Let's break down the problem step by step: --- ## **Given Data** - Distance from source, \( r = 100 \) cm - Shield thickness (water), \( d = 10 \) cm - Source emission rate, \( S = 2 \times 10^7 \) n/sec - Neutron energy, \( E_n = 2.5 \) MeV - Water’s neutron removal coefficient, \( \Sigma_{R,water} = 0.103 \) cm\(^{-1}\) - Hydrogen elastic scattering cross-section, \( \sigma = 2.5 \) barns (\(2.5 \times 10^{-24}\) cm\(^2\)) - Only hydrogen in the body considered for dose calculation - Ignore air attenuation - **First collision dose rate** (to be found, in mrad/hr) --- ## **Step 1: Attenuation of Neutron Flux by Water Shield** The neutron flux at 100 cm after passing through 10 cm water: \[ \text{Attenuation factor} = e^{-\Sigma_{R,water} \cdot d} \] \[ = e^{-0.103 \times 10} = e^{-1.03} \approx 0.357 \] --- ## **Step 2: Flux at 100 cm from the Source** Without shielding (point source): \[ \text{Flux at 100 cm, no shield} = \frac{S}{4\pi r^2} \] \[ = \frac{2 \times 10^7}{4\pi (100)^2} \] \[ = \frac{2 \times 10^7}{125,664} \approx 159.2 \text{ n/cm}^2\text{/s} \] With shielding: \[ \text{Flux at 100 cm, with shield} = 159.2 \times 0.357 = 56.87 \text{ n/cm}^2\text{/s} \] --- ## **Step 3: First Collision Dose Rate in the Body** - Only consider neutron interactions with hydrogen atoms in the body. - The **first collision dose rate** is due to neutrons that make their first collision in the body (assuming the body is a slab at 100 cm). **First collision rate per cm\(^3\) in the body:** \[ \text{First collision rate} = \phi \cdot N_H \cdot \sigma \] Where: - \( \phi \) = neutron flux at 100 cm (\( 56.87 \) n/cm\(^2\)/s) - \( N_H \) = number density of H atoms in tissue (\(\approx 6.7 \times 10^{22}\) atoms/cm\(^3\)) - \( \sigma \) = 2.5 barns = \( 2.5 \times 10^{-24} \) cm\(^2\) \[ \text{First collision rate} = 56.87 \times (6.7 \times 10^{22}) \times (2.5 \times 10^{-24}) \] \[ = 56.87 \times 0.1675 \] \[ \approx 9.53 \text{ n/cm}^3\text{/s} \] --- ## **Step 4: Energy Deposited per Collision** - Each neutron deposits its kinetic energy in the first collision (2.5 MeV per neutron). Total energy deposited per cm\(^3\) per second: \[ E = (\text{collision rate}) \times E_n \] \[ = 9.53 \text{ n/cm}^3\text{/s} \times 2.5 \text{ MeV} \] \[ = 23.83 \text{ MeV/cm}^3\text{/s} \] Convert MeV to ergs: - \(1 \text{ MeV} = 1.602 \times 10^{-6} \) ergs \[ E = 23.83 \times 1.602 \times 10^{-6} = 3.818 \times 10^{-5} \text{ ergs/cm}^3\text{/s} \] --- ## **Step 5: Dose Rate Calculation** - 1 rad = \( 100 \) ergs/g (or \(\approx 1\) erg/cm\(^3\) for tissue density 1g/cm\(^3\)) - Dose rate (rad/s): \[ \text{Dose rate (rad/s)} = \frac{E}{1} \] \[ = 3.818 \times 10^{-5} \text{ rad/s} \] - Convert to mrad/hr: - \(1 \text{ rad} = 1000 \text{ mrad}\) - \(1 \text{ hr} = 3600 \text{ s}\) \[ \text{Dose rate (mrad/hr)} = 3.818 \times 10^{-5} \times 1000 \times 3600 \] \[ = 0.1375 \text{ mrad/hr} \] --- ## **Final Answer** \[ \boxed{0.14 \text{ mrad/hr}} \] **First collision dose rate to the person from fast neutrons at 100 cm is approximately \(0.14\) mrad/hr.**