Texas Industrial Radiography Gamma Practice Test

What is the approximate reduction in dose rate at a location with a ½ inch thick lead shield in the path of Co60 radiation?

1. 2/3

2. ½

3. ¾

4. None of the above

The correct answer is: ½

To understand why the dose rate is approximately halved with a ½ inch thick lead shield in the path of Co60 radiation, it’s essential to consider how lead functions as a shielding material. Cobalt-60 (Co60) is a radionuclide that emits gamma radiation, which is highly penetrating and requires substantial protection to reduce dose rates effectively. The effectiveness of lead in attenuating this radiation can be quantified using its half-value layer (HVL), which is the thickness of the material required to reduce the intensity of the radiation by half. For Co60, the HVL in lead is around 0.5 inches, meaning that when you place a ½ inch thick lead barrier between the source and the measurement point, it reduces the radiation exposure by approximately 50%. Therefore, if the original dose rate was 100%, after passing through the ½ inch of lead, it would be reduced to about 50%. This principle highlights the significance of using adequate shielding to protect against high-energy gamma rays, emphasizing that selecting materials and their respective thicknesses are critical components in radiation safety practices. Thus, in this scenario, the approximate reduction in dose rate due to the lead shield effectively is about half of the original rate.