1. Problem Statement

We need to find the alpha particle flux density (particles per unit area per unit time) near the surface of a large plate of $^{239}\text{Pu}$ with thickness $d = 1 \text{ mm}$.

Given:

• Half-life $\tau = 2.4 \times 10^4$ years
• Density $\rho = 19800 \text{ kg/m}^3$
• Mass of atom $m_0 = 3.84 \times 10^{-25} \text{ kg}$

2. Derivation

Consider a surface area $A$ of the plate. The volume of this section is $V = A \cdot d$.

The total number of nuclei $N$ in this volume is:

$$N = \frac{\text{Total Mass}}{m_0} = \frac{\rho V}{m_0} = \frac{\rho A d}{m_0}$$

The rate of decay (Activity) is given by:

$$\frac{dN}{dt} = \lambda N = \left( \frac{\ln 2}{\tau} \right) \left( \frac{\rho A d}{m_0} \right)$$

Since the plate is thin and emission is isotropic, on average, half of the particles will escape through the top surface and half through the bottom surface (neglecting edge effects and self-absorption for thin films).

The Flux Density $J$ is the number of particles escaping per unit area per second:

$$J = \frac{1}{2A} \left( \frac{dN}{dt} \right)$$ $$J = \frac{1}{2A} \cdot \frac{\ln 2}{\tau} \cdot \frac{\rho A d}{m_0}$$ $$J = \frac{\rho d \ln 2}{2 \tau m_0}$$

3. Calculation

First, convert the half-life into seconds:

$$\tau = 2.4 \times 10^4 \text{ years} \approx 2.4 \times 10^4 \times 3.15 \times 10^7 \text{ s} \approx 7.57 \times 10^{11} \text{ s}$$

Substitute the values:

$$J = \frac{(19800)(10^{-3})(0.693)}{2(7.57 \times 10^{11})(3.84 \times 10^{-25})}$$ $$J \approx \frac{13.72}{5.81 \times 10^{-13}}$$ $$J \approx 2.36 \times 10^{13} \text{ m}^{-2}\text{s}^{-1}$$