Why does the intensity of an uncollided beam of neutrons post-target drop off exponentially with the thickness of the target? I guess I would have thought it was linear. The exponential dropoff surprised me and I don't understand why it's the case--can someone explain?
 A: the answer is simple.
we divide the target thickness into discrete slabs of equal thickness. Imagine that after traversing the first slab element, 50% of the neutrons have been absorbed. Then, after the second slab, we have 50% of 50% or 25% of the original beam strength remaining. After three slabs we have 50% of 25% or 12.5%; four slabs is 50% of 12.5 or 6.25%, five slabs is 3.125% and so on.
If you plot these percentages as a function of the number of slabs traversed, you get a decaying exponential.
A: Consider a monoenergetic beam of neutrons all traveling in the x direction through a slab target.  Let $I_0$ be the intensity (neutrons per square cm per sec) incident on the slab, denoted as the flux.  Let $\Sigma$ represent the macroscopic cross section of the target (units are 1/cm).  See any physics/nuclear engineering test for a discussion of the macroscopic cross section, such as Lamarsh, Introduction to Nuclear Reactor Theory.  $\Sigma$ dx is the probability of an interaction within the differential length dx; therefore, $\Sigma$ is the probability of an interaction per unit path length. The change in intensity as a function of the distance in the absorber is $dI(x)/dx = - \Sigma \enspace I(x) \enspace dx$ , which yields $I(x) = I_0 \enspace e^{-\Sigma x}$ for the uncollided flux at a distance x into the slab.
Note: some of the neutrons that scattered (as opposed to being absorbed) are still present but they are not counted as part of the uncollided beam.  To evaluate the total distribution of neutrons at a given point in a medium is not so simple as to evaluate the uncollided neutrons; transport theory (or if applicable diffusion theory) is required. Also the energy distribution of the neutrons must be considered.
