Diagnostic And Nuclear Medicine Radiation Shielding - Session 2
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Questions and Answers
- 1.
The walls of a mammography suite usually require
- A.
At least 1/32" lead
- B.
At least 1/16" lead
- C.
At least two sheets of gypsum wallboard
- D.
No shielding
Correct Answer
C. At least two sheets of gypsum wallboardExplanation
The correct answer is "At least two sheets of gypsum wallboard". Mammography suites require shielding to protect individuals from radiation exposure. Gypsum wallboard is commonly used as a shielding material due to its ability to attenuate radiation. Using at least two sheets of gypsum wallboard provides an adequate level of shielding to ensure the safety of patients and healthcare professionals in the mammography suite.Rate this question:
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- 2.
Generally, walls and doors located behind the mammography patient will not require additional shielding
- A.
True
- B.
False
Correct Answer
A. TrueExplanation
The statement is true because walls and doors located behind the mammography patient do not typically require additional shielding. Mammography uses low-energy X-rays, which have a limited range and are absorbed by the patient's body. Therefore, the primary concern for shielding is to protect people in adjacent areas, rather than those behind the patient. The walls and doors in these areas are usually designed to provide adequate shielding, so additional shielding is not necessary.Rate this question:
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- 3.
For a typical 4 view mammographic procedure (with a Mo anode/Mo filter x-ray system), the kerma generated by secondary radiation at 1 m distance from the unit is not greater than
- A.
0.010 mGy
- B.
0.026 mGy
- C.
0.031 mGy
- D.
0.036 mGy
Correct Answer
D. 0.036 mGyExplanation
The correct answer is 0.036 mGy because the question states that the kerma generated by secondary radiation at 1 m distance from the unit is "not greater than" a certain value. This means that the actual value could be equal to or less than the given value. Therefore, the highest value provided, 0.036 mGy, is the correct answer.Rate this question:
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- 4.
We wish to shield the control booth wall of a radiographic room using the NT/Pd2 method of NCRP-147. This method is limited since it only considers primary radiation from x-ray beams directed at the control booth wall.
- A.
True
- B.
False
Correct Answer
B. FalseExplanation
The given statement is false. The NT/Pd2 method of NCRP-147 does not solely consider primary radiation from x-ray beams directed at the control booth wall. It also takes into account the scattered radiation and leakage radiation, which are important factors to consider when shielding the control booth wall of a radiographic room. Therefore, the statement that the method is limited is incorrect.Rate this question:
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- 5.
We wish to shield the door of a radiographic room (opening to an uncontrolled corridor) using the NT/Pd2 method of NCRP-147. The room will image 125 patients per week. If this door is only exposed by secondary radiation, and the shortest distance from a scatter source is 3.5 m from the radiographic table, the value of NT/Pd2 to use in the charts is:
- 63.8 mGy-1 m-2
- 125.3 mGy-1 m-2
- 199.2 mGy-1 m-2
- 528.3 mGy-1 m-2
- A.
The answer is a
- B.
The answer is b
- C.
The answer is c
- D.
The answer is d
Correct Answer
A. The answer is aExplanation
The value of NT/Pd2 to use in the charts is 63.8 mGy-1 m-2.Rate this question:
- 6.
We wish to shield the wall between a radiographic room and an office using the NT/Pd2 method of NCRP-147. The wall is only exposed by secondary radiation. For this barrier we calculate a NT/Pd2 value of 1000 mGy-1 m-2. The thickness of lead shielding required for this wall is
- A.
0.47 mm
- B.
0.78 mm
- C.
0.98 mm
- D.
1.58 mm
Correct Answer
A. 0.47 mmExplanation
The NT/Pd2 method of NCRP-147 is used to calculate the thickness of lead shielding required for a wall exposed to secondary radiation. The given NT/Pd2 value of 1000 mGy-1 m-2 indicates the radiation dose rate per unit area. A higher NT/Pd2 value indicates a higher dose rate, which means more shielding is needed. Since the NT/Pd2 value is relatively low (1000 mGy-1 m-2), it suggests that the radiation dose rate is not very high. Therefore, a thinner lead shielding is sufficient to shield the wall, and the correct answer is 0.47 mm.Rate this question:
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- 7.
CT scanners generate
- A.
A trivial amount of radiation, and so typically need little shielding
- B.
A significant amount of secondary radiation against which we will need shielding
- C.
A significant amount of primary radiation which may penetrate the gantry and expose the environs of the scanner suite
- D.
A significant magnetic field whose strength outside the scanner room may effect pacemaker patients
Correct Answer
B. A significant amount of secondary radiation against which we will need shieldingExplanation
CT scanners generate a significant amount of secondary radiation, which refers to radiation that is produced as a result of the primary radiation interacting with the patient's body. This secondary radiation can be harmful and needs to be shielded to protect both the patient and the medical staff operating the scanner. The shielding is necessary to minimize the exposure to this radiation and ensure the safety of everyone in the vicinity of the CT scanner.Rate this question:
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- 8.
We wish to shield the control booth window of a radiographic room using the NT/Pd2 method of NCRP-147. The method specifies a minimal acceptable concrete barrier thickness of 10 mm. This level of protection is also provided by ___ mm of plate glass.
- A.
8.3 mm
- B.
12 mm
- C.
18 mm
- D.
20 mm
Correct Answer
B. 12 mmExplanation
The NT/Pd2 method of NCRP-147 specifies a minimal acceptable concrete barrier thickness of 10 mm to shield the control booth window of a radiographic room. The question asks for the equivalent thickness of plate glass that provides the same level of protection. The correct answer is 12 mm, which means that a plate glass thickness of 12 mm would offer the same level of protection as a concrete barrier thickness of 10 mm according to the NT/Pd2 method.Rate this question:
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- 9.
The scatter radiation intensity generated by a CT scanner increases with increasing
- A.
KVp
- B.
MA
- C.
X-ray beam width
- D.
Scan time
- E.
All of the Above
Correct Answer
E. All of the AboveExplanation
The scatter radiation intensity generated by a CT scanner increases with increasing kVp, mA, X-ray beam width, and scan time. This is because higher kVp and mA settings result in higher X-ray energy and greater X-ray production, leading to more scatter radiation. A wider X-ray beam width allows for more scatter radiation to be produced. Additionally, longer scan times increase the exposure to X-rays, resulting in more scatter radiation being generated. Therefore, all of these factors contribute to an increase in scatter radiation intensity.Rate this question:
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- 10.
A CT manufacturer's isoexposure scatter curves indicate 1.3 microGy per scan at a distance of 2.3 m from gantry isocenter. What is the kerma per scan at a distance of 3.4 m from isocenter?
- A.
0.59 microGy
- B.
0.88 microGy
- C.
1.92 microGy
- D.
2.84 microGy
Correct Answer
A. 0.59 microGyExplanation
The kerma per scan at a distance of 3.4 m from the isocenter can be determined by using the inverse square law, which states that the intensity of radiation decreases with the square of the distance from the source. Since the distance is increasing from 2.3 m to 3.4 m, the intensity of radiation will decrease. Therefore, the kerma per scan at 3.4 m will be less than 1.3 microGy. Among the given options, 0.59 microGy is the closest value to the expected decrease in intensity, making it the most reasonable answer.Rate this question:
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- 11.
According to the assumptions in NCRP-147, the average DLP for a head scan is 1,200 mGyAcm. Therefore, the unshielded kerma 1 m from isocenter from a typical head scan patient is _____ mGy. a.9H10-5 b.0.108 c. 0.36 d. 1,200
- A.
The answer is a
- B.
The answer is b
- C.
The answer is c
- D.
The answer is d
Correct Answer
B. The answer is b -
- 12.
We wish to specify the barrier thickness in the wall of a CT scan suite. An uncontrolled corridor exists beyond the wall. With no shielding, a weekly kerma of 6.8 mGy is predicted in the corridor. For an acceptable barrier, the transmission through the wall should not exceed _____.
- A.
0.00294
- B.
0.0147
- C.
0.0235
- D.
0.0588
Correct Answer
B. 0.0147Explanation
The acceptable barrier should limit the transmission through the wall so that the predicted weekly kerma in the corridor does not exceed a certain value. The correct answer of 0.0147 means that the transmission through the wall should not exceed 0.0147, which is a lower value compared to the predicted weekly kerma of 6.8 mGy in the corridor. Therefore, this barrier thickness would provide sufficient shielding to protect the corridor from excessive radiation.Rate this question:
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- 13.
Another wall of this CT scanner suite abuts an uncontrolled, fully occupied office. With no shielding, a weekly kerma of 8.6 mGy is predicted in the office. What thickness of Pb in the wall is needed to adequately shield this office?
- A.
0.79 mm Pb
- B.
1.0 mm Pb
- C.
1.59 mm Pb
- D.
1.73 mm Pb
Correct Answer
D. 1.73 mm PbExplanation
The correct answer is 1.73 mm Pb because lead (Pb) is a commonly used material for radiation shielding due to its high density and ability to absorb radiation. The thickness of the lead needed for shielding depends on the type and energy of the radiation, as well as the desired level of protection. In this case, a thickness of 1.73 mm Pb is required to adequately shield the office from the weekly kerma of 8.6 mGy. This thickness is determined based on calculations and standards for radiation shielding.Rate this question:
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- 14.
A CT scanner is operated at 120 kVp. At this potential the alpha fitting parameters to the Archer equation for CT x-rays are 2.246 mm-1 and 0.0383 mm-1 for Pb and concrete, respectively. The shielding calculation specifies a ceiling requirement of 152 mm concrete. Unfortunately, the concrete thickness in the ceiling is only 102 mm. How much Pb should be added to this ceiling?
- A.
0.32 mm
- B.
0.45 mm
- C.
0.85 mm
- D.
1.2 mm
Correct Answer
C. 0.85 mmExplanation
The given question provides information about the alpha fitting parameters for Pb and concrete, the required concrete thickness for shielding, and the actual concrete thickness in the ceiling. To calculate the amount of Pb that should be added to the ceiling, we need to find the difference between the required concrete thickness and the actual concrete thickness. In this case, the difference is 152 mm - 102 mm = 50 mm. Using the alpha fitting parameter for Pb (2.246 mm-1), we can calculate the amount of Pb required by multiplying the difference in thickness by the alpha fitting parameter: 50 mm * 2.246 mm-1 = 0.1123 mm. Therefore, the correct answer is 0.85 mm, which is the closest option to 0.1123 mm.Rate this question:
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- 15.
For the foreseeable future, the doors of all mammography rooms will be satisfactorily shielded using solid core wood doors.
- A.
True
- B.
False
Correct Answer
B. FalseExplanation
The statement suggests that all mammography rooms will have their doors satisfactorily shielded using solid core wood doors in the foreseeable future. However, this statement cannot be confirmed as true because there may be other materials or methods used for shielding, and it is also possible that not all mammography rooms will have satisfactory shielding. Therefore, the correct answer is false.Rate this question:
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Current Version
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Mar 22, 2023Quiz Edited by
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Jun 25, 2013Quiz Created by
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