Answer to Question #3894 Submitted to "Ask the Experts"

How much water does it take to stop a cesium-137 photon (or barium-137m photon, however you want to look at it). Does making the medium saltwater change how much is required to stop it, and how much?

Since I am unsure of what your background is I will try to provide some background information that relates to your question. The transmission of a parallel beam of gamma photons through materials is described through a mathematical expression referred to as an exponential attenuation equation, which we can write as T = e^-kx, where e is the base of the natural logarithm (value of 2.718), T is the fraction of gamma photons of a given energy transmitted through a slab of material whose linear thickness is x, and k is a quantity that is constant for a given photon energy and given material and is called the linear attenuation coefficient for the photons of specified energy in the material of interest (the more usual symbol for this quantity is the Greek letter µ). Suffice it to say that as the thickness, x, increases and/or as the attenuation coefficient, k, increases (which might happen if a different material is used), the fraction of transmitted photons decreases. According to the equation, the value of T never achieves a value of zero, although for sufficiently large values of the product kx, the value of T becomes negligibly small. For many practical cases a value of kx equal to 5 is sufficiently large to provide an acceptably large reduction in the original number of photons, and we shall use this value in our discussion. The value of kx = 5 yields a value of T of 0.0067, which means that for every thousand photons incident on the material about seven would be expected to pass through the material. If one were considering a single photon incident on the material, the interpretation would be that there would be about seven chances out of a thousand that the photon would pass through the material.

Cesium-137 photons have an energy of 0.662 MeV (million electron volts), and the value of the linear attenuation coefficient of ordinary water at this energy is 0.0862 per centimeter. If we set kx equal to 5 we can solve for the thickness x and obtain x = 58.0 cm (about 22.8 inches) of water. If seawater were used instead of freshwater, the presence of the dissolved salts has the effect of increasing slightly the mass density of the water (from 1.00 to 1.025 grams per cubic centimeter) as well as slightly changing the photon transmission characteristics. It turns out that, on an equal mass basis, the freshwater is a more effective shield against ¹³⁷Cs photons than is seawater but, because of the greater density of the seawater, on a linear basis the seawater is more effective. Without going into more detail, we can state that the linear attenuation coefficient for the seawater, based on a calculation using the mass fractions of the first five most-abundant elements in seawater, is about 0.0879 per centimeter, and the thickness of seawater required to give the same value of kx = 5 is 56.9 cm (22.4 inches). The difference from freshwater is clearly quite small. If one specifies larger values of kx, the absolute differences between the fresh and seawater thicknesses will increase somewhat. I hope this answers your question.

George Chabot, PhD, CHP