER-1 half-value layer
, The best material to use as shielding would be? - ANSWER-lead
When using lead to shield an Ir-192 source, you would need ____________to
reduce the original exposure to one-half. - ANSWER-0.2 in. (5.08 mm)
Standing in a 100 mR/h (1000 J-lSv/h) field, the radiographer will make four 10
min exposures. The radiographer's total exposure for the day would be: -
ANSWER-66.7 mR (677 μSv/h)
With 30 Ci of iridium with a gamma constant of
5.2 R/h/Ci at 1 ft (52 mSv/h/Ci at 0.3 m), what would your exposure rate be at 1 ft
(0.3 m) from the source? - ANSWER-156 R per hour
If you have 73 Ci of Co-60, your exposure rate at 2 ft (0.6 m) from the source
would be? - ANSWER-255.5 R
You have an intensity of 50 mR/h at 100 ft (50 μSvv/hat 30.48 m) from the source.
Moving to 50 ft (15.24 m) would increase your exposure rate by a factor of what? -
ANSWER-4 times
If the radiographer has 100 mR/h at 25ft (1000 μSv at 7.63), the exposure rate at 50
ft (15.24 m) would be ________ with 1 half-value of shielding between the
radiographer and the source. - ANSWER-12.5 mR/h
In the early years of radiography, what caused personnel to be overexposed? -
ANSWER-Poor equipment design, lack of understanding, lack of and poor safety
practices.
RADIATION SAFETY EXAM 2026/2027
WITH 220 QUESTIONS AND EXPERT-
VERIFIED CORRECT ANSWERS |
ALREADY GRADED A+ |
GUARANTEED PASS
Exposure devices using depleted uranium (DU) must be leak tested every: -
ANSWER-12 months
A leak test would be considered failed should the sample tested return a value
greater than: - ANSWER-0.005 μCi (185 Bq)
Should a source or exposure devise return a wipe resulting in more than 0.005 μCi
(185 Bq) the licensee must do what? - ANSWER-Immediately remove it from
service and have it decontaminated or disposed of
Gamma alarms (audible & visible), also known as area alarms, must be tested
every - ANSWER-day prior to use
,There are basically three methods of protection from ionizing radiation What are
they? - ANSWER-Dense atomic structures, time and distance
To determine the total dose received - ANSWER-Multiply the rate of dose by the
total time exposed
At a dose rate of 60 mR/h (600 11Sv/h), 1h of exposure would be? - ANSWER-60
mR (600 11Sv)
The specific gamma ray constant for Co-60 can be
expressed as: - ANSWER-14 R/h (140 mSv/h)
The specific gamma ray constant for Ir-192 can be
expressed as: - ANSWER-5.3 R/h/Ci at 1 ft (140 mSv/Ci at 1 ft)
A specific gamma ray constant can be stated as: - ANSWER-dose rate per curie at
a specific distance
The inverse square law is written as: - ANSWER-the intensity is indirectly
proportional to the distance squared
Mathematically, the inverse square law is written as: - ANSWER-I (sub 1) x I (sub
2) = D (sub 2) squared x D (sub 1) squared
Shielding can be expressed by the amount of shilding provided, so that the amount
of shielding that would reduce the original exposure rate to one-half would be
stated as: - ANSWER-1 half-value layer
, The best material to use as shielding would be? - ANSWER-lead
When using lead to shield an Ir-192 source, you would need ____________to
reduce the original exposure to one-half. - ANSWER-0.2 in. (5.08 mm)
Standing in a 100 mR/h (1000 J-lSv/h) field, the radiographer will make four 10
min exposures. The radiographer's total exposure for the day would be: -
ANSWER-66.7 mR (677 μSv/h)
With 30 Ci of iridium with a gamma constant of
5.2 R/h/Ci at 1 ft (52 mSv/h/Ci at 0.3 m), what would your exposure rate be at 1 ft
(0.3 m) from the source? - ANSWER-156 R per hour
If you have 73 Ci of Co-60, your exposure rate at 2 ft (0.6 m) from the source
would be? - ANSWER-255.5 R
You have an intensity of 50 mR/h at 100 ft (50 μSvv/hat 30.48 m) from the source.
Moving to 50 ft (15.24 m) would increase your exposure rate by a factor of what? -
ANSWER-4 times
If the radiographer has 100 mR/h at 25ft (1000 μSv at 7.63), the exposure rate at 50
ft (15.24 m) would be ________ with 1 half-value of shielding between the
radiographer and the source. - ANSWER-12.5 mR/h
In the early years of radiography, what caused personnel to be overexposed? -
ANSWER-Poor equipment design, lack of understanding, lack of and poor safety
practices.
