Diagnostic and Nuclear Medicine Radiation Shielding - Session 2

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.

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.

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.

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.

The value of NT/Pd2 to use in the charts is 63.8 mGy-1 m-2.

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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.

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.

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.

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.

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.

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.

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.

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.

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.