Let's break down the problem step-by-step. ### **Given Data:** - **X-ray output:** 25 microSieverts/pulse at 30 cm - **Tube loading:** 200 pulses/hour - **Distance:** 30 cm from the source - **Questions:** 1. Advice on the size of the controlled area required. 2. If mobile lead screens are used, what thickness is required to limit the controlled area? --- ## **Step 1: Calculate the Hourly Dose Rate at 30 cm** \[ \text{Dose rate per hour at 30cm} = 25 \, \mu Sv/\text{pulse} \times 200 \, \text{pulses/hour} \] \[ = 5000 \, \mu Sv/\text{hour} = 5 \, mSv/\text{hour} \] --- ## **Step 2: Dose Limits for Controlled Area** - **Controlled area:** Dose rate should not exceed **7.5 μSv/hour** (as recommended by ICRP and regulatory bodies). - **Design goal:** Dose rate outside the controlled area should be < 7.5 μSv/hour. --- ## **Step 3: Calculate the Distance for Controlled Area Boundary** The dose falls off by the **inverse square law**: \[ D_1 = D_0 \left(\frac{r_0}{r_1}\right)^2 \] Where: - \( D_0 = 5 \, mSv/\text{hour} \) at \( r_0 = 0.3 \, m \) - \( D_1 = 7.5 \, \mu Sv/\text{hour} = 0.0075 \, mSv/\text{hour} \) - \( r_1 = ? \) \[ 0.0075 = 5 \left(\frac{0.3}{r_1}\right)^2 \] \[ \frac{0.0075}{5} = \left(\frac{0.3}{r_1}\right)^2 \] \[ 0.0015 = \left(\frac{0.3}{r_1}\right)^2 \] \[ \sqrt{0.0015} = \frac{0.3}{r_1} \] \[ r_1 = \frac{0.3}{\sqrt{0.0015}} \] \[ \sqrt{0.0015} \approx 0.0387 \] \[ r_1 \approx \frac{0.3}{0.0387} \approx 7.75 \text{ meters} \] --- ### **1. Advice on the Size of the Controlled Area** > The controlled area should extend to a radius of approximately **7.75 meters** from the X-ray source to ensure dose rates do not exceed 7.5 μSv/hour. --- ## **Step 4: Lead Shielding Calculation** Assume you want to use **lead screens** to **reduce the controlled area** (i.e., allow people to be closer than 7.75 m). ### **Attenuation Formula:** \[ I = I_0 e^{-\mu x} \] Where: - \( I \) = required intensity outside the shield (7.5 μSv/hr) - \( I_0 \) = incident intensity (5 mSv/hr = 5000 μSv/hr) - \( \mu \) = linear attenuation coefficient of lead at 300 kVp - \( x \) = required thickness (in cm or mm) \[ \frac{I}{I_0} = e^{-\mu x} \] \[ \frac{7.5}{5000} = e^{-\mu x} \] \[ 0.0015 = e^{-\mu x} \] \[ \ln(0.0015) = -\mu x \] \[ -6.5 \approx -\mu x \] \[ \mu x = 6.5 \] #### **Find μ for 300 kVp X-rays in Lead:** - At 300 kVp, \(\mu\) for lead ≈ **0.56 mm\(^{-1}\)** (check NIST tables or physics data). \[ x = \frac{6.5}{0.56} \approx 11.6 \text{ mm} \] --- ### **2. Required Thickness of Lead Screen** > **A lead thickness of approximately 12 mm (rounded up for safety) is required to reduce the dose rate from 5 mSv/hr to 7.5 μSv/hr at 30 cm from the source.** --- ## **Final Answers:** ### **1. Controlled Area Size:** - The controlled area should extend to a radius of **7.75 meters** from the source. ### **2. Lead Thickness:** - **12 mm of lead** is required to limit the dose rate to below 7.5 μSv/hr at 30 cm from the source. --- ### **Summary Table** | Requirement | Value | |---------------------------|-------------| | Controlled area radius | 7.75 m | | Lead thickness (at 30 cm) | 12 mm | --- **Let me know if you need references or further clarification!**