# Answer to Question #4025 Submitted to "Ask the Experts"

*Category: Doses and Dose Calculations — Basic dose information, dose quantities, units*

The following question was answered by an expert in the appropriate field:

*Is the absorbed dose in Gy (that is, number of joules/kg of material) independent of the shape, density, and absorption properties of the material? **For example, in a gamma chamber of, let us say, 5 kGy/hr dose rate, will we get 5 kGy after one hour of exposure into a material of given mass, shaped in to (**a) a compact sphere and (**b) a thin wire (assume that the chamber is cylindrical in shape and both occasions, that is, sphere and wire were kept along/near cylinder axis). **Will the absorbed dose be different for different materials (say, silica glass and iron)?*

The direct answers to your question are "no, no, no, and yes," although some elaboration may be helpful. The absorbed dose is measured by the energy absorbed per unit mass of material. In the discussion that follows I shall assume that a dosimetric condition referred to as secondary charged particle equilibrium exists at the surface of the objects being irradiated. A detailed discussion of this condition is probably not warranted here; stated briefly it implies that gamma rays have interacted to the extent that an equilibrium distribution of charged particles exists in which the types, energies, number, and directional distribution of charged particles entering any given small volume of material are equal to the types, energies, number, and directional distribution of charged particles leaving the same volume.

For small masses of a given material and for a uniform and constant gamma radiation field it is often true that the energy absorbed per unit mass is nearly constant throughout the mass because there is little gamma radiation attenuation in the material; then the shape, size, and density of the irradiated material are inconsequential. As the mass of the material increases, however, there is greater likelihood that the gamma radiation will be attenuated as it passes through the material, the dose per unit mass will change from one location to another within the material, and the average dose throughout the entire mass will likely be smaller than when a smaller mass of the same material in a similar geometric shape is irradiated. Since material density affects the mass for a given volume of material, the density also has an effect on gamma radiation attenuation and dose.

The shape of the object also affects the average dose since radiation that travels along a long dimension of the object will be attenuated to a greater extent than radiation that traverses a shorter dimension. The primary gamma rays are attenuated in a generally exponential fashion with path length traversed, according to an equation of the form, T = exp((-µ/ρ)(ρ)(x)), where T refers to the fractional transmission of gamma rays when traversing a path length x in a material whose mass attenuation coefficient is µ/ρ, and whose mass density is ρ (an example of use of this expression is shown later).

For the two geometries you cite, if the radiation source is arranged more or less uniformly and symmetrically around the walls of the cylinder, we would expect the attenuation in the sphere to be greater than that in the elongated receptor, since gamma radiation entering from the walls of the cylinder would most likely travel greater path lengths through the sphere than through the more linear geometry.

If the source were at one or both ends of the cylinder rather than around the walls, it is possible that the elongated source, depending on how thin it is, may experience the lower dose because of increased path lengths along the length of the cylinder. You should note also that this argument assumes that the gamma radiation field, independent of attenuation, is constant over the volume of the chamber to be occupied by the object.

If the field is nonuniform in the cylindrical chamber, the doses to objects of differing sizes and/or shapes placed in the chamber will also be nonuniform. The manufacturer of closed irradiators of the type you describe normally provides data to show how the dose uniformity throughout the irradiation volume varies because of source geometry effects, and this must be considered when you perform irradiations of various objects.

Finally, notwithstanding the size and shape of the object being irradiated, the atomic composition of the material comprising the object also affects the delivered dose. Thus, the 5 kGy/hr value you cite for the irradiator has meaning for only a specified material. Often such rates are specified for air in the chamber, and one must employ appropriate corrections to translate to dose rate in another material. The gamma radiation may interact more readily in some materials than others and may transfer more energy on average to some materials than to others.

The extent to which materials differ in this regard may be estimated from their mass energy absorption coefficients, usually symbolized µ_{en}ρ. Given two different materials irradiated in the same gamma fields, and assuming no significant attenuation in either, the dose in material 2 relative to that in material 1 is given by the ratio (µ_{en}/ρ)_{2}/(µ_{en}/ρ)_{1}. Values of these coefficients may be found in various references, including the Web site of the National Institute of Standards and Technology. The values vary with material composition and with photon energy. You didn't specify what your source material was; for typical commercial irradiators the source is likely either ^{137}Cs (662 keV) or ^{60}Co (1.25 MeV). I looked up the values of µ_{en}/ρ for iron and for pyrex glass (which value wouldn’t be very different from that for soda glass) at the above Web site. For ^{137}Cs the ratio of the mass energy absorption coefficient for iron divided by the coefficient for glass is 0.0280/0.0292 = 0.959; for ^{60}Co the comparable ratio is 0.02472/0.02650 = 0.933.

The ratios for both energies are less than unity, implying that, for the same gamma radiation field, the dose to iron will be somewhat less than that to glass. You should also note that for the same shaped objects of the same volumes, iron would produce more photon attenuation than would the glass—for example, for the ^{60}Co photons the fraction of photons transmitted through 0.5 cm of iron would be approximately (using the equation for T noted earlier):

T = exp(-0.0535 cm^{2}/g)(7.87 g/cm^{3})(0.5 cm) = 0.81;

similarly, transmission through 0.5 cm of glass would be:

T = exp(-0.0567 cm^{2}/g)(2.23 g/cm^{3})(0.5 cm) = 0.94.

The values for µ/ρ and ρ were obtained from the same Web site noted above.

I hope this answer satisfies your needs. Thanks for the question.

George Chabot, PhD, CHP