Let's break down the problem **step by step**: --- ### **Given Data** - **Source:** \( ^{210}\text{Po-B} \) point source - **Neutron energy:** \( E_n = 2.5 \) MeV - **Source emission rate:** \( S = 2 \times 10^7 \) n/sec - **Distance from source to person:** \( d = 100 \) cm - **Water shield thickness:** \( x = 10 \) cm - **Hydrogen elastic scattering cross-section (\(2.5\,\text{MeV}\) n):** \(\sigma_{s,H} = 2.5\, \text{barns} = 2.5 \times 10^{-24} \text{cm}^2\) - **Water's neutron removal coefficient:** \(\Sigma_{rem} = 0.103\, \text{cm}^{-1}\) - **Ignore air attenuation; only consider neutron interactions with H in the body** - **First collision dose rate (in mrad/hr) is required** --- ## **Step 1: Calculate Neutron Flux at 100 cm (Without Shielding)** For a **point source**, the neutron flux at distance \( r \) (if there is no attenuation) is: \[ \phi_0 = \frac{S}{4\pi r^2} \] \[ \phi_0 = \frac{2 \times 10^7}{4\pi (100)^2} \] \[ = \frac{2 \times 10^7}{4\pi \times 10^4} \] \[ = \frac{2 \times 10^7}{1.256 \times 10^5} \] \[ \approx 159.2 \text{ n/cm}^2\text{/s} \] --- ## **Step 2: Attenuation Through Water Shield** Attenuation by the shield (water, 10 cm thick): \[ \phi = \phi_0 e^{-\Sigma_{rem} x} \] \[ \Sigma_{rem} = 0.103\, \text{cm}^{-1}, \quad x = 10\, \text{cm} \] \[ \phi = \phi_0 \cdot e^{-0.103 \times 10} \] \[ = \phi_0 \cdot e^{-1.03} \] \[ \approx \phi_0 \cdot 0.357 \] \[ \phi = 159.2 \times 0.357 \approx 56.9 \text{ n/cm}^2\text{/s} \] --- ## **Step 3: First Collision Rate in the Body** First collision rate per cm³ is: \[ \text{Collision Rate} = \phi \cdot N_H \cdot \sigma_{s,H} \] But for **first collision dose**, we consider the **collision density** in the body, **per hydrogen atom**. #### **Number Density of Hydrogen in Body** Assume **soft tissue ≈ water**: - Water's density \( \rho = 1\,\text{g/cm}^3 \) - 1 mole \( \text{H}_2\text{O} \): 18 g - 2 moles H per mole water Number density of H in water: \[ n_H = \frac{2 \times N_A \times \rho}{18} \] \[ = \frac{2 \times 6.022 \times 10^{23} \times 1}{18} \] \[ = \frac{12.044 \times 10^{23}}{18} \] \[ = 0.669 \times 10^{23} = 6.69 \times 10^{22} \text{ H atoms/cm}^3 \] #### **Macroscopic Cross-Section for H in Water** \[ \Sigma_{s,H} = n_H \cdot \sigma_{s,H} \] \[ = (6.69 \times 10^{22}) \times (2.5 \times 10^{-24}) = 0.167 \text{ cm}^{-1} \] #### **First Collision Density (per cm³ per s):** \[ \text{First Collision Rate} = \phi \cdot \Sigma_{s,H} \] \[ = 56.9 \times 0.167 \approx 9.51 \text{ n/cm}^3\text{/s} \] --- ## **Step 4: First Collision Dose Rate Calculation** ### **Energy Deposited per Collision** - For 2.5 MeV neutrons, **elastic scattering with hydrogen** can transfer up to 2.5 MeV per collision (worst case), but the average energy transfer is **half** (since neutron and proton have same mass): \[ \text{Average energy per collision} = \frac{E_n}{2} = 1.25 \text{ MeV} \] \[ = 1.25 \times 1.602 \times 10^{-6} \text{ erg} \] \[ = 2.003 \times 10^{-6} \text{ erg} \] ### **Energy Deposited per cm³ per sec:** \[ \text{Energy/sec/cm}^3 = (\text{collision rate}) \times (\text{energy per collision}) \] \[ = 9.51 \times 2.003 \times 10^{-6} \text{ erg/cm}^3\text{/s} \] \[ = 1.904 \times 10^{-5} \text{ erg/cm}^3\text{/s} \] ### **Convert to Dose Rate (rad/s):** - 1 rad = 100 erg/g - Assume tissue density = 1 g/cm³ \[ \text{Dose rate (rad/s)} = \frac{1.904 \times 10^{-5} \text{ erg/cm}^3\text{/s}}{100 \text{ erg/g}} = 1.904 \times 10^{-7} \text{ rad/s} \] ### **Convert to mrad/hr:** \[ 1 \text{ rad/s} = 3600 \text{ rad/hr} \] \[ 1 \text{ rad} = 1000 \text{ mrad} \] So, \[ \text{Dose rate (mrad/hr)} = 1.904 \times 10^{-7} \times 3600 \times 1000 \] \[ = 1.904 \times 10^{-7} \times 3.6 \times 10^6 \] \[ = 1.904 \times 3.6 \times 10^{-1} \] \[ = 6.854 \times 10^{-1} \] \[ = 0.69 \text{ mrad/hr} \] --- ## **Final Answer** \[ \boxed{0.69 \ \text{mrad/hr}} \] is the **first collision dose rate at 100 cm** from the source, **with 10 cm water shielding**.