Let's solve the problem step-by-step. --- ## **Given Data:** - **Mass activity, \( A \) = 127 mCi/g** - **Uniform distribution in water sphere** - **Isotopes: \( ^{137}\mathrm{Cs} \), \( ^{58}\mathrm{Co} \), \( ^{241}\mathrm{Am} \)** - **Find dose rate at center of sphere of radius:** 1. Large (radiation equilibrium attained) 2. Small (charged particle equilibrium just reached) 3. \( r = 2 \) cm --- ## **Step 1: Convert Activity to SI Units** \[ 1\, \text{mCi} = 3.7 \times 10^7\, \text{Bq} \] \[ A = 127\, \text{mCi/g} = 127 \times 3.7 \times 10^7\, \text{Bq/g} = 4.699 \times 10^9\, \text{Bq/g} \] --- ## **Step 2: Find Dose Rate General Formula** The dose rate at the center of a sphere, assuming **equilibrium** (all energy is absorbed): \[ \text{Dose Rate} = A \cdot E_\gamma \cdot y \cdot f \] Where: - \( A \): Activity (Bq/g) - \( E_\gamma \): Mean photon energy per disintegration (MeV) - \( y \): Yield (photons/disintegration) - \( f \): Conversion factor from MeV/g to Gy (1 MeV/g = \( 1.602 \times 10^{-13} \) J/g; 1 Gy = 1 J/kg) \[ \text{Dose Rate (Gy/s)} = \frac{A \cdot E_\gamma \cdot y \cdot 1.602 \times 10^{-13}}{1} \] If the source is uniformly distributed and equilibrium is reached, **all emitted energy is deposited**. --- ## **Step 3: Isotope Data** | Isotope | \( E_\gamma \) (MeV) | \( y \) (photons/dis) | |--------------|----------------------|-----------------------| | \( ^{137}\mathrm{Cs} \) | 0.662 | 0.85 | | \( ^{58}\mathrm{Co} \) | 1.215 | 2.1 | | \( ^{241}\mathrm{Am} \) | 0.060 | 0.36 | --- ## **Step 4: Dose Rate Calculation** **Dose rate in Gy/s per gram:** \[ \text{Dose Rate} = A \cdot E_\gamma \cdot y \cdot 1.602 \times 10^{-13} \] ### **(1) Radiation equilibrium (large sphere):** #### **(a) \( ^{137}\mathrm{Cs} \):** \[ = 4.699 \times 10^9 \cdot 0.662 \cdot 0.85 \cdot 1.602 \times 10^{-13} \] \[ = 4.699 \times 10^9 \cdot 0.5627 \cdot 1.602 \times 10^{-13} \] \[ = 2.642 \times 10^9 \cdot 1.602 \times 10^{-13} \] \[ = 4.235 \times 10^{-4}\ \text{Gy/s} \] #### **(b) \( ^{58}\mathrm{Co} \):** \[ = 4.699 \times 10^9 \cdot 1.215 \cdot 2.1 \cdot 1.602 \times 10^{-13} \] \[ = 4.699 \times 10^9 \cdot 2.5515 \cdot 1.602 \times 10^{-13} \] \[ = 11.98 \times 10^9 \cdot 1.602 \times 10^{-13} \] \[ = 1.919 \times 10^{-3}\ \text{Gy/s} \] #### **(c) \( ^{241}\mathrm{Am} \):** \[ = 4.699 \times 10^9 \cdot 0.060 \cdot 0.36 \cdot 1.602 \times 10^{-13} \] \[ = 4.699 \times 10^9 \cdot 0.0216 \cdot 1.602 \times 10^{-13} \] \[ = 101.5 \times 10^6 \cdot 1.602 \times 10^{-13} \] \[ = 1.627 \times 10^{-5}\ \text{Gy/s} \] --- ## **Step 5: Charged Particle Equilibrium** For **charged particle equilibrium (CPE)**, the sphere radius must be at least the **range of secondary electrons** (gamma, beta, alpha): - For gamma emitters (\( ^{137}\mathrm{Cs}, ^{58}\mathrm{Co} \)), electron range in water is a few mm. - For alpha (\( ^{241}\mathrm{Am} \)), range is about 40 microns. **So, at CPE radius, dose rate ≈ same as equilibrium above**, as almost all energy is deposited. --- ## **Step 6: For Radius = 2 cm** At 2 cm, most photons will escape unless their mean free path is much less than 2 cm. For water, linear attenuation coefficient for 0.6 MeV gamma ~0.09 cm\(^{-1}\): \[ I = I_0 e^{-\mu x} \] \[ e^{-\mu x} = e^{-0.09 \times 2} = e^{-0.18} \approx 0.84 \] So about 16% is attenuated, 84% escapes. Only a fraction of energy is deposited. So, **fraction deposited**: \[ \text{Fraction deposited} = 1 - e^{-\mu r} \] For r = 2 cm, μ = 0.09 cm\(^{-1}\): \[ 1 - e^{-0.18} = 1 - 0.836 = 0.164 \] Only about 16.4% of emitted energy is deposited. So, **dose rate at r = 2 cm**: \[ \text{Dose rate} = \text{Dose rate (equilibrium)} \times 0.164 \] --- ## **Step 7: Final Answers (Summary Table)** | Isotope | Dose Rate (equilibrium) (Gy/s) | Dose Rate at r = 2 cm (Gy/s) | |---------|-------------------------------|-------------------------------| | \( ^{137}\mathrm{Cs} \) | \( 4.24 \times 10^{-4} \) | \( 6.95 \times 10^{-5} \) | | \( ^{58}\mathrm{Co} \) | \( 1.92 \times 10^{-3} \) | \( 3.15 \times 10^{-4} \) | | \( ^{241}\mathrm{Am} \) | \( 1.63 \times 10^{-5} \) | \( 2.67 \times 10^{-6} \) | --- ## **Summary** 1. **Radiation equilibrium (large sphere):** - \( ^{137}\mathrm{Cs} \): \( 4.24 \times 10^{-4} \) Gy/s - \( ^{58}\mathrm{Co} \): \( 1.92 \times 10^{-3} \) Gy/s - \( ^{241}\mathrm{Am} \): \( 1.63 \times 10^{-5} \) Gy/s 2. **Charged particle equilibrium (CPE):** - Nearly the same values as above (since CPE occurs at a small radius where almost all energy is deposited). 3. **Radius = 2 cm:** - \( ^{137}\mathrm{Cs} \): \( 6.95 \times 10^{-5} \) Gy/s - \( ^{58}\mathrm{Co} \): \( 3.15 \times 10^{-4} \) Gy/s - \( ^{241}\mathrm{Am} \): \( 2.67 \times 10^{-6} \) Gy/s --- **If you want dose rates in Gy/hr, multiply by 3600.** Let me know if you need the calculation in a different unit or more details!