Probability of a radioactive particle decaying after passing through a surface

A radioactive particle with mean lifetime $\tau$ is a distance $z$ from a planar surface in 3D space. It begins traveling in a constant, uniform random direction at constant velocity $v$. What is the chance it passes through the plane before decaying?

Let the angle of $v$ from the normal to the surface be $\theta$ and the azimuthal angle $\phi$.

The chance that the particle has not decayed at time $t$ is $e^{-t/\tau}$.

The length to the plane along the direction $\vec{v}$ is $z/\cos\theta$. Thus the chance that it escapes is

\[ e^{-z/v\tau\cos\theta} \]

Let $b$ be the normalized distance $b=z/v \tau$.

We’d like to average over all possible directions of travel. \[ \frac{1}{4\pi} \int_0^{2 \pi} \int_0^{\pi/2} e^{-b/\cos\theta} \sin \theta \; d\theta \; d\phi \]

Notice that the $\theta$ integral is only to $\pi/2$: if the particle travels downward it will never reach the plane, so contributes nothing.

The answer is \[\boxed{\frac{1}{2}Ei_2(b)}\] where $Ei_2$ is an exponential integral function.

In order to do the integral:

First do the $\phi$ integral.

\[\frac{1}{2} \int_0^{\pi/2} e^{-b/\cos\theta} \sin \theta \; d\theta\]

(From this point on we drop the leading $1 / 2$ for brevity.)

Let $x=\cos\theta$, $dx = - \sin\theta d\theta$.

\[ -\int_1^0 e^{-b/x}\;dx = \int_0^1 e^{-b/x}\;dx \]

Let $y=1/x$, $dy=-1/x^2 dx$ or $dy/y^2 = -dx$

\[ -\int_\infty^1 e^{-b y} y^{-2} \;dy = \int_1^\infty e^{-b y} y^{-2} \;dy \]

This is the definition of the second exponential integral Ei$_2(b)$.

Putting back the $1 / 2$,

\[\boxed{\frac{1}{2}Ei_2(b)}\]

More about $Ei$

The definition of the exponential integral is

\[ E_n(z) = \int_0^\infty e^{-z t} t^{-n} \; d t \] In Mathematica it is ExpIntegralE[n, z].

It happens that $Ei_2(b) = e^{-b} - b \Gamma(0,b)$ where $\Gamma(s,z)$ is the Upper Incomplete Gamma function.

Written on January 15th, 2016