Let's break down and solve **Question 8** step by step: --- ## **Step 1: Understanding the Problem** We have an annular region (2D cross-section), filled with a homogeneous absorbing fluid. A monochromatic UV source irradiates the system, and the quantum yield (reaction initiation) depends on the local photon absorption rate, which decays exponentially due to absorption by the medium. We are to derive an **integral expression for the volume-averaged apparent kinetic rate constant** across the annulus, using: - **Material extinction coefficient:** \( \alpha \) - **Incident central radiant intensity:** \( I_ \) - **Radial boundaries:** \( R_i \) (inner), \( R_o \) (outer) - **Other parameters:** \( \Phi \) (quantum yield factor) - The law: Intensity decays as \( I(r) = I_ e^{-\alpha (r - R_i)} \) We want the final answer in the form: \[ \langle k \rangle = \frac{2}{R_o^2 - R_i^2} \int_{R_i}^{R_o} \Phi \cdot I_ \cdot e^{-\alpha (r - R_i)} \cdot r \, dr \] --- ## **Step 2: Expressing the Local Rate Constant** At radius \( r \), the local volumetric photon absorption rate (proportional to the photochemical rate) is: \[ \text{rate}(r) = \Phi \cdot I(r) = \Phi \cdot I_ \cdot e^{-\alpha (r - R_i)} \] --- ## **Step 3: Volume-Averaged Apparent Kinetic Rate Constant** The average over the annular cross-section is: \[ \langle k \rangle = \frac{\text{Total absorption rate over volume}}{\text{Total volume}} \] For a 2D annulus of thickness \( dr \), the area element is \( dA = 2\pi r \, dr \) (but since we want per unit area, we normalize by total area). But here, the question asks for the average over the area (ignoring \( 2\pi \)), so we use: \[ \langle k \rangle = \frac{1}{A} \int_{A} \Phi \cdot I_ e^{-\alpha(r - R_i)} dA \] In polar coordinates for an annulus: \[ A = \pi (R_o^2 - R_i^2) \] \[ dA = 2\pi r \, dr \] But since the provided equation uses \( r \, dr \), let's keep it as the average per \( r \, dr \). So, for the area average: \[ \langle k \rangle = \frac{1}{R_o^2 - R_i^2} \int_{R_i}^{R_o} \Phi \cdot I_ e^{-\alpha(r - R_i)} \cdot 2r \, dr \] \[ = \frac{2}{R_o^2 - R_i^2} \int_{R_i}^{R_o} \Phi \cdot I_ e^{-\alpha(r - R_i)} \cdot r \, dr \] --- ## **Step 4: Final Integral Equation** This matches the required form: \[ \boxed{ \langle k \rangle = \frac{2}{R_o^2 - R_i^2} \int_{R_i}^{R_o} \Phi \cdot I_ \cdot e^{-\alpha (r - R_i)} \cdot r \, dr } \] --- ## **Step 5: Explanation** - \( \Phi \) = quantum yield factor (proportionality for kinetic initiation) - \( I_ e^{-\alpha(r - R_i)} \) = local photon intensity at radius \( r \) - \( r \, dr \) = radial area element (in 2D polar coordinates) - The denominator \( R_o^2 - R_i^2 \) normalizes over the annular cross-sectional area --- ## **Step 6: (Optional) Evaluate the Integral** If asked to evaluate the integral: \[ \int_{R_i}^{R_o} r e^{-\alpha (r - R_i)} dr \] Let \( u = r - R_i \), \( r = u + R_i \), \( dr = du \), limits: \( u = \) to \( u = R_o - R_i \): \[ \int_^{R_o - R_i} (u + R_i) e^{-\alpha u} du \] \[ = \int_^{R_o - R_i} u e^{-\alpha u} du + R_i \int_^{R_o - R_i} e^{-\alpha u} du \] Each can be integrated by parts or using integral tables. But as per the question, **the boxed integral is the required final answer.** --- # **Final Boxed Answer** \[ \boxed{ \left\langle k \right\rangle = \frac{2}{R_o^2 - R_i^2} \int_{R_i}^{R_o} \Phi \cdot I_ \cdot e^{-\alpha (r - R_i)} \cdot r \; dr } \] This is the rigorous integral mathematical expression for the volume-averaged apparent kinetic rate constant across the annular cross-section, as requested.